Chapter 2 Simple Comparative Experiments Solutions

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter Simple Comparative Experiments Solutions - The breaking strength of a fiber is required to be at least 5 psi. Past experience has indicated that the standard deviation of breaking strength is = 3 psi. A random sample of four specimens is tested. The results are y =45, y =53, y 3 =5 and y 4 =47. (a) State the hypotheses that you think should be tested in this experiment. H : = 5 H : > 5 (b) Test these hypotheses using =.5. What are your conclusions? n = 4, = 3, y = /4 (45 + 53 + 5 + 47) = 48.75 y o 48.75 5.5 zo.8333 3 3 n 4 Since z.5 =.645, do not reject. (c) Find the P-value for the test in part (b). From the z-table: P. 7967 3. 7995. 7967. 4 (d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is y z y z n 48.75 n.963 48.75.963 45. 8 5. 69 - The viscosity of a liquid detergent is supposed to average 8 centistokes at 5C. A random sample of 6 batches of detergent is collected, and the average viscosity is 8. Suppose we know that the standard deviation of viscosity is = 5 centistokes. (a) State the hypotheses that should be tested. H : = 8 H : 8 (b) Test these hypotheses using =.5. What are your conclusions? -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY y o 8 8 zo.9 5 5 n 6 4 Since z / = z.5 =.96, do not reject. (c) What is the P-value for the test? P (. 74). 549 (d) Find a 95 percent confidence interval on the mean. The 95% confidence interval is y z y z n n 8. 965 4 8. 965 4 8. 5 8. 5 799. 75 84. 5-3 The diameters of steel shafts produced by a certain manufacturing process should have a mean diameter of.55 inches. The diameter is known to have a standard deviation of =. inch. A random sample of shafts has an average diameter of.545 inches. (a) Set up the appropriate hypotheses on the mean. H : =.55 H :.55 (b) Test these hypotheses using =.5. What are your conclusions? n =, =., y =.545 Since z.5 =.96, reject H. y o.545.55 zo 5.8. n (c) Find the P-value for this test. P=.6547x -56 (d) Construct a 95 percent confidence interval on the mean shaft diameter. The 95% confidence interval is y z y z n n..545.96.545.96.. 54438. 5456-4 A normally distributed random variable has an unknown mean and a known variance = 9. Find the sample size required to construct a 95 percent confidence interval on the mean, that has total width of.. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Since y N(,9), a 95% two-sided confidence interval on is yz yz n n 3 y (. 96) y (. 96) n 3 n If the total interval is to have width., then the half-interval is.5. Since z / = z.5 =.96,. 963 n n n. 5. 963. 5. 76. 76 38. 3 39-5 The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested, and the following results are obtained: Days 8 38 4 63 4 59 6 34 5 39 (a) We would like to demonstrate that the mean shelf life exceeds days. Set up appropriate hypotheses for investigating this claim. H : = H : > (b) Test these hypotheses using =.. What are your conclusions? y = 3 s = [ (8-3) + (4-3) + (4-3) + (6-3) + (5-3) + (38-3) + (63-3) + (59-3) + (34-3) + ( 39-3) ] / ( - ) s = 3438 / 9 = 38 s 38 9. 54 t o y o 3 s n 9. 54 78. since t.,9 =.8; do not reject H Minitab Output T-Test of the Mean Test of mu =. vs mu >. Variable N Mean StDev SE Mean T P Shelf Life 3. 9.54 6.8.78.54 T Confidence Intervals -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Variable N Mean StDev SE Mean 99. % CI Shelf Life 3. 9.54 6.8 (.9, 5.9) (c) Find the P-value for the test in part (b). P=.54 (d) Construct a 99 percent confidence interval on the mean shelf life. s s The 95% confidence interval is y t y t, n, n n n 954 3 3.5 3 3.5 954. 9 59. -6 Consider the shelf life data in Problem -5. Can shelf life be described or modeled adequately by a normal distribution? What effect would violation of this assumption have on the test procedure you used in solving Problem -5? A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of the normality assumption. If shelf life is not normally distributed, then the impact of this on the t-test in problem -5 is not too serious unless the departure from normality is severe. Normal Probability Plot for Shelf Life ML Estimates Percent 99 95 9 8 7 6 5 4 3 ML Estimates Mean 3 StDev 8.548 Goodness of Fit AD*.9 5 86 96 6 6 6 Data 36 46 56 66 76-7 The time to repair an electronic instrument is a normally distributed random variable measured in hours. The repair time for 6 such instruments chosen at random are as follows: Hours 59 8 4 379 79 64 36 68 5 49 6 485 7-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) You wish to know if the mean repair time exceeds 5 hours. Set up appropriate hypotheses for investigating this issue. H : = 5 H : > 5 (b) Test the hypotheses you formulated in part (a). What are your conclusions? Use =.5. y = 47.5 s =46 / (6 - ) = 9746.8 s 9746. 8 98. 73 y o 4.5 5 to.67 s 98.73 n 6 since t.5,5 =.753; do not reject H Minitab Output T-Test of the Mean Test of mu = 5. vs mu > 5. Variable N Mean StDev SE Mean T P Hours 6 4.5 98.7 4.7.67.6 T Confidence Intervals Variable N Mean StDev SE Mean 95. % CI Hours 6 4.5 98.7 4.7 ( 88.9, 94.) (c) Find the P-value for this test. P=.6 (d) Construct a 95 percent confidence interval on mean repair time. The 95% confidence interval is y t s n y t, n, n s n 98.73 4.5.3 4.5.3 98.73 6 6 88. 9 94. -8 Reconsider the repair time data in Problem -7. Can repair time, in your opinion, be adequately modeled by a normal distribution? The normal probability plot below does not reveal any serious problem with the normality assumption. -5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot for Hours ML Estimates Percent 99 95 9 8 7 6 5 4 3 ML Estimates Mean 4.5 StDev 95.599 Goodness of Fit AD*.85 5 5 5 5 Data 35 45-9 Two machines are used for filling plastic bottles with a net volume of 6. ounces. The filling processes can be assumed to be normal, with standard deviation of =.5 and =.8. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is 6. ounces. An experiment is performed by taking a random sample from the output of each machine. Machine Machine 6.3 6. 6. 6.3 6.4 5.96 5.97 6.4 6.5 5.98 5.96 6. 6.5 6. 6. 6. 6. 5.99 5.99 6. (a) State the hypotheses that should be tested in this experiment. H : = H : (b) Test these hypotheses using =.5. What are your conclusions? y n 6. 5. 5 y n 6. 5. 8 z o y y n n 6. 56. 8. 5. 8 35. z.5 =.96; do not reject (c) What is the P-value for the test? P=.77 (d) Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines. -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The 95% confidence interval is (6.56.5) (9.6) y y z y y z n n n n.5.8 (6.5 6.5) (9.6).45.45.5.8 - Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that = =. psi. From random samples of n = and n = we obtain y = 6.5 and y = 55.. The company will not adopt plastic unless its breaking strength exceeds that of plastic by at least psi. Based on the sample information, should they use plastic? In answering this questions, set up and test appropriate hypotheses using =.. Construct a 99 percent confidence interval on the true mean difference in breaking strength. H : - = H : - > y 6. 5 n y n 55. z o y y 6. 555. 585. n n z. =.5; do not reject The 99 percent confidence interval is (6.5 55.) (.575) y y z y y z n n n n 6.4 (6.5 55.) (.575) 8.6 - The following are the burning times of chemical flares of two different formulations. The design engineers are interested in both the means and variance of the burning times. Type Type 65 8 64 56 8 67 7 69 57 59 83 74 66 75 59 8 8 7 65 79 (a) Test the hypotheses that the two variances are equal. Use =.5. -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY F 599.,, 43. F H: H: F 97599.,, S S 9. 64 9. 367 S 858. 98. S 87. 73. 48 Do not reject. F 43 599.,,. (b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use =.5. What is the P-value for this test? ( n ) S ( n ) S 5695. S p 86. 775 n n 8 S p 93. y y 7. 4 7. t S p 93. 48. n n t 58.,. Do not reject. From the computer output, t=.5; do not reject. Also from the computer output P=.96 Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Type vs Type N Mean StDev SE Mean Type 7.4 9.6.9 Type 7. 9.37 3. 95% CI for mu Type - mu Type : ( -8.6, 9.) T-Test mu Type = mu Type (vs not =): T =.5 P =.96 DF = 8 Both use Pooled StDev = 9.3 (c) Discuss the role of the normality assumption in this problem. Check the assumption of normality for both types of flares. The assumption of normality is required in the theoretical development of the t-test. However, moderate departure from normality has little impact on the performance of the t-test. The normality assumption is more important for the test on the equality of the two variances. An indication of nonnormality would be of concern here. The normal probability plots shown below indicate that burning time for both formulations follow the normal distribution. -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot for Type ML Estimates Percent 99 95 9 8 7 6 5 4 3 ML Estimates Mean 7.4 StDev 8.78863 Goodness of Fit AD*.387 5 5 6 7 Data 8 9 Normal Probability Plot for Type ML Estimates Percent 99 95 9 8 7 6 5 4 3 ML Estimates Mean 7. StDev 8.88594 Goodness of Fit AD*.7 5 5 6 7 Data 8 9 - An article in Solid State Technology, "Orthogonal Design of Process Optimization and Its Application to Plasma Etching" by G.Z. Yin and D.W. Jillie (May, 987) describes an experiment to determine the effect of C F 6 flow rate on the uniformity of the etch on a silicon wafer used in integrated circuit manufacturing. Data for two flow rates are as follows: C F 6 Uniformity Observation (SCCM) 3 4 5 6 5.7 4.6.6 3. 3. 3.8 4.6 3.4.9 3.5 4. 5. (a) Does the C F 6 flow rate affect average etch uniformity? Use =.5. No, C F 6 flow rate does not affect average etch uniformity. Minitab Output Two Sample T-Test and Confidence Interval -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Two sample T for Uniformity Flow Rat N Mean StDev SE Mean 5 6 3.37.76.3 6 3.933.8.34 95% CI for mu (5) - mu (): ( -.63,.4) T-Test mu (5) = mu () (vs not =): T = -.35 P =. DF = Both use Pooled StDev =.79 (b) What is the P-value for the test in part (a)? From the computer printout, P=. (c) Does the C F 6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use =.5. H: H: F555.,, 55. 5776. F 86.. 674 Do not reject; C F 6 flow rate does not affect wafer-to-wafer variability. (d) Draw box plots to assist in the interpretation of the data from this experiment. The box plots shown below indicate that there is little difference in uniformity at the two gas flow rates. Any observed difference is not statistically significant. See the t-test in part (a). 5 Uniformity 4 3 5 Flow Rate -3 A new filtering device is installed in a chemical unit. Before its installation, a random sample yielded the following information about the percentage of impurity: y =.5, S =.7, and n = 8. After installation, a random sample yielded y =., S = 94.73, n = 9. (a) Can you concluded that the two variances are equal? Use =.5. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY. 5, 7, 8 4. 53 S. 7 F S 94. 73 Do Not Reject. Assume that the variances are equal. H F : H :. 7 (b) Has the filtering device reduced the percentage of impurity significantly? Use =.5. H : H : ( n )S ( n )S ( 8 )(. 7 ) ( 9 )( 94. 73) S p 97. 74 n n 8 9 S p 9. 89 y y. 5. t. 479 S p 9. 89 n n 8 9 t. 5, 5. 753 Do not reject. There is no evidence to indicate that the new filtering device has affected the mean -4 Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: 5.34 6.65 4.76 5.98 7.5 6. 7.55 5.54 5.6 6. 5.97 7.35 5.44 4.39 4.98 5.5 6.35 4.6 6. 5.3 (a) Construct a 95 percent confidence interval estimate of. n S n S, n ( ), n.8897.8897 3.85 8.97.457.686 (b) Test the hypothesis that =.. Use =.5. What are your conclusions? H 59.,. H : : SS 59. 385 9759., 8. 97 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Do not reject. There is no evidence to indicate that (c) Discuss the normality assumption and its role in this problem. The normality assumption is much more important when analyzing variances then when analyzing means. A moderate departure from normality could cause problems with both statistical tests and confidence intervals. Specifically, it will cause the reported significance levels to be incorrect. (d) Check normality by constructing a normal probability plot. What are your conclusions? The normal probability plot indicates that there is not any serious problem with the normality assumption. Normal Probability Plot for Uniformity ML Estimates Percent 99 95 9 8 7 6 5 4 3 ML Estimates Mean 5.88 StDev.86656 Goodness of Fit AD*.835 5 3.8 4.8 5.8 Data 6.8 7.8-5 The diameter of a ball bearing was measured by inspectors, each using two different kinds of calipers. The results were: Inspector Caliper Caliper Difference Difference^.65.64...65.65. 3.66.64..4 4.67.66.. 5.67.67. 6.65.68 -.3.9 7.67.64.3.9 8.67.65..4 9.65.65..68.67...68.68..65.69 -.4.6 3.. 45 (a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use =.5. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY H : H : or equivalently H H : : d d Minitab Output Paired T-Test and Confidence Interval Paired T for Caliper - Caliper N Mean StDev SE Mean Caliper.665.5.35 Caliper.66.758.58 Difference.5.6.579 95% CI for mean difference: (-.4,.54) T-Test of mean difference = (vs not = ): T-Value =.43 P-Value =.674 (b) Find the P-value for the test in part (a). P=.674 (c) Construct a 95 percent confidence interval on the difference in the mean diameter measurements for the two types of calipers. Sd Sd d t, D d t n, n n n...5. d.5...5 d -6 An article in the Journal of Strain Analysis (vol.8, no., 983) compares several procedures for predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows: Girder Karlsruhe Method Lehigh Method Difference Difference^ S/.86.6.5.565 S/.5.99.59.58 S3/.3.63.59.678 S4/.339.6.77.7679 S5/..65.35.85 S/.4.78.4.576 S/.365.37.38.7584 S/3.537.86.45.34 S/4.559.5.57.5749 Sum =.465.85 Average =.74 (a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? Use =.5. H : H : or equivalently H H : : d d n d di.465.74 9 n i -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY s d n n di d i.85 (.465) i n i 9.35 n 9 Minitab Output Paired T-Test and Confidence Interval Paired T for Karlsruhe - Lehigh N Mean StDev SE Mean Karlsruh 9.34.46.487 Lehigh 9.66.494.65 Difference 9.739.35.45 t, n t. 5, 9 95% CI for mean difference: (.7,.3777) T-Test of mean difference = (vs not = ): T-Value = 6.8 P-Value =. d.74 t 6.8 Sd.35 n 9. 36, reject the null hypothesis. (b) What is the P-value for the test in part (a)? P=. (c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load. S d d t,n n. 35. 74. 36 9. 73 d d d S d d t,n n. 35. 74. 36 9. 37777 (d) Investigate the normality assumption for both samples. Normal Probability Plot.999 Probability.99.95.8.5..5.. Av erage:.34 StDev :.463 N: 9.5.5.35 Karlsruhe.45.55 Anderson-Darling Normality Test A-Squared:.86 P-Value:.537-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot.999.99.95 Probability.8.5..5....5..5 Lehigh Av erage:.66 StDev :.49386 N: 9 Anderson-Darling Normality Test A-Squared:.77 P-Value:.8 (e) Investigate the normality assumption for the difference in ratios for the two methods. Normal Probability Plot.999.99.95 Probability.8.5..5.....3.4.5 Difference Av erage:.73889 StDev :.3599 N: 9 Anderson-Darling Normality Test A-Squared:.38 P-Value:.464 (f) Discuss the role of the normality assumption in the paired t-test. As in any t-test, the assumption of normality is of only moderate importance. In the paired t-test, the assumption of normality applies to the distribution of the differences. That is, the individual sample measurements do not have to be normally distributed, only their difference. -7 The deflection temperature under load for two different formulations of ABS plastic pipe is being studied. Two samples of observations each are prepared using each formulation, and the deflection temperatures (in F) are reported below: Formulation Formulation 99 98 77 76 98 94 3 6 97 85 88 9 6 89 93 95 84 97 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Construct normal probability plots for both samples. Do these plots support assumptions of normality and equal variance for both samples? Normal Probability Plot.999 Probability.99.95.8.5..5.. Av erage:.5 StDev :.757 N: 85 95 Form 5 5 Anderson-Darling Normality Test A-Squared:.45 P-Value:.7 Normal Probability Plot Probability.999.99.95.8.5..5.. Av erage: 93.83 StDev : 9.94949 N: 75 85 Form 95 5 Anderson-Darling Normality Test A-Squared:.443 P-Value:.36 (b) Do the data support the claim that the mean deflection temperature under load for formulation exceeds that of formulation? Use =.5. Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Form vs Form N Mean StDev SE Mean Form.5..9 Form 93.8 9.95.9 95% CI for mu Form - mu Form : ( -., 5.9) T-Test mu Form = mu Form (vs >): T =.8 P =.4 DF = Both use Pooled StDev =. (c) What is the P-value for the test in part (a)? P =.4-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -8 Refer to the data in problem -7. Do the data support a claim that the mean deflection temperature under load for formulation exceeds that of formulation by at least 3 F? Yes, formulation exceeds formulation by at least 3 F. Minitab Output Two-Sample T-Test and CI: Form, Form Two-sample T for Form vs Form N Mean StDev SE Mean Form.5..9 Form 93.8 9.95.9r Difference = mu Form - mu Form Estimate for difference: 7.4 95% lower bound for difference:.36 T-Test of difference = 3 (vs >): T-Value =.8 P-Value =.47 DF = Both use Pooled StDev =. -9 In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two different etching solutionsare being evaluated. Eight randomly selected wafers have been etched in each solution and the observed etch rates (in mils/min) are shown below: Solution Solution 9.9.6..6 9.4.3... 9.3.7.4.3 9.8.5.3 (a) Do the data indicate that the claim that both solutions have the same mean etch rate is valid? Use =.5 and assume equal variances. See the Minitab output below. Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Solution vs Solution N Mean StDev SE Mean Solution 8 9.95.465.6 Solution 8.36.33.8 95% CI for mu Solution - mu Solution: ( -.83, -.43) T-Test mu Solution = mu Solution (vs not =):T = -.38 P =.3 DF = 4 Both use Pooled StDev =.368 (b) Find a 95% confidence interval on the difference in mean etch rate. From the Minitab output, -.83 to.43. (c) Use normal probability plots to investigate the adequacy of the assumptions of normality and equal variances. -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot Probability.999.99.95.8.5..5.. Av erage: 9.95 StDev :.4659 N: 8 9.5. Solution.5 Anderson-Darling Normality Test A-Squared:. P-Value:.743 Normal Probability Plot.999 Probability.99.95.8.5..5.. Av erage:.365 StDev :.369 N: 8....3.4 Solution.5.6.7 Anderson-Darling Normality Test A-Squared:.58 P-Value:.99 Both the normality and equality of variance assumptions are valid. - Two popular pain medications are being compared on the basis of the speed of absorption by the body. Specifically, tablet is claimed to be absorbed twice as fast as tablet. Assume that and are known. Develop a test statistic for H : = H : 4 y y ~ N,, assuming that the data is normally distributed. n n y y The test statistic is: zo, reject if zo z 4 n n - Suppose we are testing H : = H : -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY where and are known. Our sampling resources are constrained such that n + n = N. How should we allocate the N observations between the two populations to obtain the most powerful test? The most powerful test is attained by the n and n that maximize z o for given y Thus, we chose n and n to y y max z, subject to n + n = N. o n n This is equivalent to min L, subject to n n n n N n + n = N. Now dl, implies that n / n = /. dn n N n y. Thus n and n are assigned proportionally to the ratio of the standard deviations. This has intuitive appeal, as it allocates more observations to the population with the greatest variability. - Develop Equation -46 for a ( - ) percent confidence interval for the variance of a normal distribution. SS ~. Thus, SS n P, n, n. Therefore, SS SS, P, n, n so SS SS, is the ( - )% confidence interval on., n, n -3 Develop Equation -5 for a ( - ) percent confidence interval for the ratio / and are the variances of two normal distributions., where S S ~ F n, n S PF, n, n F or S, n, n S S P F, n, n F S, n, n S -4 Develop an equation for finding a ( - ) percent confidence interval on the difference in the means of two normal distributions where. Apply your equation to the portland cement experiment data, and find a 95% confidence interval. -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY y y S n S n ~ t, S S S S t y y t,, n n n n S S S S y y t y y t,, n n n n where S S n n S S n n n n Using the data from Table - n y S 6. 764 38. n y S 7. 343. 646.38.646.38.646 6.764 7.343. 6.764 7.343. This agrees with the result in Table -.. 38. 646 where 7. 4 7. 38. 646.46.889-5 Construct a data set for which the paired t-test statistic is very large, but for which the usual twosample or pooled t-test statistic is small. In general, describe how you created the data. Does this give you any insight regarding how the paired t-test works? A B delta 7.66 8.46.754.359.4555.965 9.9977.8.4.977.34.4339-5.934-5.3.94-6.7-5.594.4788 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output Paired T-Test and Confidence Interval 9.95.69.7485 -.944 -.358.95854-4.697-3.3446.3465-6.699-5.933.7656 Paired T for A - B N Mean StDev SE Mean A.59.6 3.8 B.48. 3. Difference -.89.398.6 95% CI for mean difference: (-.74, -.65) T-Test of mean difference = (vs not = ): T-Value = -7.7 P-Value =. Two Sample T-Test and Confidence Interval Two sample T for A vs B N Mean StDev SE Mean A.6. 3. B.5. 3. 95% CI for mu A - mu B: ( -.4, 8.6) T-Test mu A = mu B (vs not =): T = -. P =.85 DF = 8 Both use Pooled StDev =. These two sets of data were created by making the observation for A and B moderately different within each pair (or block), but making the observations between pairs very different. The fact that the difference between pairs is large makes the pooled estimate of the standard deviation large and the two-sample t-test statistic small. Therefore the fairly small difference between the means of the two treatments that is present when they are applied to the same experimental unit cannot be detected. Generally, if the blocks are very different, then this will occur. Blocking eliminates the variabiliy associated with the nuisance variable that they represent. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 3 Experiments with a Single Factor: The Analysis of Variance Solutions 3- The tensile strength of portland cement is being studied. Four different mixing techniques can be used economically. The following data have been collected: Mixing Technique Tensile Strength (lb/in ) 39 3 865 89 3 33 975 35 3 8 9 985 35 4 6 7 6 765 (a) Test the hypothesis that mixing techniques affect the strength of the cement. Use =.5. Design Expert Output Response: Tensile Strengthin lb/in^ ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4.897E+5 3.63E+5.73.5 significant A 4.897E+5 3.63E+5.73.5 Residual.539E+5 85.69 Lack of Fit. Pure Error.539E+5 85.69 Cor Total 6.436E+5 5 The Model F-value of.73 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 97. 56.63-356.5 56.63 3-3 933.75 56.63 4-4 666.5 56.63 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -85.5 8.8 -.3.39 vs 3 37.5 8.8.47.65 vs 4 34.75 8.8 3.8.5 vs 3.5 8.8.78.67 vs 4 49. 8.8 6. <. 3 vs 4 67.5 8.8 3.34.59 The F-value is.73 with a corresponding P-value of.5. Mixing technique has an effect. (b) Construct a graphical display as described in Section 3-5.3 to compare the mean tensile strengths for the four mixing techniques. What are your conclusions? S y i. MS n E 85. 7 4 56. 65 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (4) (3) () () 7 8 9 3 3 Tensile Strength Based on examination of the plot, we would conclude that and 3 are the same; that 4 differs from and 3, that differs from and 3, and that and 4 are different. (c) Use the Fisher LSD method with =.5 to make comparisons between pairs of means. LSD t LSD t,n a. 5, 64 LSD. 79 MS n E ( 85. 7 ) 4 64. 85 74. 495 Treatment vs. Treatment 4 = 356.5-666.5 = 49. > 74.495 Treatment vs. Treatment 3 = 356.5-933.75 =.5 > 74.495 Treatment vs. Treatment = 356.5-97. = 85.5 > 74.495 Treatment vs. Treatment 4 = 97. - 666.5 = 34.75 > 74.495 Treatment vs. Treatment 3 = 97. - 933.75 = 37.5 < 74.495 Treatment 3 vs. Treatment 4 = 933.75-666.5 = 67.5 > 74.495 The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree with graphical method for this experiment. (d) Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of the normality assumption? There is nothing unusual about the normal probability plot of residuals. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 99 Norm al % probability 95 9 8 7 5 3 5-8.5-96.4375 -.65 73.875 58 Residual (e) Plot the residuals versus the predicted tensile strength. Comment on the plot. There is nothing unusual about this plot. Residuals vs. Predicted 58 73.875 Residuals -.65-96.4375-8.5 666.5 788.75 9.5 333.75 356.5 Predicted (f) Prepare a scatter plot of the results to aid the interpretation of the results of this experiment. Design-Expert automatically generates the scatter plot. The plot below also shows the sample average for each treatment and the 95 percent confidence interval on the treatment mean. 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 33 One Factor Plot 39.75 Tensile Strength 939.5 759.6 579. 3 4 Technique 3- Rework part (b) of Problem 3- using the Duncan's multiple range test. Does this make any difference in your conclusions? R R R 3 4 S y i. r r r. 5. 5. 5 MS n, S E 85. 7 56. 65 4 3. 8 56. 65 74. 46 y i. 3, S y 3. 356. 65 8. 9 i. 4, S 3. 3356. 65 88. 56 yi. Treatment vs. Treatment 4 = 356.5-666.5 = 49. > 88.56 (R 4 ) Treatment vs. Treatment 3 = 356.5-933.75 =.5 > 8.9 (R 3 ) Treatment vs. Treatment = 356.5-97. = 85.5 > 74.46 (R ) Treatment vs. Treatment 4 = 97. - 666.5 = 34.75 > 8.9 (R 3 ) Treatment vs. Treatment 3 = 97. - 933.75 = 37.5 < 74.46 (R ) Treatment 3 vs. Treatment 4 = 933.75-666.5 = 67.5 > 74.46 (R ) Treatment 3 and Treatment are not different. All other pairs of means differ. This is the same result obtained from the Fisher LSD method and the graphical method. (b) Rework part (b) of Problem 3- using Tukey s test with =.5. Do you get the same conclusions from Tukey s test that you did from the graphical procedure and/or Duncan s multiple range test? Minitab Output Tukey's pairwise comparisons Family error rate =.5 Individual error rate =.7 Critical value = 4. Intervals for (column level mean) - (row level mean) -43 3 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 53 3 - -5 75 46 4 67 5 3 543 78 55 No, the conclusions are not the same. The mean of Treatment 4 is different than the means of Treatments,, and 3. However, the mean of Treatment is not different from the means of Treatments and 3 according to the Tukey method, they were found to be different using the graphical method and Duncan s multiple range test. (c) Explain the difference between the Tukey and Duncan procedures. A single critical value is used for comparison with the Tukey procedure where a critical values are used with the Duncan procedure. Tukey s test has a type I error rate of for all pairwise comparisons where Duncan s test type I error rate varies based on the steps between the means. Tukey s test is more conservative and has less power than Duncan s test. 3-3 Reconsider the experiment in Problem 3-. Find a 95 percent confidence interval on the mean tensile strength of the portland cement produced by each of the four mixing techniques. Also find a 95 percent confidence interval on the difference in means for techniques and 3. Does this aid in interpreting the results of the experiment? MSE MSE yi. t i yi. t,n a,n a n n Treatment - Treatment 3: Treatment : 97. 79 8837 4 97 3. 387 847. 63 394. 387 Treatment : 356.53.387 33. 863 379. 637 Treatment 3: 933.753.387 8. 363 3 357. 37 Treatment 4: 666.53.387 54. 863 4 789. 637 y y i. j. t,n a MS n E 85. 7 97. 933. 75. 79 4 37. 45 3. 745 i j y y i. j. t,n a MS n 3-4 An experiment was run to determine whether four specific firing temperatures affect the density of a certain type of brick. The experiment led to the following data: Temperature Density.8.9.7.6.7 5.7.4.5.4 5.9.8.8.6.5 75.9.7.8.4 E 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Does the firing temperature affect the density of the bricks? Use =.5. No, firing temperature does not affect the density of the bricks. Refer to the Design-Expert output below. Design Expert Output Response: Density ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.6 3.5..569 not significant A.6 3.5..569 Residual.36 4.6 Lack of Fit. Pure Error.36 4.6 Cor Total.5 7 The "Model F-value" of. implies the model is not significant relative to the noise. There is a 5.69 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -.74.7-5.5.8 3-5.7.7 4-75.7.8 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.4..3.45 vs 3....8465 vs 4.4..37.756 vs 3 -.. -.5.6 vs 4 -.. -.76.996 3 vs 4...9.855 (b) Is it appropriate to compare the means using Duncan s multiple range test in this experiment? The analysis of variance tells us that there is no difference in the treatments. There is no need to proceed with Duncan s multiple range test to decide which mean is difference. (c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There is nothing unusual about the residual plots. 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals. Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5 Residuals.75 -.5 -.75 -.3 -.3 -.75 -.5.75..5.56.6.68.74 Residual Predicted (d) Construct a graphical display of the treatments as described in Section 3-5.3. Does this graph adequately summarize the results of the analysis of variance in part (b). Yes. Scaled t Distribution (5) (75,5,)..3.4.5.6.7.8 M ean D ensity 3-5 Rework Part (d) of Problem 3-4 using the Fisher LSD method. What conclusions can you draw? Explain carefully how you modified the procedure to account for unequal sample sizes. When sample sizes are unequal, the appropriate formula for the LSD is LSD t,n a MS E ni n j Treatment vs. Treatment =.74.5 =.4 >.3 Treatment vs. Treatment 3 =.74.7 =. <.87 Treatment vs. Treatment 4 =.74.7 =.4 <.3 Treatment 3 vs. Treatment =.7.5 =. <.3 Treatment 4 vs. Treatment =.7.5 =. <.446 Treatment 3 vs. Treatment 4 =.7.7 =. <.3 3-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Treatment, temperature of, is different than Treatment, temperature of 5. All other pairwise comparisons do not identify differences. Notice something very interesting has happened here. The analysis of variance indicated that there were no differences between treatment means, yet the LSD procedure found a difference; in fact, the Design-Expert output indicates that the P-value if slightly less that.5. This illustrates a danger of using multiple comparison procedures without relying on the results from the analysis of variance. Because we could not reject the hypothesis of equal means using the analysis of variance, we should never have performed the Fisher LSD (or any other multiple comparison procedure, for that matter). If you ignore the analysis of variance results and run multiple comparisons, you will likely make type I errors. The LSD calculations utilized Equation 3-3, which accommodates different sample sizes. Equation 3-3 simplifies to Equation 3-33 for a balanced design experiment. 3-6 A manufacturer of television sets is interested in the effect of tube conductivity of four different types of coating for color picture tubes. The following conductivity data are obtained: Coating Type Conductivity 43 4 5 46 5 49 37 43 3 34 36 3 7 4 9 7 3 9 (a) Is there a difference in conductivity due to coating type? Use =.5. Yes, there is a difference in means. Refer to the Design-Expert output below.. Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 844.69 3 8.56 4.3.3 significant A 844.69 3 8.56 4.3.3 Residual 36.5 9.69 Lack of Fit. Pure Error 36.5 9.69 Cor Total 8.94 5 The Model F-value of 4.3 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 45.. - 45.5. 3-3 3.5. 4-4 9.5. Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.5 3.4 -.8.9378 vs 3.75 3.4 4.6.6 vs 4 5.75 3.4 5..3 vs 3 3. 3.4 4.4.4 vs 4 6. 3.4 5..3 3 vs 4 3. 3.4.96.3578 (b) Estimate the overall mean and the treatment effects. 3-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ 7 / 6 37. 9375 ˆ y. y.. 45. 37. 9375 7. 65 ˆ y. y.. 45. 5 37. 9375 7. 35 ˆ 3 y3. y.. 3. 537. 9375 5. 6875 ˆ y y 9. 5 37. 9375 8. 6875 4 4... (c) Compute a 95 percent interval estimate of the mean of coating type 4. Compute a 99 percent interval estimate of the mean difference between coating types and 4. 9. 69 Treatment 4: 9. 5. 79 4 4. 455 4 34. 845 Treatment - Treatment 4: 45 9. 5 3. 55 6 4.. 64 5 336. 4 9 69 (d) Test all pairs of means using the Fisher LSD method with =.5. Refer to the Design-Expert output above. The Fisher LSD procedure is automatically included in the output. The means of Coating Type and Coating Type are not different. The means of Coating Type 3 and Coating Type 4 are not different. However, Coating Types and produce higher mean conductivity that does Coating Types 3 and 4. (e) Use the graphical method discussed in Section 3-5.3 to compare the means. Which coating produces the highest conductivity? S y i. MS E 6. 96. 9 Coating types and produce the highest conductivity. n 4 Scaled t Distribution (4) (3) ()() 3 35 4 45 5 Conductivity 3-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (f) Assuming that coating type 4 is currently in use, what are your recommendations to the manufacturer? We wish to minimize conductivity. Since coatings 3 and 4 do not differ, and as they both produce the lowest mean values of conductivity, use either coating 3 or 4. As type 4 is currently being used, there is probably no need to change. 3-7 Reconsider the experiment in Problem 3-6. Analyze the residuals and draw conclusions about model adequacy. There is nothing unusual in the normal probability plot. A funnel shape is seen in the plot of residuals versus predicted conductivity indicating a possible non-constant variance. Normal plot of residuals Residuals vs. Predicted 6.75 99 Normal % probability 95 9 8 7 5 3 5 Residuals 3 -.75-4.5-8.5-8.5-4.5 -.75 3 6.75 9.5 33.5 37.5 4.5 45.5 Residual Predicted Residuals vs. Coating Type 6.75 3 Residuals -.75-4.5-8.5 3 4 Coating Type 3-8 An article in the ACI Materials Journal (Vol. 84, 987. pp. 3-6) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3 x 6 cylinder was used, and the 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in the following table. Rodding Level Compressive Strength 53 53 44 5 6 65 5 56 73 53 5 5 49 5 (a) Is there any difference in compressive strength due to the rodding level? Use =.5. There are no differences. Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 8633.33 3 9544.44.87.38 not significant A 8633.33 3 9544.44.87.38 Residual 4933.33 8 56.67 Lack of Fit. Pure Error 4933.33 8 56.67 Cor Total 69566.67 The "Model F-value" of.87 implies the model is not significant relative to the noise. There is a.38 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 5. 4.3-5 586.67 4.3 3-66.67 4.3 4-5 5. 4.3 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -86.67 58.4 -.48.76 vs 3-6.67 58.4 -.83.5 vs 4. 58.4.. vs 3 -. 58.4 -.34.748 vs 4 86.67 58.4.48.76 3 vs 4 6.67 58.4.83.5 (b) Find the P-value for the F statistic in part (a). From computer output, P=.38. (c) Analyze the residuals from this experiment. What conclusions can you draw about the underlying model assumptions? There is nothing unusual about the residual plots. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 3.333 99 Normal % probability 95 9 8 7 5 3 5 Residuals 7.8333 8.3333-34.667-86.6667-86.6667-34.667 8.3333 7.8333 3.333 5. 56.67 553.33 58. 66.67 Residual Predicted Residuals vs. Rodding Level 3.333 7.8333 Residuals 8.3333-34.667-86.6667 3 4 Rodding Level (d) Construct a graphical display to compare the treatment means as describe in Section 3-5.3. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (, 5) (5) () 48 459 5 54 58 63 664 Mean Compressive Strength 3-9 An article in Environment International (Vol. 8, No. 4, 99) describes an experiment in which the amount of radon released in showers was investigated. Radon enriched water was used in the experiment and six different orifice diameters were tested in shower heads. The data from the experiment are shown in the following table. Orifice Dia. Radon Released (%).37 8 83 83 85.5 75 75 79 79.7 74 73 76 77. 67 7 74 74.4 6 6 67 69.99 6 6 64 66 (a) Does the size of the orifice affect the mean percentage of radon released? Use =.5. Yes. There is at least one treatment mean that is different. Design Expert Output Response: Radon Released in % ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 33.38 5 6.68 3.85 <. significant A 33.38 5 6.68 3.85 <. Residual 3.5 8 7.35 Lack of Fit. Pure Error 3.5 8 7.35 Cor Total 65.63 3 The Model F-value of 3.85 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) EstimatedStandard Mean Error -.37 8.75.36 -.5 77..36 3-.7 75..36 4-. 7.75.36 5-.4 65..36 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6-.99 6.75.36 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs 5.75.9 3..77 vs 3 7.75.9 4.4.8 vs 4..9 5.74 <. vs 5 7.75.9 9.6 <. vs 6..9.43 <. vs 3..9.4.35 vs 4 5.5.9.74.35 vs 5..9 6.6 <. vs 6 4.5.9 7.43 <. 3 vs 4 3.5.9.7.7 3 vs 5..9 5. <. 3 vs 6.5.9 6.39 <. 4 vs 5 6.75.9 3.5.4 4 vs 6 9..9 4.7. 5 vs 6.5.9.7.557 (b) Find the P-value for the F statistic in part (a). P=3.6 x -8 (c) Analyze the residuals from this experiment. There is nothing unusual about the residuals. Normal plot of residuals Residuals vs. Predicted 4 99 Normal % probability 95 9 8 7 5 3 5 Residuals.85 -.375 -.565-4.75-4.75 -.565 -.375.85 4 6.75 67.75 7.75 77.75 8.75 Residual Predicted 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 Residuals vs. Orifice Diameter.85 Residuals -.375 -.565-4.75 3 4 5 6 Orifice Diameter (d) Find a 95 percent confidence interval on the mean percent radon released when the orifice diameter is.4. 7. 35 Treatment 5 (Orifice =.4): 6. 4 6. 5 67. 848 (e) Construct a graphical display to compare the treatment means as describe in Section 3-5.3. What conclusions can you draw? Scaled t Distribution (6) (5) (4) (3) () () 6 65 7 75 8 Conductivity Treatments 5 and 6 as a group differ from the other means;, 3, and 4 as a group differ from the other means, differs from the others. 3- The response time in milliseconds was determined for three different types of circuits that could be used in an automatic valve shutoff mechanism. The results are shown in the following table. 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Circuit Type Response Time 9 8 5 3 7 3 3 6 5 8 6 7 (a) Test the hypothesis that the three circuit types have the same response time. Use =.. From the computer printout, F=6.8, so there is at least one circuit type that is different. Design Expert Output Response: Response Time in ms ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 543.6 7.8 6.8.4 significant A 543.6 7.8 6.8.4 Residual.8 6.9 Lack of Fit. Pure Error.8 6.9 Cor Total 746.4 4 The Model F-value of 6.8 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -.8.84 -..84 3-3 8.4.84 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.4.6-4.38.9 vs 3.4.6.9.374 vs 3 3.8.6 5.3. (b) Use Tukey s test to compare pairs of treatment means. Use =.. S y i. MS E 69. 8385 n 5 q., 3, 5. 4 t. 8385 5. 4 9 66. vs. :.8-.=.4 > 9.66 vs. 3:.8-8.4=.4 < 9.66 vs. 3:.-8.4=3.8 > 9.66 and are different. and 3 are different. Notice that the results indicate that the mean of treatment differs from the means of both treatments and 3, and that the means for treatments and 3 are the same. Notice also that the Fisher LSD procedure (see the computer output) gives the same results. (c) Use the graphical procedure in Section 3-5.3 to compare the treatment means. What conclusions can you draw? How do they compare with the conclusions from part (a). The scaled-t plot agrees with part (b). In this case, the large difference between the mean of treatment and the other two treatments is very obvious. 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (3) () () 5 5 5 Tensile Strength (d) Construct a set of orthogonal contrasts, assuming that at the outset of the experiment you suspected the response time of circuit type to be different from the other two. H 3 H 3 C y. y. y3. C 54 4 6 SS F C C 6 56 59. 59. 3.3 6.9 Type differs from the average of type and type 3. (e) If you were a design engineer and you wished to minimize the response time, which circuit type would you select? Either type or type 3 as they are not different from each other and have the lowest response time. (f) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? The normal probability plot has some points that do not lie along the line in the upper region. This may indicate potential outliers in the data. 3-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 7.8 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5 Residuals 4.55.3 -.95-5. -5. -.95.3 4.55 7.8 8.4.85 5.3 8.75. Residual Predicted Residuals vs. Circuit Type 7.8 4.55 Residuals.3 -.95-5. 3 Circuit Type 3- The effective life of insulating fluids at an accelerated load of 35 kv is being studied. Test data have been obtained for four types of fluids. The results were as follows: Fluid Type Life (in h) at 35 kv Load 7.6 8.9 6.3 7.4..6 6.9 5.3 8.6 7. 9.5.3 3.4 3.6 9.4 8.5.5.3 4 9.3. 6.9 7.5 8.3 9.8 (a) Is there any indication that the fluids differ? Use =.5. At =.5 there are no difference, but at since the P-value is just slightly above.5, there is probably a difference in means. Design Expert Output Response: Life in in h 3-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.7 3.6 3.5.55 not significant A 3.6 3.5 3.5.55 Residual 65.99 3.3 Lack of Fit. Pure Error 65.99 3.3 Cor Total 96.6 3 The Model F-value of 3.5 implies there is a 5.5% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 8.65.74-7.95.74 3-3.95.74 4-4 8.8.74 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.7.5.67.5 vs 3 -.3.5 -.9.43 vs 4 -.7.5 -.6.8753 vs 3-3..5 -.86.97 vs 4 -.87.5 -.83.483 3 vs 4.3.5.3.554 (b) Which fluid would you select, given that the objective is long life? Treatment 3. The Fisher LSD procedure in the computer output indicates that the fluid 3 is different from the others, and it s average life also exceeds the average lives of the other three fluids. (c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? There is nothing unusual in the residual plots. Normal plot of residuals.95 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5 Residuals.55.5 -.5 -.65 -.65 -.5.5.55.95 7.95 8.7 9.45..95 Residual Predicted 3-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.95 Residuals vs. Fluid Type.55 Residuals.5 -.5 -.65 3 4 Fluid Type 3- Four different designs for a digital computer circuit are being studied in order to compare the amount of noise present. The following data have been obtained: Circuit Design Noise Observed 9 9 3 8 8 6 73 56 8 3 47 6 5 35 5 4 95 46 83 78 97 (a) Is the amount of noise present the same for all four designs? Use =.5. No, at least one treatment mean is different. Design Expert Output Response: Noise ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4. 3 44..78 <. significant A 4. 3 44..78 <. Residual 948.8 6 84.3 Lack of Fit. Pure Error 948.8 6 84.3 Cor Total 499.8 9 The Model F-value of.78 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 9. 6.7-7. 6.7 3-3 36.6 6.7 4-4 79.8 6.7 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -5.8 8.59-5.9 <. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY vs 3-7.4 8.59 -.3.597 vs 4-6.6 8.59-7.6 <. vs 3 33.4 8.59 3.89.3 vs 4-9.8 8.59 -.4.75 3 vs 4-43. 8.59-5.3. (b) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? There is nothing unusual about the residual plots. Normal plot of residuals Residuals vs. Predicted 7. 99 Normal % probability 95 9 8 7 5 3 5 Residuals 4.45-8.3 -.5-33.8-33.8 -.5-8.3 4.45 7. 9. 34.35 49.5 64.65 79.8 Residual Predicted Residuals vs. Circuit Design 7. 4.45 Residuals -8.3 -.5-33.8 3 4 Circuit Design (c) Which circuit design would you select for use? Low noise is best. From the Design Expert Output, the Fisher LSD procedure comparing the difference in means identifies Type as having lower noise than Types and 4. Although the LSD procedure comparing Types and 3 has a P-value greater than.5, it is less than.. Unless there are other reasons for choosing Type 3, Type would be selected. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3-3 Four chemists are asked to determine the percentage of methyl alcohol in a certain chemical compound. Each chemist makes three determinations, and the results are the following: Chemist Percentage of Methyl Alcohol 84.99 84.4 84.38 85.5 85.3 84.88 3 84.7 84.48 85.6 4 84. 84. 84.55 (a) Do chemists differ significantly? Use =.5. There is no significant difference at the 5% level, but chemists differ significantly at the % level. Design Expert Output Response: Methyl Alcohol in % ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.4 3.35 3.5.83 not significant A.4 3.35 3.5.83 Residual.86 8. Lack of Fit. Pure Error.86 8. Cor Total.9 The Model F-value of 3.5 implies there is a 8.3% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 84.47.9-85.5.9 3-3 84.79.9 4-4 84.8.9 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.58.7 -.8.67 vs 3 -.3.7 -.8.73 vs 4.9.7.7.549 vs 3.7.7..3479 vs 4.77.7.88.5 3 vs 4.5.7.88.966 (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals.85.45 -.95 -.43 -.43 -.95.45.85.5 84.8 84.48 84.67 84.86 85.5 Residual Predicted Residuals vs. Chemist.5.85 Residuals.45 -.95 -.43 3 4 Chemist (c) If chemist is a new employee, construct a meaningful set of orthogonal contrasts that might have been useful at the start of the experiment. Chemists Total C C C3 53.4-55.6-3 3 54.36-4 5.85 Contrast Totals: -4.86.39 -.5 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY SS SS SS C C C3 4. 86 3. 39 36. 5 3. 656. 8. 38. 656 FC 6. 5*. 77. 8 FC. 75. 77. 38 FC3 3. 54. 77 Only contrast is significant at 5%. 3-4 Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the three brands are different. Five batteries of each brand are tested with the following results: Weeks of Life Brand Brand Brand 3 76 8 96 8 9 75 96 96 84 98 9 8 (a) Are the lives of these brands of batteries different? Yes, at least one of the brands is different. Design Expert Output Response: Life in Weeks ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 96.3 598.7 38.34 <. significant A 96.3 598.7 38.34 <. Residual 87. 5.6 Lack of Fit. Pure Error 87. 5.6 Cor Total 383.33 4 The Model F-value of 38.34 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 95..77-79.4.77 3-3.4.77 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs 5.8.5 6.33 <. vs 3-5..5 -.8.594 vs 3 -..5-8.4 <. (b) Analyze the residuals from this experiment. There is nothing unusual about the residuals. 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 7.6 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5 Residuals 4.6.6 -.4-4.4-4.4 -.4.6 4.6 7.6 79.4 84.65 89.9 95.5.4 Residual Predicted Residuals vs. Brand 7.6 4.6 Residuals.6 -.4-4.4 3 Brand (c) Construct a 95 percent interval estimate on the mean life of battery brand. Construct a 99 percent interval estimate on the mean difference between the lives of battery brands and 3. y i. t,n a MS 5. 6 Brand : 79. 4. 79 5 79. 4 3. 849 75. 55 83. 49 Brand - Brand 3: y i. n E y j. t,n a 5. 6 79. 4. 4 3. 55 5 8. 63 3 3. 369 MS n E 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (d) Which brand would you select for use? If the manufacturer will replace without charge any battery that fails in less than 85 weeks, what percentage would the company expect to replace? Chose brand 3 for longest life. Mean life of this brand in.4 weeks, and the variance of life is estimated by 5.6 (MSE). Assuming normality, then the probability of failure before 85 weeks is: 85. 4 5. 6 3. 9. 5 That is, about 5 out of, batteries will fail before 85 week. 3-5 Four catalysts that may affect the concentration of one component in a three component liquid mixture are being investigated. The following concentrations are obtained: Catalyst 3 4 58. 56.3 5. 5.9 57. 54.5 54. 49.9 58.4 57. 55.4 5. 55.8 55.3 5.7 54.9 (a) Do the four catalysts have the same effect on concentration? No, their means are different. Design Expert Output Response: Concentration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 85.68 3 8.56 9.9.4 significant A 85.68 3 8.56 9.9.4 Residual 34.56.88 Lack of Fit. Pure Error 34.56.88 Cor Total.4 5 The Model F-value of 9.9 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 56.9.76-55.77.85 3-3 53.3.98 4-4 5.3.85 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.3.4.99.346 vs 3 3.67.4.96. vs 4 5.77.4 5.7.3 vs 3.54.3.96.735 vs 4 4.65. 3.87. 3 vs 4..3.63.98 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. Normal plot of residuals.6667 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5.84667 Residuals -.483333 -.8833-3.3333-3.3333 -.8833 -.483333.84667.6667 5.3 5.57 54. 55.46 56.9 Residual Predicted Residuals vs. Catalyst.6667.84667 Residuals -.483333 -.8833-3.3333 3 4 Catalyst (c) Construct a 99 percent confidence interval estimate of the mean response for catalyst. y i. t,n a MS. 88 Catalyst : 56. 9 3. 55 5 56. 9. 386 54. 584 59. 86 n E 3-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3-6 An experiment was performed to investigate the effectiveness of five insulating materials. Four samples of each material were tested at an elevated voltage level to accelerate the time to failure. The failure times (in minutes) is shown below. Material Failure Time (minutes) 57 94 78 4 8 3 88 56 576 4355 4 495 74 537 5 5 7 5 9 (a) Do all five materials have the same effect on mean failure time? No, at least one material is different. Design Expert Output Response: Failure Time in Minutes ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3E+8 4.58E+7 6.9.38 significant A.3E+8 4.58E+7 6.9.38 Residual 6.5E+75 4.67E+6 Lack of Fit. Pure Error 6.5E+75 4.67E+6 Cor Total.657E+89 The Model F-value of 6.9 implies the model is significant. There is only a.38% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 59.75.67-6.5.67 3-3 94.75.67 4-4 573..67 5-5.75.67 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs 53.5 443.44..967 vs 3-78. 443.44 -.93.73 vs 4-5563.5 443.44-3.85.6 vs 5 49. 443.44..99 vs 3-935.5 443.44 -.3.6 vs 4-576.75 443.44-3.96.3 vs 5-4.5 443.44-3.8E-3.9976 3 vs 4-78.5 443.44 -.93.73 3 vs 5 93. 443.44.3.64 4 vs 5 57.5 443.44 3.96.3 (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. What information do these plots convey? 3-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 437 Residuals vs. Predicted Normal plot of residuals 99 Residuals 938.5-45.5-839.5 Normal % probability 95 9 8 7 5 3 5-58 6.5 435.44 864.6 493.8 573. -58-839.5-45.5 938.5 437 Predicted Residual The plot of residuals versus predicted has a strong outward-opening funnel shape, which indicates the variance of the original observations is not constant. The residuals plotted in the normal probability plot also imply that the normality assumption is not valid. A data transformation is recommended. (c) Based on your answer to part (b) conduct another analysis of the failure time data and draw appropriate conclusions. A natural log transformation was applied to the failure time data. The analysis identifies that there exists at least one difference in treatment means. Design Expert Output Response: Failure Time in Minutes Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 65.6 4 4.6 37.66 <. significant A 65.6 4 4.6 37.66 <. Residual 6.44 5. Lack of Fit. Pure Error 6.44 5. Cor Total 8.49 9 The Model F-value of 37.66 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 5.5.5 -.4.5 3-3 7.7.5 4-4 8..5 5-5.9.5 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs 3.8.74 5.5. vs 3 -.66.74-3.6.6 vs 4-3.6.74-4.7.7 vs 5 3.5.74 4.5.7 vs 3-6.47.74-8.75 <. 3-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY vs 4-6.97.74-9.4 <. vs 5 -.66.74 -.89.3856 3 vs 4 -.5.74 -.67.56 3 vs 5 5.8.74 7.85 <. 4 vs 5 6.3.74 8.5 <. There is nothing unusual about the residual plots when the natural log transformation is applied. Normal plot of residuals.6479 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5.733576 Residuals -.8766 -.95 -.945 -.945 -.95 -.8766.733576.6479.4.99 4.73 6.47 8. Residual Predicted Residuals vs. Material.6479.733576 Residuals -.8766 -.95 -.945 3 4 5 Material 3-7 A semiconductor manufacturer has developed three different methods for reducing particle counts on wafers. All three methods are tested on five wafers and the after-treatment particle counts obtained. The data are shown below. Method Count 3 4 6 4 4 3 35 3 58 7 97 68 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Do all methods have the same effect on mean particle count? No, at least one method has a different effect on mean particle count. Design Expert Output Response: Count ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 8963.73 448.87 7.9.64 significant A 8963.73 448.87 7.9.64 Residual 6796. 566.33 Lack of Fit. Pure Error 6796. 566.33 Cor Total 5759.73 4 The Model F-value of 7.9 implies the model is significant. There is only a.64% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 3.4.64-38..64 3-3 73..64 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -4.8 5.5 -.65.53 vs 3-59.6 5.5-3.96.9 vs 3-34.8 5.5 -.3.393 (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. Are there potential concerns about the validity of the assumptions? The plot of residuals versus predicted appears to be funnel shaped. This indicates the variance of the original observations is not constant. The residuals plotted in the normal probability plot do not fall along a straight line, which suggests that the normality assumption is not valid. A data transformation is recommended. 47 Residuals vs. Predicted Normal plot of residuals 99 Residuals 3.75.5 -.75 Normal % probability 95 9 8 7 5 3 5-46 3.4 8.3 43. 58. 73. -46 -.75.5 3.75 47 Predicted Residual 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Based on your answer to part (b) conduct another analysis of the particle count data and draw appropriate conclusions. For count data, a square root transformation is often very effective in resolving problems with inequality of variance. The analysis of variance for the transformed response is shown below. The difference between methods is much more apparent after applying the square root transformation. Design Expert Output Response: Count Transform: Square root Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 63.9 3.95 9.84.3 significant A 63.9 3.95 9.84.3 Residual 38.96 3.5 Lack of Fit. Pure Error 38.96 3.5 Cor Total.86 4 The Model F-value of 9.84 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 3.6.8-6..8 3-3 8.3.8 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.84.4 -.49.85 vs 3-5.4.4-4.4.8 vs 3 -..4 -.94.767 3-8 Consider testing the equality of the means of two normal populations, where the variances are unknown but are assumed to be equal. The appropriate test procedure is the pooled t test. Show that the pooled t test is equivalent to the single factor analysis of variance. S p y. y. t ~ tn assuming n = n = n S p n n n n y j y. y j y. yij y. j j n i j n MS E for a= n yi. y.. Furthermore, y. y., which is exactly the same as SS Treatments in a one-way n n i classification with a=. Thus we have shown that SSTreatments t. In general, we know that tu F, u MS so that t ~ F, n. Thus the square of the test statistic from the pooled t-test is the same test statistic that results from a single-factor analysis of variance with a=. E 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3-33 3-9 Show that the variance of the linear combination a i c i y i. is a i n i c i. i i n j ij. a i i a i n j ij i a i i. i a i i. i y V c y V c y c V y c V, y ij V a i c i n i 3- In a fixed effects experiment, suppose that there are n observations for each of four treatments. Let 3 Q,,Q Q be single-degree-of-freedom components for the orthogonal contrasts. Prove that 3 Q Q Q SS Treatments. 3 3 4 3 3 4 3 4 3 3 Q SS y y C Q SS y y y C Q SS y y y y C C.. C... C.... n ) y y ( Q n ) y y y ( Q n ) y y y y ( Q......... 6 3 4 3 3 4 3 4 3 n y y y Q Q Q i j i j. i. i. 6 9 4 3 and since 4 i i... j i j.. i y y y y, we have Treatments.. i i. i.. i. SS n y n y n y y Q Q Q 4 3 4 4 3 for a=4. 3- Use Bartlett's test to determine if the assumption of equal variances is satisfied in Problem 3-4. Use =.5. Did you reach the same conclusion regarding the equality of variance by examining the residual plots? c q.36, where

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY S S S 3. 4. 8. 8 q N alog S n c 3a a a i p i log n N a i a i ni S i S p N a 5. 5 4. 8 5. 8 S p 5 3 5. 5 4. 8 5. 8 S p 5. 6 5 3 a c 5 5 3 33 i 3 c. 389 33 4 i S q q N alog S n 5 3log 5. 6 4log. log 4. 8 log. 8 i q 4. 3754. 5. 675 p a i log q. 675. 36. 36. 3386. 5, 4 9. 49 c. 389 Cannot reject null hypothesis; conclude that the variance are equal. This agrees with the residual plots in Problem 3-6. S i i 3- Use the modified Levene test to determine if the assumption of equal variances is satisfied on Problem 3-4. Use =.5. Did you reach the same conclusion regarding the equality of variances by examining the residual plots? The absolute value of Battery Life brand median is: y ij yi Brand Brand Brand 3 4 4 8 4 5 4 4 4 The analysis of variance indicates that there is not a difference between the different brands and therefore the assumption of equal variances is satisfired. Design Expert Output Response: Mod Levine ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] 3-34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Sum of Mean F Source Squares DF Square Value Prob > F Model.93.47.7.938 A.93.47.7.938 Pure Error 8. 6.67 Cor Total 8.93 4 3-3 Refer to Problem 3-. If we wish to detect a maximum difference in mean response times of milliseconds with a probability of at least.9, what sample size should be used? How would you obtain a preliminary estimate of? nd a, use MS E from Problem 3- to estimate n 3 6. 9. 986n. Letting. 5, P(accept) =., a Trial and Error yields: n P(accept) 5..7 6 5.43.9 7 8.6.4 Choose n 6, therefore N 8 Notice that we have used an estimate of the variance obtained from the present experiment. This indicates that we probably didn t use a large enough sample (n was 5 in problem 3-) to satisfy the criteria specified in this problem. However, the sample size was adequate to detect differences in one of the circuit types. When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert will be able to bound the variability in the response, by statements such as the standard deviation is going to be at least or the standard deviation shouldn t be larger than. 3-4 Refer to Problem 3-4. (a) If we wish to detect a maximum difference in mean battery life of.5 percent with a probability of at least.9, what sample size should be used? Discuss how you would obtain a preliminary estimate of for answering this question. Use the MS E from Problem 3-4. nd n. 59. 6667. 44n a 3 5. 6 Letting 5., P(accept) =., a 3-35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Trial and Error yields: n P(accept) 4 7.895.8 45 3.3. 5 47.369.5 Choose n 45, therefore N 35 See the discussion from the previous problem about the estimate of variance. (b) If the difference between brands is great enough so that the standard deviation of an observation is increased by 5 percent, what sample size should be used if we wish to detect this with a probability of at least.9? a N a 5 3. 5 P( accept ).. P n. 5. n n 565 Trial and Error yields: n P(accept) 4 7 4.84.3 45 3 5.3. 5 47 5.4. Choose n 5, therefore N 5 3-5 Consider the experiment in Problem 3-6. If we wish to construct a 95 percent confidence interval on the difference in two mean battery lives that has an accuracy of weeks, how many batteries of each brand must be tested?.5 MS E 5. 6 width t. 5,N a MS n Trial and Error yields: n t width 5.79 5.44 7.5 3.6 3 9.99.996 3 93.99.96 Choose n 3, therefore N 93 E 3-6 Suppose that four normal populations have means of =5, =6, 3 =5, and 4 =6. How many observations should be taken from each population so that the probability or rejecting the null hypothesis 3-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY of equal population means is at least.9? Assume that =.5 and that a reasonable estimate of the error variance is =5.,i,, 3, 4 i 4 5, 4 i 4 i i i i 4 5, 3 55 5, 4 5 n i a n n n 4 5 n, 5 3, 4., From the O.C. curves we can construct the following: n - 4..8.8 5.4 6.8.9 Therefore, select n=5 3-7 Refer to Problem 3-6. (a) How would your answer change if a reasonable estimate of the experimental error variance were = 36? n, 5 3, 4. n i n. 6944n a 436. 6944n, From the O.C. curves we can construct the following: n - 5.863 6.4.76 6.4.5.85 7.5 4.9.9 Therefore, select n=7 (b) How would your answer change if a reasonable estimate of the experimental error variance were = 49? n, 5 3, 4. n i n. 5n a 449. 5n, From the O.C. curves we can construct the following: 3-37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY n - 7.89 4.6.84 8. 8..89 9.4 3.9.9 Therefore, select n=9 (c) Can you draw any conclusions about the sensitivity of your answer in the particular situation about how your estimate of affects the decision about sample size? As our estimate of variability increases the sample size must increase to ensure the same power of the test. (d) Can you make any recommendations about how we should use this general approach to choosing n in practice? When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert will be able to bound the variability in the response, by statements such as the standard deviation is going to be at least or the standard deviation shouldn t be larger than. 3-8 Refer to the aluminum smelting experiment described in Section 4-. Verify that ratio control methods do not affect average cell voltage. Construct a normal probability plot of residuals. Plot the residuals versus the predicted values. Is there an indication that any underlying assumptions are violated? Design Expert Output Response: Cell Average ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.746E-3 3 9.53E-4..89 not significant A.746E-3 3 9.53E-4..89 Residual.9 4.48E-3 Lack of Fit. Pure Error.9 4.48E-3 Cor Total.9 3 The "Model F-value" of. implies the model is not significant relative to the noise. There is a 89. % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 4.86.7-4.83.7 3-3 4.85.7 4-4 4.84.7 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.7.39.69.498 vs 3.3.39.35.7337 vs 4.5.39.65.55 vs 3 -.3.39 -.35.7337 vs 4 -.667E-3.39 -.43.966 3 vs 4..39.3.7659 The following residual plots are satisfactory. 3-38

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals.55 -.5 -.565 3 -. -. -.565 -.5.55.5 4.833 4.84 4.847 4.853 4.86 Residual Predicted Residuals vs. Algorithm.5.55 Residuals -.5 3 -.565 -. 3 4 Algorithm 3-9 Refer to the aluminum smelting experiment in Section 3-8. Verify the analysis of variance for pot noise summarized in Table 3-3. Examine the usual residual plots and comment on the experimental validity. Design Expert Output Response: Cell StDev Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.7 3.6.96 <. significant A 6.7 3.6.96 <. Residual.87.94 Lack of Fit. Pure Error.87.94 Cor Total 8.4 3 The Model F-value of.96 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. 3-39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - -3.9. - -3.5. 3-3 -.. 4-4 -3.36. Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.4.8.38.7 vs 3 -.89.8-5.3 <. vs 4.7.8.5.445 vs 3 -.3.8-7.4 <. vs 4 -.5.8 -.86.3975 3 vs 4.6.8 6.55 <. The following residual plots identify the residuals to be normally distributed, randomly distributed through the range of prediction, and uniformly distributed across the different algorithms. This validates the assumptions for the experiment. Normal plot of residuals.5896 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5.45645 Residuals -.669 -.88858 3 -.556 -.556 -.88858 -.669.45645.5896-3.5-3.8 -.85 -.53 -. Residual Predicted Residuals vs. Algorithm.5896.45645 Residuals -.669 3 -.88858 -.556 3 4 Algorithm 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3-3 Four different feed rates were investigated in an experiment on a CNC machine producing a component part used in an aircraft auxiliary power unit. The manufacturing engineer in charge of the experiment knows that a critical part dimension of interest may be affected by the feed rate. However, prior experience has indicated that only dispersion effects are likely to be present. That is, changing the feed rate does not affect the average dimension, but it could affect dimensional variability. The engineer makes five production runs at each feed rate and obtains the standard deviation of the critical dimension (in -3 mm). The data are shown below. Assume that all runs were made in random order. Feed Rate Production Run (in/min) 3 4 5.9..3.8.7.6.9..7. 4..8.8.5.6 6.9.3.5.. (a) Does feed rate have any effect on the standard deviation of this critical dimension? Because the residual plots were not acceptable for the non-transformed data, a square root transformation was applied to the standard deviations of the critical dimension. Based on the computer output below, the feed rate has an effect on the standard deviation of the critical dimension. Design Expert Output Response: Run StDev Transform: Square root Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.4 3.3 7.5.3 significant A.4 3.3 7.5.3 Residual.3 6.93E-3 Lack of Fit. Pure Error.3 6.93E-3 Cor Total.7 9 The Model F-value of 7.5 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -.3. -.3. 3-4.7. 4-6.39. Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs 4.37E-3.8.6.876 vs 3.3.8.5.68 vs 4 -.88.8-3.8.58 vs 3.7.8.99.3373 vs 4 -.9.8-3.34.4 3 vs 4 -..8-4.33.5 (b) Use the residuals from this experiment of investigate model adequacy. Are there any problems with experimental validity? The residual plots are satisfactory. 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals.58487 Residuals vs. Predicted 99 Normal % probability 95 9 8 7 5 3 5.8646 Residuals -.8983 -.356 -.6864 -.6864 -.356 -.8983.8646.58487.7.3.33.36.39 Residual Predicted Residuals vs. Feed Rate.58487.8646 Residuals -.8983 -.356 -.6864 3 4 Feed Rate 3-3 Consider the data shown in Problem 3-. (a) Write out the least squares normal equations for this problem, and solve them for and i, using the usual constraint 3 3 i ˆ i. Estimate. 5 ˆ 5ˆ 5ˆ 5ˆ 3 =7 5 ˆ 5ˆ =54 5 ˆ 5ˆ = 5 ˆ 5ˆ 3 =4 Imposing ˆ, therefore ˆ 3. 8, ˆ 3, ˆ 8 4, ˆ 5 4 i i.. 3. 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ ˆ 3. 8. 4 4. (b) Solve the equations in (a) using the constraint ˆ 3. Are the estimators ˆ i and ˆ the same as you found in (a)? Why? Now estimate and compare your answer with that for (a). What statement can you make about estimating contrasts in the i? Imposing the constraint, ˆ 3 we get the following solution to the normal equations: ˆ 8. 4, ˆ. 4, ˆ 3. 8, and ˆ 3. These estimators are not the same as in part (a). However, ˆ ˆ. 4 3. 8 4, is the same as in part (a). The contrasts are estimable.. (c) Estimate, 3 and using the two solutions to the normal equations. Compare the results obtained in each case. Contrast Estimated from Part (a) Estimated from Part (b).8.8 3 3-9. -9. 9. 4.6 Contrasts and are estimable, 3 is not estimable. 3-3 Apply the general regression significance test to the experiment in Example 3-. Show that the procedure yields the same results as the usual analysis of variance. From Table 3-3: y.. 376 from Example 3-, we have: ˆ 5. 4 ˆ. 56 3 5 i 5 j R, SS E y ij ˆ 5. 4 ˆ 6. 56 4 ˆ. 36 ˆ 4. 4 5 69, with 5 degrees of freedom. ˆ y 5 i.. 5 i ˆy i. 5. 4376 5. 449. 3677. 5688) 6. 568 4. 454 63,. 8 with 5 degrees of freedom. 5 j y ij R, 69 63. 8 6. with 5-5 degrees of freedom. This is identical to the SS E found in Example 3-. 3-43

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The reduced model: ˆy 5. 4376 5655. 4 R.. R, with degree of freedom. R, R 63. 8 5655. 4 475. 76 Note: R SSTreatment Finally, F from Example 3-. R t 4 SSE 8. 94 4 76 8. 6. which is the same as computed in Example 3-., with 5-=4 degrees of freedom. 3-33 Use the Kruskal-Wallis test for the experiment in Problem 3-. Are the results comparable to those found by the usual analysis of variance? From Design Expert Output of Problem 3- Response: Life in in h ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.7 3.6 3.5.55 not significant A 3.6 3.5 3.5.55 Residual 65.99 3.3 Lack of Fit. Pure Error 65.99 3.3 Cor Total 96.6 3 H NN a i R n i. i 3N 4. 5, 3. 44. 34 4 5 5. 8 7 8 Accept the null hypothesis; the treatments are not different. This agrees with the analysis of variance. 3-34 Use the Kruskal-Wallis test for the experiment in Problem 3-. Compare conclusions obtained with those from the usual analysis of variance? From Design Expert Output of Problem 3- Response: Noise ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4. 3 44..78 <. significant A 4. 3 44..78 <. Residual 948.8 6 84.3 Lack of Fit. 3-44

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Pure Error 948.8 6 84.3 Cor Total 499.8 9 H NN a i R i. 3N ni. 5, 4. 69. 3 6 3. 9 84 Reject the null hypothesis because the treatments are different. This agrees with the analysis of variance. 3-35 Consider the experiment in Example 3-. Suppose that the largest observation on tensile strength is incorrectly recorded as 5. What effect does this have on the usual analysis of variance? What effect does is have on the Kruskal-Wallis test? The incorrect observation reduces the analysis of variance F from 4.76 to 5.44. It does not change the value of the Kruskal-Wallis test. 3-45

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 4 Randomized Blocks, Latin Squares, and Related Designs Solutions 4- A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze the data from this experiment (use =.5) and draw appropriate conclusions. Bolt Chemical 3 4 5 73 68 74 7 67 73 67 75 7 7 3 75 68 78 73 68 4 73 7 75 75 69 Design Expert Output Response: Strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 57. 4 39.5 Model.95 3 4.3.38. not significant A.95 3 4.3.38. Residual.8.8 Cor Total 9.75 9 The "Model F-value" of.38 implies the model is not significant relative to the noise. There is a. % chance that a "Model F-value" this large could occur due to noise. Std. Dev..35 R-Squared.377 Mean 7.75 Adj R-Squared.58 C.V..88 Pred R-Squared -.746 PRESS 6.56 Adeq Precision.558 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 7.6.6-7.4.6 3-3 7.4.6 4-4 7.6.6 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.8.85 -.94.3665 vs 3 -.8.85 -..564 vs 4 -..85 -.35.37 vs 3 -..85 -.7.635 vs 4 -..85 -.4.846 3 vs 4 -..85 -.3.885 There is no difference among the chemical types at =.5 level. 4- Three different washing solutions are being compared to study their effectiveness in retarding bacteria growth in five-gallon milk containers. The analysis is done in a laboratory, and only three trials 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY can be run on any day. Because days could represent a potential source of variability, the experimenter decides to use a randomized block design. Observations are taken for four days, and the data are shown here. Analyze the data from this experiment (use =.5) and draw conclusions. Days Solution 3 4 3 8 39 6 4 7 44 3 5 4 Design Expert Output Response: Growth ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 6.9 3 368.97 Model 73.5 35.75 4.7.3 significant A 73.5 35.75 4.7.3 Residual 5.83 6 8.64 Cor Total 86.5 The Model F-value of 4.7 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..94 R-Squared.934 Mean 8.75 Adj R-Squared.985 C.V. 5.68 Pred R-Squared.755 PRESS 7.33 Adeq Precision 9.687 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 3..47-5.5.47 3-3 8..47 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.5.8 -.8.36 vs 3 5..8 7..4 vs 3 7.5.8 8.3. There is a difference between the means of the three solutions. The Fisher LSD procedure indicates that solution 3 is significantly different than the other two. 4-3 Plot the mean tensile strengths observed for each chemical type in Problem 4- and compare them to a scaled t distribution. What conclusions would you draw from the display? 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution () () (3,4) 7. 7. 7. 73. Mean Strength S y i. MS b E. 8 5. 63 There is no obvious difference between the means. This is the same conclusion given by the analysis of variance. 4-4 Plot the average bacteria counts for each solution in Problem 4- and compare them to an appropriately scaled t distribution. What conclusions can you draw? Scaled t Distribution (3) () () 5 5 5 Bacteria Growth S y i. MS b E 8. 64 4. 47 There is no difference in mean bacteria growth between solutions and. However, solution 3 produces significantly lower mean bacteria growth. This is the same conclusion reached from the Fisher LSD procedure in Problem 4-4. 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-5 An article in the Fire Safety Journal ( The Effect of Nozzle Design on the Stability and Performance of Turbulent Water Jets, Vol. 4, August 98) describes an experiment in which a shape factor was determined for several different nozzle designs at six levels of efflux velocity. Interest focused on potential differences between nozzle designs, with velocity considered as a nuisance variable. The data are shown below: Jet Efflux Velocity (m/s) Nozzle Design.73 4.37 6.59.43 3.46 8.74.78.8.8.75.77.78.85.85.9.86.8.83 3.93.9.95.89.89.83 4.4.97.98.88.86.83 5.97.86.78.76.76.75 (a) Does nozzle design affect the shape factor? Compare nozzles with a scatter plot and with an analysis of variance, using =.5. Design Expert Output Response: Shape ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.63 5.3 Model. 4.6 8.9.3 significant A. 4.6 8.9.3 Residual.57.865E-3 Cor Total. 9 The Model F-value of 8.9 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..54 R-Squared.647 Mean.86 Adj R-Squared.5688 C.V. 6.3 Pred R-Squared.96 PRESS.3 Adeq Precision 9.438 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -.78. -.85. 3-3.9. 4-4.94. 5-5.8. Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.7.3 -.3.3 vs 3 -..3-3.88.9 vs 4 -.6.3-5.3 <. vs 5 -.3.3 -..377 vs 3 -.48.3 -.56.335 vs 4 -.9.3 -.9.86 vs 5.4.3.9.3 3 vs 4 -.4.3 -.35.96 3 vs 5.88.3.86.97 4 vs 5.3.3 4..4 Nozzle design has a significant effect on shape factor. 4-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.4 One Factor Plot.436 Shape.94478.84776.749435 3 4 5 Nozzle Design (b) Analyze the residual from this experiment. The plots shown below do not give any indication of serious problems. Thre is some indication of a mild outlier on the normal probability plot and on the plot of residualks versus the predicted velocity. Normal plot of residuals Residuals vs. Predicted.333 99 Normal % probability 95 9 8 7 5 3 5.73333 Residuals.3333 -.86667 -.786667 -.786667 -.86667.3333.73333.333.73.8.87.95. Residual Predicted 4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.333 Residuals vs. Nozzle Design.73333 Residuals.3333 -.86667 -.786667 3 4 5 Nozzle Design (c) Which nozzle designs are different with respect to shape factor? Draw a graph of average shape factor for each nozzle type and compare this to a scaled t distribution. Compare the conclusions that you draw from this plot to those from Duncan s multiple range test. S y i. MS b E. 865 6. 85 R = r.5 (,) S yi. = (.95)(.85)=.6446 R 3 = r.5 (3,) S yi. = (3.)(.85)=.6774 R 4 = r.5 (4,) S yi. = (3.8)(.85)=.6949 R 5 = r.5 (5,) S yi. = (3.5)(.85)=.7 Mean Difference R vs 4.667 >.7 different vs 3. >.6949 different vs.767 >.6774 different vs 5.367 <.6446 5 vs 4.3 >.6949 different 5 vs 3.8833 >.6774 different 5 vs.4 <.6446 vs 4.9 >.6774 different vs 3.4833 <.6446 3 vs 4.467 <.6446 4-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution () (5) () (3) (4 ).75.8.85.9.95 Shape Factor 4-6 Consider the ratio control algorithm experiment described in Chapter 3, Section 3-8. The experiment was actually conducted as a randomized block design, where six time periods were selected as the blocks, and all four ratio control algorithms were tested in each time period. The average cell voltage and the standard deviation of voltage (shown in parentheses) for each cell as follows: Ratio Control Time Period Algorithms 3 4 5 6 4.93 (.5) 4.86 (.4) 4.75 (.5) 4.95 (.6) 4.79 (.3) 4.88 (.5) 4.85 (.4) 4.9 (.) 4.79 (.3) 4.85 (.5) 4.75 (.3) 4.85 (.) 3 4.83 (.9) 4.88 (.3) 4.9 (.) 4.75 (.5) 4.8 (.8) 4.9 (.) 4 4.89 (.3) 4.77 (.4) 4.94 (.5) 4.86 (.5) 4.79 (.3) 4.76 (.) (a) Analyze the average cell voltage data. (Use =.5.) Does the choice of ratio control algorithm affect the cell voltage? Design Expert Output Response: Average ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.7 5 3.487E-3 Model.746E-3 3 9.53E-4.9.94 not significant A.746E-3 3 9.53E-4.9.94 Residual.7 5 4.8E-3 Cor Total.9 3 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a 9.4 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..69 R-Squared.366 Mean 4.84 Adj R-Squared -.56 C.V..43 Pred R-Squared -.466 PRESS.8 Adeq Precision.688 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 4.86.8 4-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - 4.83.8 3-3 4.85.8 4-4 4.84.8 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.7.4.67.556 vs 3.3.4.33.7438 vs 4.5.4.6.549 vs 3 -.3.4 -.33.7438 vs 4 -.667E-3.4 -.4.9674 3 vs 4..4.9.7748 The ratio control algorithm does not affect the mean cell voltage. (b) Perform an appropriate analysis of the standard deviation of voltage. (Recall that this is called pot noise. ) Does the choice of ratio control algorithm affect the pot noise? Design Expert Output Response: StDev Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.94 5.9 Model 6.7 3.6 33.6 <. significant A 6.7 3.6 33.6 <. Residual.93 5.6 Cor Total 8.4 3 The Model F-value of 33.6 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..5 R-Squared.8693 Mean -3.4 Adj R-Squared.843 C.V. -8.8 Pred R-Squared.6654 PRESS.37 Adeq Precision.446 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - -3.9. - -3.5. 3-3 -.. 4-4 -3.36. Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.4.4.93.3 vs 3 -.89.4-6.9 <. vs 4.7.4.87.83 vs 3 -.3.4-9. <. vs 4 -.5.4 -.6.34 3 vs 4.6.4 8.6 <. A natural log transformatio was applied to the pot noise data. The ratio control algorithm does affect the pot noise. (c) Conduct any residual analyses that seem appropriate. 4-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.88958 99 Normal % probability 95 9 8 7 5 3 5.6945 Residuals -.35673 -.978 -.35993 -.35993 -.978 -.35673.6945.88958-3.73-3.6 -.78 -.3 -.84 Residual Predicted Residuals vs. Algorithm.88958.6945 Residuals -.35673 -.978 -.35993 3 4 Algorithm The normal probability plot shows slight deviations from normality; however, still acceptable. (d) Which ratio control algorithm would you select if your objective is to reduce both the average cell voltage and the pot noise? Since the ratio control algorithm has little effect on average cell voltage, select the algorithm that minimizes pot noise, that is algorithm #. 4-7 An aluminum master alloy manufacturer produces grain refiners in ingot form. This company produces the product in four furnaces. Each furnace is known to have its own unique operating characteristics, so any experiment run in the foundry that involves more than one furnace will consider furnace a nuisance variable. The process engineers suspect that stirring rate impacts the grain size of the product. Each furnace can be run at four different stirring rates. A randomized block design is run for a particular refiner and the resulting grain size data is shown below. Furnace 4-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Stirring Rate 3 4 5 8 4 5 6 4 5 6 9 5 4 6 9 7 9 3 6 (a) Is there any evidence that stirring rate impacts grain size? Design Expert Output Response: Grain Size ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 65.9 3 55.6 Model.9 3 7.4.85.4995 not significant A.9 3 7.4.85.4995 Residual 78.6 9 8.67 Cor Total 65.44 5 The "Model F-value" of.85 implies the model is not significant relative to the noise. There is a 49.95 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..95 R-Squared.3 Mean 7.69 Adj R-Squared -.38 C.V. 38.3 Pred R-Squared -.46 PRESS 46.7 Adeq Precision 5.39 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -5 5.75.47-8.5.47 3-5 7.75.47 4-8.75.47 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs -.75.8 -.3.93 vs 3 -..8 -.96.36 vs 4-3..8 -.44.836 vs 3.75.8.36.77 vs 4 -.5.8 -..97 3 vs 4 -..8 -.48.645 The analysis of variance shown above indicates that there is no difference in mean grain size due to the different stirring rates. (b) Graph the residuals from this experiment on a normal probability plot. Interpret this plot. 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 99 Normal % probability 95 9 8 7 5 3 5-3.85 -.65 -.35.4375 3.875 Residual The plot indicates that normality assumption is valid. (c) Plot the residuals versus furnace and stirring rate. Does this plot convey any useful information? Residuals vs. Stirring Rate 3.875.4375 Residuals -.35 -.65-3.85 3 4 Stirring Rate The variance is consistent at different stirring rates. Not only does this validate the assumption of uniform variance, it also identifies that the different stirring rates do not affect variance. (d) What should the process engineers recommend concerning the choice of stirring rate and furnace for this particular grain refiner if small grain size is desirable? There really isn t any effect due to the stirring rate. 4-8 Analyze the data in Problem 4- using the general regression significance test. : : : ˆ 4ˆ ˆ 4ˆ 3 3 4 4ˆ 4ˆ 4 3 4 4 3 4 ˆ 4ˆ 3ˆ ˆ ˆ 3ˆ ˆ ˆ 3ˆ ˆ ˆ 3ˆ =5 ˆ =9 ˆ = 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3 : : : : 3 : 4 ˆ 4 3 3 4 ˆ ˆ ˆ 3 3 ˆ ˆ ˆ 3 3 ˆ ˆ ˆ 4ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ =3 3ˆ =34 3ˆ =5 3ˆ =36 3ˆ =5 3 3 3 3 3 4 ˆ ˆ ˆ ˆ Applying the constraints ˆ ˆ, we obtain: i j 5 ˆ, ˆ 5, ˆ 78, 9 89 5 8 ˆ 3, ˆ, ˆ, ˆ 3, ˆ 95 4 5 5 78 9 89 5 R,, 5 9 3 34 5 ij 68 y, SS y R,, Model Restricted to i : E 8 68 69. 7 5. 83 ij 95 36 5 69. 7 : : : : 3 : 4 ˆ 3ˆ 3ˆ 3ˆ 3 3ˆ 4 =5 3 ˆ 3ˆ =34 3 ˆ 3ˆ =5 3 ˆ 3ˆ 3 =36 ˆ 3ˆ =5 3 4 Applying the constraint ˆ j, we obtain: 5 ˆ, ˆ 5 8 89 /, ˆ, ˆ 3, ˆ 95 4. Now: 5 89 5 8 95 R, 5 34 5 36 5 R 535. 67, R,, R, 69. 7 535. 67 73. 5 SSTreatments Model Restricted to : j : : : : 3 ˆ 4ˆ 4ˆ 4ˆ 3 =5 4 ˆ 4ˆ =9 4 ˆ 4ˆ = ˆ 4ˆ =3 4 3 Applying the constraint ˆ i, we obtain: 5 ˆ, ˆ 5, ˆ 78, 9 ˆ 3 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY R R 5 5 78 9, 5 9 3 49. 5, R,, R, 69. 7 49. 5 6. 9 SSBlocks 4-9 Assuming that chemical types and bolts are fixed, estimate the model parameters i and j in Problem 4-. Using Equations 4-4, Applying the constraints, we obtain: 35 3 7 3 7 ˆ, ˆ, ˆ, ˆ 3, ˆ 4, 35 ˆ, 65 ˆ, 75 ˆ 3, ˆ 4, ˆ 5 65 4- Draw an operating characteristic curve for the design in Problem 4-. Does this test seem to be sensitive to small differences in treatment effects? Assuming that solution type is a fixed factor, we use the OC curve in appendix V. Calculate i b a 4 i 3 8. 69 using MS E to estimate. We have: a a b 3 6. If ˆ i MSE, then: 4. 5 and. 7 3 If ˆ i MS E, then: 4. 63 and. 55, etc. 3 This test is not very sensitive to small differences. 4- Suppose that the observation for chemical type and bolt 3 is missing in Problem 4-. Analyze the problem by estimating the missing value. Perform the exact analysis and compare the results. y 3 is missing. ŷ a b 8 57 43 ' ' ' ay. by. 3 y.. 4 3. 36 75 5 Thus, y. =357.5, y.3 =3.5, and y.. =435.5 Source SS DF MS F Chemicals.7844 3 4.65.54 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Bolts 58.8875 4 Error.765.9784 Total 93.4344 8 F.,3, =.66, Chemicals are not significant. 4- Two missing values in a randomized block. Suppose that in Problem 4- the observations for chemical type and bolt 3 and chemical type 4 and bolt 4 are missing. (a) Analyze the design by iteratively estimating the missing values as described in Section 4-.3. ŷ 3 ' ' 4y. 5y. 3 y.. and ' ŷ 44 ' ' ' 4y4. 4 y.. 5y. Data is coded y-7. As an initial guess, set y 3 equal to the average of the observations available for chemical. Thus, y3. 5. Then, 4 4 8 5 6 5. 5 ŷ44 3. 4 4 57 8. 4 ŷ3 5. 4 4 8 5 6 3. 4 ŷ44. 63 4 57 7. 63 ŷ44 5. 44 4 8 5 6 3. 44 ŷ44. 63 ŷ 5 44 ŷ 63 3. 44. Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 56.83 4 39. Model 9.59 3 3..8.56 not significant A 9.59 3 3..8.56 Residual 8.4.53 Cor Total 84.83 9 (b) Differentiate SS E with respect to the two missing values, equate the results to zero, and solve for estimates of the missing values. Analyze the design using these two estimates of the missing values. SS SSE SSE From, we obtain: y y 3 44 E yij y 5 i. y 4. j SS E. 6y3. 6y44 6. 8y3 3. 7y44. y3 y44 y.. R 4-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY. ŷ. ŷ 3 3. ŷ. ŷ 44 44 6. 8 3. 7 ŷ 5 45, ŷ 63 3. 44. These quantities are almost identical to those found in part (a). The analysis of variance using these new data does not differ substantially from part (a). (c) Derive general formulas for estimating two missing values when the observations are in different blocks. SS E y y y y y y y y y y y y i. iu k. kv.u iu.v kv.. iu iu ykv b a ab kv SSE SSE From, we obtain: y y 3 44 whose simultaneous solution is: ŷ ŷ iu kv y' i. ay' a i. ŷ ŷ iu kv a b ab ay' i. by'. j ab ( a )( b ) ay' k. by' ab ab y' a b ab y'.u b a b a b a b by'.u y'.. b a ay' k. by'.v y'.. a b.v.. ŷkv ab y'.. ŷiu ab ab y'.. aba b ay' k. by'.v y' a b ab.. (d) Derive general formulas for estimating two missing values when the observations are in the same block. Suppose that two observations y ij and y kj are missing, ik (same block j). SS E y SSE SSE From, we obtain y y 3 44 whose simultaneous solution is: y y y y y y y y y y i. ij k. kj. j ij kj.. ij ij ykj ŷ ŷ ij kj b ayi. by. j y.. ŷ a b a b kj ayk. by. j y.. ŷ ij a a b a b ab kj 4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ŷ ij b ay by y a b ay by y ayi. by. j y.. k.. j.. a b ŷ kj ayk. by. j y.. a b b a ay by y a b a b i. 4. j i.... j.. 4-3 An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect of the distance of the object from the eye on the focus time. Four different distances are of interest. He has five subjects available for the experiment. Because there may be differences among individuals, he decides to conduct the experiment in a randomized block design. The data obtained follow. Analyze the data from this experiment (use =.5) and draw appropriate conclusions. Subject Distance (ft) 3 4 5 4 6 6 6 6 6 7 6 6 6 8 5 3 3 5 6 4 4 3 Design Expert Output Response: Focus Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 36.3 4 9.7 Model 3.95 3.98 8.6.5 significant A 3.95 3.98 8.6.5 Residual 5.3.7 Cor Total 84.55 9 The Model F-value of 8.6 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..3 R-Squared.689 Mean 4.85 Adj R-Squared.636 C.V. 3.8 Pred R-Squared.9 PRESS 4.5 Adeq Precision.43 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -4 6.8.5-6 5..5 3-8 3.6.5 4-3.8.5 Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs.6.7.4.448 vs 3 3..7 4.48.8 vs 4 3..7 4.. vs 3.6.7.4.448 vs 4.4.7.96.736 3 vs 4 -..7 -.8.784 Distance has a statistically significant effect on mean focus time. 4-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-4 The effect of five different ingredients (A, B, C, D, E) on reaction time of a chemical process is being studied. Each batch of new material is only large enough to permit five runs to be made. Furthermore, each runs requires approximately / hours, so only five runs can be made in one day. The experimenter decides to run the experiment as a Latin square so that day and batch effects can be systematically controlled. She obtains the data that follow. Analyze the data from this experiment (use =.5) and draw conclusions. Day Batch 3 4 5 A=8 B=7 D= C=7 E=3 C= E= A=7 D=3 B=8 3 B=4 A=9 C= E= D=5 4 D=6 C=8 E=6 B=6 A= 5 E=4 D= B=3 A=8 C=8 Minitab Output General Linear Model Factor Type Levels Values Batch random 5 3 4 5 Day random 5 3 4 5 Catalyst fixed 5 A B C D E Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Catalyst 4 4.44 4.44 35.36.3. Batch 4 5.44 5.44 3.86.3.348 Day 4.4.4 3.6.98.455 Error 37.5 37.5 3.7 Total 4 6.64 4-5 An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the assembly time for a color television component. Four operators are selected for the study. Furthermore, the engineer knows that each assembly method produces such fatigue that the time required for the last assembly may be greater than the time required for the first, regardless of the method. That is, a trend develops in the required assembly time. To account for this source of variability, the engineer uses the Latin square design shown below. Analyze the data from this experiment ( =.5) draw appropriate conclusions. Order of Operator Assembly 3 4 C= D=4 A=7 B=8 B=7 C=8 D= A=8 3 A=5 B= C= D=9 4 D= A= B= C=4 Minitab Output General Linear Model Factor Type Levels Values Order random 4 3 4 Operator random 4 3 4 Method fixed 4 A B C D Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P 4-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Method 3 7.5 7.5 4.67 3.8.4 Order 3 8.5 8.5 6.67 3.5.89 Operator 3 5.5 5.5 7.67 9.8. Error 6.5.5.75 Total 5 53. 4-6 Suppose that in Problem 4-4 the observation from batch 3 on day 4 is missing. Estimate the missing value from Equation 4-4, and perform the analysis using this value. y 354 is missing. ŷ p y y y p p y 5 4 46 3 4 5 8 i... j...k... 354. 3 58 Minitab Output General Linear Model Factor Type Levels Values Batch random 5 3 4 5 Day random 5 3 4 5 Catalyst fixed 5 A B C D E Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Catalyst 4 8.676 8.676 3.69.5. Batch 4 6.9 6.9 4.3.4.9 Day 4 8.764 8.764.9.77.567 Error 34.37 34.37.86 Total 4 87.849 4-7 Consider a p x p Latin square with rows ( i ), columns ( k ), and treatments ( j ) fixed. Obtain least squares estimates of the model parameters i, k, j. i p : p ˆ p ˆ p ˆ p ˆ y i p i j i j k p p j j k k p : pˆ pˆ p ˆ p ˆ y, i,,..., p j p : pˆ p ˆ pˆ p ˆ y, j,,..., p k i j i k p i i p j j p : pˆ p ˆ p ˆ p ˆ y k k i... j...k k..., k,,..., p There are 3p+ equations in 3p+ unknowns. The rank of the system is 3p-. Three side conditions are p necessary. The usual conditions imposed are: ˆ ˆ ˆ. The solution is then: i i p j j p k k y... ˆ y... p ˆ y y, i,,..., p i i..... 4-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ y y, j,,..., p j. j.... ˆ y y, k,,..., p k i..... 4-8 Derive the missing value formula (Equation 4-4) for the Latin square design.... SS E y y i... j. y..k y yijk p p p p Let y ijk be missing. Then SS E y ijk y y y y y y y y i.. p ijk SS E where R is all terms without y ijk.. From, we obtain: y. j. p ijk ijk..k p ijk... p ijk R y ijk p p py' i.. y'. j. y'..k p p y'..., or y ijk p y' i.. y'. j. y'..k p p y'... 4-9 Designs involving several Latin squares. [See Cochran and Cox (957), John (97).] The p x p Latin square contains only p observations for each treatment. To obtain more replications the experimenter may use several squares, say n. It is immaterial whether the squares used are the same are different. The appropriate model is y ijkh h i( h ) j k( h ) ( ) jh ijkh i,,..., p j,,..., p k,,..., p h,,...,n where y ijkh is the observation on treatment j in row i and column k of the hth square. Note that ih ( ) and k( h) are row and column effects in the hth square, and h is the effect of the hth square, and ( ) jh is the interaction between treatments and squares. (a) Set up the normal equations for this model, and solve for estimates of the model parameters. Assume that appropriate side conditions on the parameters are ˆ, ˆ, and ˆ j for each h, j j h h ˆ, ˆ for each h, and ˆ jh h i i h for each j. jh k k h 4-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ y ˆ h y ˆ y j ˆ ˆ i( h ) k( h ) ^.... j.. jh...h y y y y i..h..kh y...... y. j.h...h y...h y. j.. y...h y... (b) Write down the analysis of variance table for this design. Source SS DF y.... j.. Treatments y np np y h Squares y... p np Treatment x Squares p-... n- y. j.h y... p np SS Treatments SS Squares (p-)(n-) yi..h Rows y... h n(p-) p np Columns y.. kh y... h p np n(p-) Error subtraction n(p-)(p-) y... Total y ijkh np np - 4- Discuss how the operating characteristics curves in the Appendix may be used with the Latin square design. For the fixed effects model use: p j j p, p p p For the random effects model use: p, p p p 4- Suppose that in Problem 4-4 the data taken on day 5 were incorrectly analyzed and had to be discarded. Develop an appropriate analysis for the remaining data. Two methods of analysis exist: () Use the general regression significance test, or () recognize that the design is a Youden square. The data can be analyzed as a balanced incomplete block design with a=b=5, r=k=4 and =3. Using either approach will yield the same analysis of variance. 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output General Linear Model Factor Type Levels Values Catalyst fixed 5 A B C D E Batch random 5 3 4 5 Day random 4 3 4 Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Catalyst 4 9.8.67 3.4 7.48.8 Batch 4.667.667.97.73.598 Day 3 6.95 6.95.37.58.646 Error 8 3.33 3.33 4.7 Total 9 7.55 4- The yield of a chemical process was measured using five batches of raw material, five acid concentrations, five standing times, (A, B, C, D, E) and five catalyst concentrations (,,,, ). The Graeco-Latin square that follows was used. Analyze the data from this experiment (use =.5) and draw conclusions. Acid Concentration Batch 3 4 5 A=6 B=6 C=9 D=6 E=3 B=8 C= D=8 E= A= 3 C= D= E=6 A=5 B=3 4 D=5 E=5 A= B=4 C=7 5 E= A=4 B=7 C=7 D=4 Factor Type Levels Values Time fixed 5 A B C D E Catalyst random 5 a b c d e Batch random 5 3 4 5 Acid random 5 3 4 5 General Linear Model Analysis of Variance for Yield, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Time 4 34.8 34.8 85.7 4.65. Catalyst 4.. 3..5.79 Batch 4...5.43.785 Acid 4 4.4 4.4 6..4.443 Error 8 46.8 46.8 5.85 Total 4 436. 4-3 Suppose that in Problem 4-5 the engineer suspects that the workplaces used by the four operators may represent an additional source of variation. A fourth factor, workplace (,,, ) may be introduced and another experiment conducted, yielding the Graeco-Latin square that follows. Analyze the data from this experiment (use =.5) and draw conclusions. Order of Operator Assembly 3 4 C= B= D=4 A=8 B=8 C= A= D= 3 A=9 D= B=7 C=5 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 D=9 A=8 C=8 B=6 Minitab Output General Linear Model Factor Type Levels Values Method fixed 4 A B C D Order random 4 3 4 Operator random 4 3 4 Workplac random 4 a b c d Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Method 3 95.5 95.5 3.833 3.47.67 Order 3.5.5.67..996 Operator 3 9. 9. 6.333.69.66 Workplac 3 7.5 7.5.5.7.843 Error 3 7.5 7.5 9.67 Total 5 5. However, there are only three degrees of freedom for error, so the test is not very sensitive. 4-4 Construct a 5 x 5 hypersquare for studying the effects of five factors. Exhibit the analysis of variance table for this design. Three 5 x 5 orthogonal Latin Squares are: ABCDE BCDEA CDEAB DEABC EABCD 345 453 345 534 345 Let rows = factor, columns = factor, Latin letters = factor 3, Greek letters = factor 4 and numbers = factor 5. The analysis of variance table is: Source DF Rows 4 Columns 4 Latin Letters 4 Greek Letters 4 Numbers 4 Error 4 Total 4 4-5 Consider the data in Problems 4-5 and 4-3. Suppressing the Greek letters in 4-3, analyze the data using the method developed in Problem 4-9. Square - Operator Batch 3 4 Row Total C= D=4 A=7 B=8 (39) B=7 C=8 D= A=8 (44) 3 A=5 B= C= D=9 (35) 4 D= A= B= C=4 (46) (3) (5) (4) (36) 64=y 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Square - Operator Batch 3 4 Row Total C= B= D=4 A=8 (43) B=8 C= A= D= (4) 3 A=9 D= B=7 C=5 (4) 4 D=9 A=8 C=8 B=6 (4) (37) (4) (49) (4) 68=y Assembly Methods Totals A y... =65 B y... =68 C y.3.. =9 D y.4.. =9 Source SS DF MS F Assembly Methods Squares 59.5.5 3 53.8.5 4.* A x S Assembly Order (Rows) 8.75 9. 3 6.9 3.7.77 Operators (columns) 7.5 6.75 Error 45.5 3.79 Total 33.5 3 Significant at %. 4-6 Consider the randomized block design with one missing value in Table 4-7. Analyze this data by using the exact analysis of the missing value problem discussed in Section 4-.4. Compare your results to the approximate analysis of these data given in Table 4-8. : 5 4 3 4 3 4 4 4 4 3 3 4 4 =7 3 4 =3 4 = 3 4 =- 4 =5 : 4 4 : 3 3 3 : 4 4 3 4 : 4 4 4 : 4 : 4 3 : 3 4 : 4 3 4 4 =-4 3 4 3 =-3 =6 Applying the constraints ˆ i 3 4 4 3 3 4 4 4 ˆ, we obtain: j 3 =9 4 4 4 59 94 ˆ, ˆ, ˆ, ˆ 3, ˆ 4, 36 36 36 36 36 77 ˆ, 36 68 ˆ, 36 4 ˆ 3, 36 ˆ 4 36 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY R With 7 degrees of freedom. 4 4 ˆ.. j y. j. i j,, ˆy ˆ i yi. 45. 38 78 y ij, SS y R,, E 45. 38. 78 6. which is identical to SS E obtained in the approximate analysis. In general, the SS E in the exact and approximate analyses will be the same. To test H o : the reduced model is yij j ij. The normal equations used are: i ij : 5 ˆ 4 ˆ 4ˆ 3ˆ 3 4ˆ =7 4 : 4 4 ˆ =-4 : 4 4ˆ =-3 3 : 3 3ˆ =6 3 4 : 4 4ˆ =8 4 Applying the constraint ˆ j, we obtain: 9 ˆ, 6 35 ˆ, 6 3 ˆ, 6 3 ˆ 3, 6 4 53 ˆ 4. Now R, ˆy ˆ.. j y. j. 6 j 99 5 with 4 degrees of freedom. R, R,, R, 38. 78 99. 5 39. 53 SSTreatments with 7-4=3 degrees of freedom., R is used to test H o : i. The sum of squares for blocks is found from the reduced model used are: Model Restricted to : j yij. The normal equations i ij : 5 4 3 4 3 4 4 =7 : 4 ˆ 4 =3 : 3 ˆ 3 = 3 : 4 ˆ 4 3 =- 4 : 4 ˆ 4 4 =5 Applying the constraint ˆ, we obtain: i 3 4 9 9 3 ˆ, ˆ, ˆ, ˆ 3, ˆ 4 4-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY with 4 degrees of freedom. with 7-4=3 degrees of freedom. R R, ˆy 4.. ˆ i yi.. i 59 83, R,, R, 38. 78 59. 83 78. 95 SS Blocks Source DF SS(exact) SS(approximate) Tips 3 39.53 39.98 Blocks 3 78.95 79.53 Error 8 6. 6. Total 4 5.74 5.73 Note that for the exact analysis, SS T SS SS SS. Tips Blocks E 4-7 An engineer is studying the mileage performance characteristics of five types of gasoline additives. In the road test he wishes to use cars as blocks; however, because of a time constraint, he must use an incomplete block design. He runs the balanced design with the five blocks that follow. Analyze the data from this experiment (use =.5) and draw conclusions. Car Additive 3 4 5 7 4 3 4 4 3 3 4 3 4 9 4 3 5 8 There are several computer software packages that can analyze the incomplete block designs discussed in this chapter. The Minitab General Linear Model procedure is a widely available package with this capability. The output from this routine for Problem 4-7 follows. The adjusted sums of squares are the appropriate sums of squares to use for testing the difference between the means of the gasoline additives. Minitab Output General Linear Model Factor Type Levels Values Additive fixed 5 3 4 5 Car random 5 3 4 5 Analysis of Variance for Mileage, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Additive 4 3.7 35.7333 8.9333 9.8. Car 4 35.333 35.333 8.883 9.67. Error.67.67.96 Total 9 76.95 4-8 Construct a set of orthogonal contrasts for the data in Problem 4-7. Compute the sum of squares for each contrast. 4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY One possible set of orthogonal contrasts is: H () : 4 5 : : 4 5 : 43 4 5 The sums of squares and F-tests are: H () H (3) H (4) Brand -> 3 4 5 Q i 33/4 /4-3/4-4/4-7/4 cq i i SS F () - - -85/4 3. 39.9 () - -/4 4.3 5.3 (3) - -3/4.4.83 (4) - - 4 - - -5/4.9.5 Contrasts () and () are significant at the % and 5% levels, respectively. 4-9 Seven different hardwood concentrations are being studied to determine their effect on the strength of the paper produced. However the pilot plant can only produce three runs each day. As days may differ, the analyst uses the balanced incomplete block design that follows. Analyze this experiment (use =.5) and draw conclusions. Hardwood Days Concentration (%) 3 4 5 6 7 4 7 4 6 9 6 37 4 34 8 4 9 49 45 5 43 8 3 4 36 3 7 There are several computer software packages that can analyze the incomplete block designs discussed in this chapter. The Minitab General Linear Model procedure is a widely available package with this capability. The output from this routine for Problem 4-9 follows. The adjusted sums of squares are the appropriate sums of squares to use for testing the difference between the means of the hardwood concentrations. Minitab Output General Linear Model Factor Type Levels Values Concentr fixed 7 4 6 8 4 Days random 7 3 4 5 6 7 Analysis of Variance for Strength, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Concentr 6 37.6 37.43 9.57.4. Days 6 394. 394. 65.68 3..7 Error 8 68.57 68.57.7 Total 6.9 4-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-3 Analyze the data in Example 4-6 using the general regression significance test. : 3 3 3 3 3 4 3 3 3 3 3 4 =87 3 4 =8 3 4 =4 3 =6 4 = : 3 3 : 3 3 3 : 3 3 3 4 : 3 3 4 : 3 : 3 3 : 3 4 : 3 3 4 3 = 3 4 3 =7 3 3 3 =4 =8 4 3 4 Applying the constraints i j, we obtain: 87 /, 9/ 8, 7/ 8, 3 4/ 8, 4 / 8, 7/ 8, 3/ 8, 3 4 / 8, 4 / 8 with 7 degrees of freedom. y ij,. E ij 6356 SS y R(,, ) 6356. 635. 75 3. 5. To test H o : i the reduced model is y ij j ij. The normal equations used are: : 3 3 3 3 3 4 =87 : 3 3 = : 3 3 =7 3 : 3 3 3 =4 4 : 3 3 4 =8 Applying the constraint ˆ j, we obtain: 87 ˆ, R with 4 degrees of freedom. 7 ˆ, 6, ˆy ˆ, 6 3 ˆ 3, 6 4 ˆ.. j y. j,. j 633 ˆ 4 6 4-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY R, R,, R, 635. 75 633.. 75 SSTreatments with 7-4=3 degrees of freedom., R is used to test H o :. i The sum of squares for blocks is found from the reduced model used are: yij. The normal equations i ij Model Restricted to : j : 3 3 3 3 3 4 =87 : 3 3 =8 : 3 3 =4 3 : 3 3 3 =6 4 : 3 3 4 = The sum of squares for blocks is found as in Example 4-6. We may use the method shown above to find an adjusted sum of squares for blocks from the reduced model, y. ij i ij 4-3 Prove that a k Q i i a is the adjusted sum of squares for treatments in a BIBD. We may use the general regression significance test to derive the computational formula for the adjusted treatment sum of squares. We will need the following: and the sum of squares we need is: R R kqi ˆ i, kqi kyi. a n y b i,, ˆy ˆ i yi. The normal equation for is, from equation (4-35), ij. j a b ˆ.. j y. j i j a, ˆ y.. ˆ i yi. ˆ j y. j b i j b j y. j k : kˆ a i n ˆ k ˆ ij i j y. j and from this we have: ky a ˆ. j j y. j ky. j ˆ y. jnijˆ i i 4-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY therefore, R, ˆy.. ˆ i yi. a b i j y k a y. j nij ˆ i. j kˆy. j i k k j y. k R(, ) a ˆ y a n y i i. ij. j i k a i i i i a kq Q k a Q i SS a Treatments( adjusted ) 4-3 An experimenter wishes to compare four treatments in blocks of two runs. Find a BIBD for this experiment with six blocks. Treatment Block Block Block 3 Block 4 Block 5 Block 6 X X X X X X 3 X X X 4 X X X Note that the design is formed by taking all combinations of the 4 treatments at a time. The parameters of the design are =, a=4, b=6, k=3, and r= 4-33 An experimenter wishes to compare eight treatments in blocks of four runs. Find a BIBD with 4 blocks and = 3. The design has parameters a=8, b=4, = 3, r= and k=4. It may be generated from a 3 factorial design confounded in two blocks of four observations each, with each main effect and interaction successively confounded (7 replications) forming the 4 blocks. The design is discussed by John (97, pg. ) and Cochran and Cox (957, pg. 473). The design follows: Blocks =(I) =a 3=b 4=ab 5=c 6=ac 7=bc 8=abc X X X X X X X X 3 X X X X 4 X X X X 5 X X X X 6 X X X X 7 X X X X 8 X X X X 9 X X X X X X X X X X X X X X X X 3 X X X X 4 X X X X 4-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-34 Perform the interblock analysis for the design in Problem 4-7. The interblock analysis for Problem 4-7 uses ˆ. 77 and ˆ. 4. A summary of the interblock, intrablock and combined estimates is: Parameter Intrablock Interblock Combined. -.8.8.73..73 3 -. -5.8 -.3 4 -.93 9. -.88 5 -.8 -.8 -.8 4-35 Perform the interblock analysis for the design in Problem 4-9. The interblock analysis for problem MSBlocks( adj ) MSE b 65. 68. 76 4-9 uses ˆ. 7 and 9.. A summary of ar 7 the interblock, intrablock, and combined estimates is give below Parameter Intrablock Interblock Combined -.43 -.79 -.38-8.57-4.9-7.9 3.57-8.79.76 4.7 9..6 5 3.7. 4.67 6-5.4 -.9-6.36 7 -.86.7 -.3 4-36 Verify that a BIBD with the parameters a = 8, r = 8, k = 4, and b = 6 does not exist. These rk ( ) 83 ( ) 4 conditions imply that, which is not an integer, so a balanced design with these a 7 7 parameters cannot exist. 4-37 Show that the variance of the intra block estimators { i } is a k( a ). kqi Note that ˆ i, and Qi yi. a nij y k b j. j, and kq ky n y k i i. b j ij. j y i. b j n y ij. j y i. y i. contains r observations, and the quantity in the parenthesis is the sum of r(k-) observations, not including treatment i. Therefore, V kq k V Q rk rk i i 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY or r k Q r k k k k V i To find V ˆ i, note that: However, since a rk V k a ˆ V Q i, we have: i k a r k krk k a Furthermore, the i a k V ˆ i a V ˆ ˆ k a ˆ are not independent, this is required to show that i j 4-38 Extended incomplete block designs. Occasionally the block size obeys the relationship a < k < a. An extended incomplete block design consists of a single replicate or each treatment in each block along with an incomplete block design with k* = k-a. In the balanced case, the incomplete block design will have parameters k* = k-a, r* = r-b, and *. Write out the statistical analysis. (Hint: In the extended incomplete block design, we have = r-b+*.) As an example of an extended incomplete block design, suppose we have a=5 treatments, b=5 blocks and k=9. A design could be found by running all five treatments in each block, plus a block from the balanced incomplete block design with k* = k-a=9-5=4 and *=3. The design is: Block Complete Treatment Incomplete Treatment,,3,4,5,3,4,5,,3,4,5,,4,5 3,,3,4,5,3,4,5 4,,3,4,5,,3,4 5,,3,4,5,,3,5 Note that r=9, since the augmenting incomplete block design has r*=4, and r= r* + b = 4+5=9, and = rb+*=8-5+3=6. Since some treatments are repeated in each block it is possible to compute an error sum of squares between repeat observations. The difference between this and the residual sum of squares is due to interaction. The analysis of variance table is shown below: Source SS DF Treatments (adjusted) Qi k a a- y... j y Blocks b- k N Interaction Subtraction (a-)(b-) Error [SS between repeat observations] b(k-a) Total y.. yij N- N 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 5 Introduction to Factorial Designs Solutions 5- The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data follow: Pressure Temperature 5 3 5 9.4 9.7 9. 9. 9.6 9.4 6 9. 9.5 89.9 9.3 9.6 9. 7 9.5 9.8 9.4 9.7 9.9 9. (a) Analyze the data and draw conclusions. Use =.5. Both pressure (A) and temperature (B) are significant, the interaction is not. Design Expert Output Response:Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.4 8.4 8..6 significant A.77.38.59.4 B.3.5 8.47.85 AB.69 4.7.97.47 Residual.6 9.8 Lack of Fit. Pure Error.6 9.8 Cor Total.3 7 The Model F-value of 8. implies the model is significant. There is only a.6% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Values greater than. indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model. (b) Prepare appropriate residual plots and comment on the model s adequacy. The residuals plot show no serious deviations from the assumptions. 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals.5 99 Residuals.75 4.636E-4 -.75 Normal % probability 95 9 8 7 5 3 5 -.5 9. 9. 9.43 9.64 9.85 -.5 -.75-4.636E-4.75.5 Predicted Residual Residuals vs. Temperature Residuals vs. Pressure.5.5.75.75 3 Residuals 4.636E-4 Residuals 4.636E-4 -.75 -.75 -.5 -.5 3 3 Temperature Pressure (c) Under what conditions would you operate this process? 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Yield 9. 8 In teraction G raph Tem perature X = A: Pressure Y = B: Temperature Design Points B 5 B 6 B3 7 9.7 9 9.4 5 Yield 9. 3 7 8 9.8 4 9 5 3 P ressure Pressure set at 5 and Temperature at the high level, 7 degrees C, give the highest yield. The standard analysis of variance treats all design factors as if they were qualitative. In this case, both factors are quantitative, so some further analysis can be performed. In Section 5-5, we show how response curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor. Since both factors in this problem are quantitative and have three levels, we can fit linear and quadratic effects of both temperature and pressure, exactly as in Example 5-5 in the text. The Design-Expert output, including the response surface plots, now follows. Design Expert Output Response:Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3 5.3 6.8 <. significant A.. 7..98 B.67.67 4.83.483 A.67.67 47.74 <. B.3.3 6.7.5 AB.6.6 4.38.58 Residual.7.4 Lack of Fit 7.639E-3 3.546E-3.4.934 not significant Pure Error.6 9.8 Cor Total.3 7 The Model F-value of 6.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, A, B are significant model terms. Values greater than. indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model. Std. Dev.. R-Squared.878 Mean 9.4 Adj R-Squared.87 C.V..3 Pred R-Squared.6794 PRESS.4 Adeq Precision.968 Coefficient Standard 95% CI 95% CI 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Estimate DF Error Low High VIF Intercept 9.5.6 9.39 9.66 A-Pressure -.9.34 -.7 -.7. B-Temperature.75.34 6.594E-4.5. A -.4.59 -.54 -.8. B.4.59..37. AB -.87.4 -.8 3.548E-3. Final Equation in Terms of Coded Factors: Yield = +9.5 -.9 * A +.75 * B -.4 * A +.4 * B -.87 * A * B Final Equation in Terms of Actual Factors: Yield = +48.5463 +.86759 * Pressure -.644 * Temperature -.848E-3 * Pressure +.4667E-3 * Temperature -5.83333E-4 * Pressure * Temperature 7. 9.8 Yield 9.7 B: Temperature 65. 6. 9.6 9. 9.5 9.5 9.4 9.3 9. 9.4 9.3 Yield 9 9.8 9.6 9.4 9. 9 55. 9.6 5.. 7.5 5..5 3. A: Pressure 7.. 65. 7.5 6. 5..5 A: Pres s ure 3. B: Temperature 55. 5. 5- An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut. She selects three feed rates and four depths of cut. She then conducts a factorial experiment and obtains the following data: Depth of Cut (in) Feed Rate (in/min).5.8..5 74 79 8 99. 64 68 88 4 6 73 9 96 9 98 99 4 5-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.5 86 4 8 88 88 95 99 99 4 8 4.3 98 99 95 99 7 (a) Analyze the data and draw conclusions. Use =.5. The depth (A) and feed rate (B) are significant, as is the interaction (AB). Design Expert Output Response: Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 584.67 53.5 8.49 <. significant A 5. 3 78.37 4.66 <. B 36.5 58.5 55. <. AB 557.6 6 9.84 3.3.8 Residual 689.33 4 8.7 Lack of Fit. Pure Error 689.33 4 8.7 Cor Total 653. 35 The Model F-value of 8.49 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. (b) Prepare appropriate residual plots and comment on the model s adequacy. The residual plots shown indicate nothing unusual. 8 Residuals vs. Predicted Normal plot of residuals 99 Residuals 3.83333 -.333333-4.5 Normal % probability 95 9 8 7 5 3 5-8.66667 66. 77.7 88.33 99.5.67-8.66667-4.5 -.333333 3.83333 8 Predicted Residual 5-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 8 Residuals vs. Feed Rate 8 Residuals vs. Depth of Cut 3.83333 3.83333 Residuals -.333333 Residuals -.333333-4.5-4.5-8.66667-8.66667 3 3 4 Feed Rate Depth of Cut (c) Obtain point estimates of the mean surface finish at each feed rate. Feed Rate Average. 8.58.5 97.58.3 3.83 DESIGN-EXPERT Plot Surface Finish X = B: Feed Rate 4 O ne Factor P lot W a rn in g! F a c to r in v o lv e d in a n in te ra c tio n. Actual Factor A: Depth of Cut = Average.5 S urface Finish 8 7 7 3.5 6.. 5.3 Feed R ate (d) Find P-values for the tests in part (a). The P-values are given in the computer output in part (a). 5-3 For the data in Problem 5-, compute a 95 percent interval estimate of the mean difference in response for feed rates of. and.5 in/min. 5-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY We wish to find a confidence interval on, where is the mean surface finish for. in/min and is the mean surface finish for.5 in/min. y y t MS E n y.. y.. t, abn ) n...., ab MS (8.7) ( 8.5833 97.5833) (.64) 6 9.3 3 Therefore, the 95% confidence interval for is -6. 9.3. n E 5-4 An article in Industrial Quality Control (956, pp. 5-8) describes an experiment to investigate the effect of the type of glass and the type of phosphor on the brightness of a television tube. The response variable is the current necessary (in microamps) to obtain a specified brightness level. The data are as follows: Glass Phosphor Type Type 3 8 3 9 9 3 85 85 95 9 3 6 35 4 5 4 35 3 (a) Is there any indication that either factor influences brightness? Use =.5. Both factors, phosphor type (A) and Glass type (B) influence brightness. Design Expert Output Response: Current in microamps ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 556.67 5 33.33 58.8 <. significant A 933.33 466.67 8.84.44 B 445. 445. 73.79 <. AB 33.33 66.67.6.378 Residual 633.33 5.78 Lack of Fit. Pure Error 633.33 5.78 Cor Total65. 7 The Model F-value of 58.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. (b) Do the two factors interact? Use =.5. There is no interaction effect. (c) Analyze the residuals from this experiment. 5-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The residual plot of residuals versus phosphor content indicates a very slight inequality of variance. It is not serious enough to be of concern, however. Residuals vs. Predicted Normal plot of residuals 5 99 Residuals 8.75.5-3.75 Normal % probability 95 9 8 7 5 3 5-5. 44.7 63.33 8.5 3.67 - -3.75.5 8.75 5 Predicted Residual Residuals vs. Glass Type Residuals vs. Phosphor Type 5 5 8.75 8.75 Residuals.5 Residuals.5-3.75-3.75 - - 3 Glass Type Phosphor Type 5-5 Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences, Wiley 977) describe an experiment to investigate the warping of copper plates. The two factors studies were the temperature and the copper content of the plates. The response variable was a measure of the amount of warping. The data were as follows: Copper Content (%) Temperature ( C) 4 6 8 5 7, 6, 4, 8,7 75,9 8,3 7, 7,3 6, 8, 5,3 3,3 5,7 3, 3, 9,3 5-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Is there any indication that either factor affects the amount of warping? Is there any interaction between the factors? Use =.5. Both factors, copper content (A) and temperature (B) affect warping, the interaction does not. Design Expert Output Response: Warping ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 968. 5 64.55 9.5 <. significant A 698.34 3 3.78 34.33 <. B 56.9 3 5.3 7.67. AB 3.78 9.64.86.37 Residual 8.5 6 6.78 Lack of Fit. Pure Error 8.5 6 6.78 Cor Total 76.7 3 The Model F-value of 9.5 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. Normal plot of residuals Residuals vs. Predicted 3.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals.75.658E-4 -.75-3.5-3.5 -.75 -.658E-4.75 3.5.5 5.38.5 5.3 3. Residual Predicted 5-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3.5 Residuals vs. Copper Content 3.5 Residuals vs. Temperature Residuals.75.658E-4 -.75 Residuals.75.658E-4 -.75-3.5-3.5 3 4 3 4 Copper Content Temperature (c) Plot the average warping at each level of copper content and compare them to an appropriately scaled t distribution. Describe the differences in the effects of the different levels of copper content on warping. If low warping is desirable, what level of copper content would you specify? Design Expert Output Factor Name Level Low Level High Level A Copper Content 4 4 B Temperature Average 5 5 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping5.5.84.6 9.4 3.9 8.74.6 Factor Name Level Low Level High Level A Copper Content 6 4 B Temperature Average 5 5 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping8.88.84 4.97.78 3.9. 5.64 Factor Name Level Low Level High Level A Copper Content 8 4 B Temperature Average 5 5 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping..84 7. 4.9 3.9 4.4 7.76 Factor Name Level Low Level High Level A Copper Content 4 B Temperature Average 5 5 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping8.5.84 4.35 3.5 3.9.49 35. Use a copper content of 4 for the lowest warping. MSE 6. 785 S. 3 b 4 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution Cu=4 Cu=6 Cu=8 Cu= 5. 8.. 4. 7. Warping (d) Suppose that temperature cannot be easily controlled in the environment in which the copper plates are to be used. Does this change your answer for part (c)? Use a copper of content of 4. This is the same as for part (c). DESIGN-EXPERT Plot Warping X = A: Copper Content Y = B: Temperature Design Points 3.7 6 6.5 5 In teraction G raph Tem perature B 5 B 75 B3 B4 5 W a rp in g. 5 3 3.9 9 4 9 7.7 3 9 7 9 4 6 8 C opper C ontent 5-6 The factors that influence the breaking strength of a synthetic fiber are being studied. Four production machines and three operators are chosen and a factorial experiment is run using fiber from the same production batch. The results are as follows: Machine Operator 3 4 9 8 5 9 8 4 9 3 6 4 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 5 9 7 (a) Analyze the data and draw conclusions. Use =.5. Only the Operator (A) effect is significant. Design Expert Output Response:Stength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 7.46 9.77 5..4 significant A 6.33 8.7.4. B.46 3 4.5..3888 AB 44.67 6 7.44.96.57 Residual 45.5 3.79 Lack of Fit. Pure Error 45.5 3.79 Cor Total 6.96 3 The Model F-value of 5. implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms aresignificant. In this case A are significant model terms. (b) Prepare appropriate residual plots and comment on the model s adequacy. The residual plot of residuals versus predicted shows that variance increases very slightly with strength. There is no indication of a severe problem. Residuals vs. Predicted Normal plot of residuals.5 99 Residuals.5 7.8597E-4 -.5 Normal % probability 95 9 8 7 5 3 5 -.5 8.5. 3.5 6. 8.5 -.5 -.5-7.8597E-4.5.5 Predicted Residual 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.5 Residuals vs. Operator.5 3 Residuals 7.8597E-4 -.5 3 -.5 3 Operator 5-7 A mechanical engineer is studying the thrust force developed by a drill press. He suspects that the drilling speed and the feed rate of the material are the most important factors. He selects four feed rates and uses a high and low drill speed chosen to represent the extreme operating conditions. He obtains the following results. Analyze the data and draw conclusions. Use =.5. (A) Feed Rate (B) Drill Speed.5.3.45.6 5.7.45.6.75.78.49.7.86.83.85.86.94.86.8.87.88 Design Expert Output Response: Force ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 7.4 5.53.5 significant A.5.5 57. <. B.9 3.3.86.6 AB.4 3.4 5.37.56 Residual. 8.6E-3 Lack of Fit. Pure Error. 8.6E-3 Cor Total.3 5 The Model F-value of 5.53 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. The factors speed and feed rate, as well as the interaction is important. 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Force.9 6 8 7 9 In teraction G raph D rill S p e e d X = B: Feed Rate Y = A: Drill Speed Design Points.8 9 4 A 5 A F o rc e.6 9.5 5 6.4. 5. 3. 4 5. 6 Feed R ate The standard analysis of variance treats all design factors as if they were qualitative. In this case, both factors are quantitative, so some further analysis can be performed. In Section 5-5, we show how response curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor. Since both factors in this problem are quantitative and have three levels, we can fit linear and quadratic effects of both temperature and pressure, exactly as in Example 5-5 in the text. The Design-Expert output, including the response surface plots, now follows. Design Expert Output Response: Force ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3 4.57 8.5.7 significant A.5.5..8 B.9.9.74.6 B.58.58 8..54 AB.5E-3.5E-3.6.6966 Residual.77 7.E-3 Lack of Fit.56 3.9 7.3.5 significant Pure Error. 8.6E-3 Cor Total.3 5 The Model F-value of 8.5 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Values greater than. indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model. Std. Dev..84 R-Squared.7455 Mean.77 Adj R-Squared.659 C.V. 3.3 Pred R-Squared.465 PRESS.6 Adeq Precision 7.835 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.69.34.6.76 A-Drill Speed.96..5.4. B-Feed Rate.47.8 -.5.. B.3.47.3.4. AB -..8 -.73.5. 5-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Final Equation in Terms of Coded Factors: Force = +.69 +.96 * A +.47 * B +.3 * B -. * A * B Final Equation in Terms of Actual Factors: Force = +.4897 +3.6667E-3 * Drill Speed -5.76667 * Feed Rate +66.66667 * Feed Rate -.3333 * Drill Speed * Feed Rate.6 Force.9.85.8.5 3.9 B: Feed Rate.4.75.7.65.6 Force.8.7.6.5.3.8.85. 5. 43.75 6.5 8.5. A: Drill Speed.6.5.4 B: Feed Rate.3.. 8.5 6.5 43.75 A: Drill Speed 5. 5-8 An experiment is conducted to study the influence of operating temperature and three types of faceplate glass in the light output of an oscilloscope tube. The following data are collected: Temperature Glass Type 5 5 58 9 39 568 87 38 57 85 386 55 7 38 53 35 3 579 99 546 45 867 3 575 53 94 599 66 889 5-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Use =.5 in the analysis. Is there a significant interaction effect? Does glass type or temperature affect the response? What conclusions can you draw? Use the method discussed in the text to partition the temperature effect into its linear and quadratic components. Break the interaction down into appropriate components. Design Expert Output Response: Light Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.4E+6 8 3.5E+5 84.77 <. significant A.59E+5 7543.6 6.37 <. B.97E+6 9.85E+5 695.6 <. AB.96E+5 4 7637.93 98.73 <. Residual 6579.33 8 365.5 Lack of Fit. Pure Error 6579.33 8 365.5 Cor Total.48E+6 6 The Model F-value of 84.77 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. Both factors, Glass Type (A) and Temperature (B) are significant, as well as the interaction (AB). For glass types and the response is fairly linear, for glass type 3, there is a quadratic effect. DESIGN-EXPERT Plot Light Output 4.4 In teraction G raph G lass Type X = B: Temperature Y = A: Glass Type Design Points A A A3 3 8 4.3 L ig h t O u tp u t 9 6 6. 9 9 7 4 8. 9 9 5 3 5 5 Tem perature Design Expert Output Response: Light Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.4E+6 8 3.5E+5 84.77 <. significant A.59E+5 7543.6 6.37 <. B.78E+6.78E+6 4869.3 <. B.96E+5.96E+5 5.39 <. AB.6E+5.3E+5 39.39 <. AB 64373.93 386.96 88.6 <. Pure Error 6579.33 8 365.5 5-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Cor Total.48E+6 6 The Model F-value of 84.77 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, B, AB, AB are significant model terms. Values greater than. indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model. Std. Dev. 9. R-Squared.9973 Mean 94.9 Adj R-Squared.996 C.V..3 Pred R-Squared.9939 PRESS 483.5 Adeq Precision 75.466 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59. 6.37 45.6 7.39 A[] 8.33 9. 9.4 47.7 A[] -4. 9. -4.93-5.7 B-Temperature 34.44 4.5 34.98 33.9. B -78. 7.8-94.6-6.8. A[]B 9. 6.37 78.83 5.6 A[]B 65.56 6.37 5.7 78.94 A[]B 7..4 47.3 93.4 A[]B 76..4 53.3 99.4 Final Equation in Terms of Coded Factors: Light Output = +59. +8.33 * A[] -4. * A[] +34.44 * B -78. * B +9. * A[]B +65.56 * A[]B +7. * A[]B +76. * A[]B Final Equation in Terms of Actual Factors: Glass Type Light Output = -3646. +59.46667 * Temperature -.78 * Temperature Glass Type Light Output = -345. +56. * Temperature -.63 * Temperature Glass Type 3 Light Output = -7845.33333 +36.3333 * Temperature -.5947 * Temperature 5-9 Consider the data in Problem 5-. Use the method described in the text to compute the linear and quadratic effects of pressure. See the alternative analysis shown in Problem 5- part (c). 5-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 5- Use Duncan s multiple range test to determine which levels of the pressure factor are significantly different for the data in Problem 5-. y 3... 9 8 y.. 9 37 y. 9 68 S MS.. 777 E.... 543 y j. an 3 r., 9 4. 6 r. 3, 9 4. 86 R 4. 6. 543 498 R 4. 86. 543 64. 3. vs. 3 =.5 >.64 (R 3 ) vs. =.3 >.498 (R ) vs. 3 =.9 <.498 (R ) Therefore, differs from and 3. 5- An experiment was conducted to determine if either firing temperature or furnace position affects the baked density of a carbon anode. The data are shown below. Temperature ( C) Position 8 85 85 57 63 565 565 8 5 583 43 59 58 988 56 547 6 538 5 4 53 Suppose we assume that no interaction exists. Write down the statistical model. Conduct the analysis of variance and test hypotheses on the main effects. What conclusions can be drawn? Comment on the model s adequacy. The model for the two-factor, no interaction model is yijk i j ijk. Both factors, furnace position (A) and temperature (B) are significant. The residual plots show nothing unusual. Design Expert Output Response: Density ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 9.55E+5 3 3.75E+5 78.4 <. significant A 76.6 76.6 6..3 B 9.453E+5 4.77E+5 69.6 <. Residual 688.78 4 44.6 Lack of Fit 88. 49.6.9.47 not significant Pure Error 537.67 447.56 Cor Total 9.587E+5 7 The Model F-value of 78.4 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. 5-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.5556 Residuals vs. Predicted 6.5556 Residuals vs. Position 6.55556 6.55556 Residuals -3.4444 Residuals -3.4444-33.4444-33.4444-53.4444-53.4444 53.56 656.5 788.75 9.35 53.94 Predicted Position 6.5556 Residuals vs. Temperature 6.55556 Residuals -3.4444-33.4444-53.4444 3 Temperature 5- Derive the expected mean squares for a two-factor analysis of variance with one observation per cell, assuming that both factors are fixed. E MS E MS E MS AB A B a i a i Degrees of Freedom b a- b a b- j b j a b ij a b i j a b ab 5-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 5-3 Consider the following data from a two-factor factorial experiment. Analyze the data and draw conclusions. Perform a test for nonadditivity. Use =.5. Column Factor Row Factor 3 4 36 39 36 3 8 3 3 37 33 34 Design Expert Output Response: data ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 69.4 5.88 5.36.6 significant A 58.5 9.5 6.4. B 8.9 3 9.64..47 Residual 8.83 6 4.8 Cor Total 638.5 The Model F-value of 5.36 implies the model is significant. There is only a.6% chance that a "Model F-Value" this large could occur due to noise. The row factor (A) is significant. The test for nonadditivity is as follows: SS SS SS SS N N N i j 44 3. 545 Error a SS b y y Re sidual 357 357 58. 5 8. 9667 y ij i.. j 43 58. 58. 9667 SS y.. SS abss N A SS A B SS B y.. ab 43 8. 8333 3. 545 5. 979 Source of Sum of Degrees of Mean Variation Squares Freedom Square F Row 58.5 9.5 57.378 Column 8.9667 3 9.63889.954 Nonadditivity 3.545 3.545.6999 Error 5.979 5 5.58558 Total 638.5 5-4 The shear strength of an adhesive is thought to be affected by the application pressure and temperature. A factorial experiment is performed in which both factors are assumed to be fixed. Analyze the data and draw conclusions. Perform a test for nonadditivity. Temperature ( F) 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Pressure (lb/in) 5 6 7 9.6.8 9. 3 9.69. 9.57 4 8.43. 9.3 5 9.98.44 9.8 Design Expert Output Response: Strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5.4 5.5.9.4 not significant A.58 3.9.54.677 B 4.66.33 6.49.36 Residual.5 6.36 Cor Total 7.39 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a.4 % chance that a "Model F-value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. Temperature (B) is a significant factor. a b y y i j SS SS SS SS N N N 7. 93 453. 777 7. 93. 58697 4. 65765. 48948 Error SS Re sidual y ij i.. j SS y.. SS abss SS SS y.. ab 43. 586974. 65765 N A A B B 43. 538833. 48948. 6644 Source of Sum of Degrees of Mean Variation Squares Freedom Square F Row.58697 3.935639.585 Column 4.65765.3885 6.996 Nonadditivity.48948.48948.474 Error.6644 5.3388 Total 7.395 5-5 Consider the three-factor model yijk i j k ijk ij jk i,,...,a j,,...,b k,,...,c 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Notice that there is only one replicate. Assuming the factors are fixed, write down the analysis of variance table, including the expected mean squares. What would you use as the experimental error in order to test hypotheses? Source Degrees of Freedom Expected Mean Square A a- a i bc a B b- C c- AB BC (a-)(b-) (b-)(c-) c a ac ab Error (AC + ABC) b(a-)(c-) Total abc- i b j b j c k c k a b ij a b i j b c jk b c j k 5-6 The percentage of hardwood concentration in raw pulp, the vat pressure, and the cooking time of the pulp are being investigated for their effects on the strength of paper. Three levels of hardwood concentration, three levels of pressure, and two cooking times are selected. A factorial experiment with two replicates is conducted, and the following data are obtained: Percentage Cooking Time 3. Hours Cooking Time 4. Hours of Hardwood Pressure Pressure Concentration 4 5 65 4 5 65 96.6 97.7 99.8 98.4 99.6.6 96. 96. 99.4 98.6.4.9 4 98.5 96. 98.4 97.5 98.7 99.6 97. 96.9 97.6 98. 98. 99. 8 97.5 95.6 97.4 97.6 97. 98.5 96.6 96. 98. 98.4 97.8 99.8 (a) Analyze the data and draw conclusions. Use =.5. Design Expert Output Response: strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59.73 7 3.5 9.6 <. significant A 7.76 3.88.6.9 B.5.5 55.4 <. C 9.37 9.69 6.5 <. AB.8.4.85.843 AC 6.9 4.5 4.7.46 5-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY BC.9. 3..75 ABC.97 4.49.35.93 Residual 6.58 8.37 Lack of Fit. Pure Error 6.58 8.37 Cor Total 66.3 35 The Model F-value of 9.6 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AC are significant model terms. All three main effects, concentration (A), pressure (C) and time (B), as well as the concentration x pressure interaction (AC) are significant at the 5% level. The concentration x time (AB) and pressure x time interactions (BC) are significant at the % level. (b) Prepare appropriate residual plots and comment on the model s adequacy. Residuals vs. Pressure Residuals vs. Cooking Time.85.85.45.45 Residuals Residuals -.45 -.45 -.85 -.85 3 Pressure Cooking Time Residuals vs. Predicted Residuals vs. Hardwood.85.85.45.45 Residuals Residuals -.45 -.45 -.85 -.85 95.9 97. 98.33 99.54.75 3 Predicted Hardwood There is nothing unusual about the residual plots. 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Under what set of conditions would you run the process? Why? DESIGN-EXPERT Plot strength.9 In teraction G raph P ressure DESIGN-EXPERT Plot strength.9 In teraction G raph H a rd w o o d X = B: Cooking Time Y = C: Pressure 99.575 C 4 C 5 C3 65 Actual Factor A: Hardwood = Average98.5 strength X = B: Cooking Time Y = A: Hardwood 9 9.5 7 5 A A 4 A3 8 Actual Factor C: Pressure = Average 9 8. 5 strength 96.95 9 6.9 5 95.6 9 5.6 3 4 3 4 C o o k in g T im e C ooking Tim e DESIGN-EXPERT Plot strength.9 In teraction G raph H a rd w o o d X = C: Pressure Y = A: Hardwood 9 9.5 7 5 A A 4 A3 8 Actual Factor B: Cooking Time = Average 9 8. 5 strength 9 6.9 5 9 5.6 4 5 6 5 P ressure For the highest strength, run the process with the percentage of hardwood at, the pressure at 65, and the time at 4 hours. The standard analysis of variance treats all design factors as if they were qualitative. In this case, all three factors are quantitative, so some further analysis can be performed. In Section 5-5, we show how response curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor. Since the factors in this problem are quantitative and two of them have three levels, we can fit linear and quadratic. The Design-Expert output, including the response surface plots, now follows. Design Expert Output Response: Strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 58. 3 4.46.85 <. significant 5-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A 7.5 7.5 8.98.3 B 3.4 3.4 9.8.64 C...58.4559 A.9.9.88.36 C 4.43 4.43.77.4 AB.6.6.8.8 AC 3.39 3.39 9..66 BC.5.5.4.535 AB.3.3 3.46.763 AC.9.9 5.8.47 AC.65.65 4.38.48 BC.8.8 5.78.5 ABC.4.4.6.336 Residual 8.9.38 Lack of Fit.7 4.43.7.3576 not significant Pure Error 6.58 8.37 Cor Total 66.3 35 The Model F-value of.85 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AC, A C, AC, BC are significant model terms. Values greater than. indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model. Std. Dev..6 R-Squared.875 Mean 98.6 Adj R-Squared.8 C.V..3 Pred R-Squared.6657 PRESS.7 Adeq Precision 4.7 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 97..6 96.67 97.74 A-Hardwood -.98.3 -.45 -.5 3.36 B-Cooking Time.78.6.4.3 6.35 C-Pressure.9.5 -.33.7 4.4 A.4.5 -.94.94.4 C.79.3.3.6.3 AB -..3 -.47.5.6 AC -.46.5 -.78 -.4.8 BC.8.3 -.8.34.4 AB.46.5 -.53.98 3.96 AC.73.3..36 3.97 AC.57.7 4.979E-3.4 3.3 BC -.55.3 -. -.75 3.3 ABC.5.5 -.6.46. Final Equation in Terms of Coded Factors: Strength = +97. -.98 * A +.78 * B +.9 * C +.4 * A +.79 * C -. * A * B -.46 * A * C +.8 * B * C +.46 * A * B +.73 * A * C +.57 * A * C -.55 * B * C +.5 * A * B * C Final Equation in Terms of Actual Factors: 5-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Strength = +9.9698 +.654 * Hardwood -.976 * Cooking Time -.4 * Pressure -.6587 * Hardwood +.34333E-4 * Pressure -.638 * Hardwood * Cooking Time -.345 * Hardwood * Pressure +.7658 * Cooking Time * Pressure +.78 * Hardwood * Cooking Time +6.486E-4 * Hardwood * Pressure +.43E-5 * Hardwood * Pressure -7.E-5 * Cooking Time * Pressure +8.338E-4 * Hardwood * Cooking Time * Pressure 65. Strength 98.5.5 6. C: Pressure 55. 5. 45. 98 99.5 99 98.5 Strength.5.5 99.5 99 98.5 98 97.5 97 65. 6. 4.. 3.5 5. 6.5 8. A: Hardwood. 3.5 5. A: Hardwood 6.5 55. 5. C: Pres s ure 45. 8. 4. Cooking Time: B = 4. 5-7 The quality control department of a fabric finishing plant is studying the effect of several factors on the dyeing of cotton-synthetic cloth used to manufacture men s shirts. Three operators, three cycle times, and two temperatures were selected, and three small specimens of cloth were dyed under each set of conditions. The finished cloth was compared to a standard, and a numerical score was assigned. The results follow. Analyze the data and draw conclusions. Comment on the model s adequacy. Temperature 3 35 Operator Operator Cycle Time 3 3 3 7 3 4 38 34 4 4 8 3 3 36 36 5 6 9 8 35 39 36 34 33 37 34 34 5 35 38 34 39 38 36 36 39 35 35 36 3 8 35 6 6 36 8 6 4 35 7 9 37 6 5-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 7 34 5 5 34 4 All three main effects, and the AB, AC, and ABC interactions are significant. There is nothing unusual about the residual plots. Design Expert Output Response: Score ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 39.33 7 7.9.4 <. significant A 436. 8. 66.5 <. B 6.33 3.67 39.86 <. C 5.7 5.7 5.8.4 AB 355.67 4 88.9 7.3 <. AC 78.8 39.4.. BC.6 5.63.7.939 ABC 46.9 4.55 3.5.59 Residual 8. 36 3.8 Lack of Fit. Pure Error 8. 36 3.8 Cor Total 357.33 53 The Model F-value of.4 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB, AC, ABC are significant model terms. DESIGN-EXPERT Plot Score 3 9 In teraction G raph Tem perature DESIGN-EXPERT Plot Score 3 9 In teraction G raph O perator X = A: Cycle Time Y = C: Temperature C 3 C 35 Actual Factor B: Operator = Average S c o re 3 5 3 X = A: Cycle Time Y = B: Operator 3 5 B B B3 3 Actual Factor C: Temperature = Average 3 S c o re 7 7 3 3 4 5 6 4 5 6 C yc le T im e C ycle Tim e 5-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Operator Residuals vs. Cycle Time 3 3.5.5 Residuals 4.636E-4 3 Residuals 4.636E-4 3 3 -.5 -.5-3 -3 3 3 Operator Cycle Time Residuals vs. Predicted Residuals vs. Temperature 3 3.5.5 Residuals 4.636E-4 Residuals 4.636E-4 4 3 -.5 -.5-3 -3 4. 7.5 3.5 33.75 37. Predicted Temperature 5-8 In Problem 5-, suppose that we wish to reject the null hypothesis with a high probability if the difference in the true mean yield at any two pressures is as great as.5. If a reasonable prior estimate of the standard deviation of yield is., how many replicates should be run? nad b n 3 3. 5.. 5n n b abn 5 5 (3)(3)().4 replications will be enough to detect the given difference. 5-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 5-9 The yield of a chemical process is being studied. The two factors of interest are temperature and pressure. Three levels of each factor are selected; however, only 9 runs can be made in one day. The experimenter runs a complete replicate of the design on each day. The data are shown in the following table. Analyze the data assuming that the days are blocks. Day Day Pressure Pressure Temperature 5 6 7 5 6 7 Low 86.3 84. 85.8 86. 85. 87.3 Medium 88.5 87.3 89. 89.4 89.9 9.3 High 89. 9. 9.3 9.7 93. 93.7 Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 3. 3. Model 9.8 8 3.73 5.84 <. significant A 5.5.75 5.8.36 B 99.85 49.93 93.98 <. AB 4.45 4...733 Residual 4.5 8.53 Cor Total 7.7 7 The Model F-value of 5.84 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Both main effects, temperature and pressure, are significant. 5- Consider the data in Problem 5-5. Analyze the data, assuming that replicates are blocks. Design Expert Output Response: Warping ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.8.8 Model 968. 5 64.55 9.96 <. significant A 698.34 3 3.78 35.9 <. B 56.9 3 5.3 8.3. AB 3.78 9.64.95.4 Residual 97. 5 6.48 Cor Total 76.7 3 The Model F-value of 9.96 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Both temperature and copper content are significant. This agrees with the analysis in Problem 5-5. 5- Consider the data in Problem 5-6. Analyze the data, assuming that replicates are blocks. 5-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design-Expert Output Response: Stength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.4.4 Model 7.46 9.77 4.89.7 significant A 6.33 8.7 9.84. B.46 3 4.5.3.479 AB 44.67 6 7.44.84.799 Residual 44.46 4.4 Cor Total 6.96 3 The Model F-value of 4.89 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A are significant model terms. Only the operator factor (A) is significant. This agrees with the analysis in Problem 5-6. 5- An article in the Journal of Testing and Evaluation (Vol. 6, no., pp. 58-55) investigated the effects of cyclic loading and environmental conditions on fatigue crack growth at a constant MPa stress for a particular material. The data from this experiment are shown below (the response is crack growth rate). Environment Frequency Air H O Salt H O.9.6.9.47.5.93.48.3.75..3.6.65 3. 3..68 3.8 3.4.6 3.96 3.98.38 3.64 3.4.4. 9.96..7...8 9.6 9.36.8.3.4 (a) Analyze the data from this experiment (use =.5). Design Expert Output Response: Crack Growth ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 376. 8 47. 34. <. significant A 9.89 4.95 5.4 <. B 64.5 3.3 59.9 <. AB.97 4 5.49 6.89 <. Residual 5.4 7. Lack of Fit. 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Pure Error 5.4 7. Cor Total 38.53 35 The Model F-value of 34. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. Both frequency and environment, as well as their interaction are significant. (b) Analyze the residuals. The residual plots indicate that there may be some problem with inequality of variance. This is particularly noticable on the plot of residuals versus predicted response and the plot of residuals versus frequency. Residuals vs. Predicted Normal plot of residuals.7 99 Residuals.5 -.4 -.97 Normal % probability 95 9 8 7 5 3 5 -.53.9 4.8 6.5 8.4.59 -.53 -.97 -.4.5.7 Predicted Residual Residuals vs. Environment Residuals vs. Frequency.7.7.5.5 Residuals -.4 Residuals -.4 -.97 -.97 -.53 -.53 3 3 Environment Frequency (c) Repeat the analyses from parts (a) and (b) using ln(y) as the response. Comment on the results. 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Crack Growth Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.46 8.68 79.57 <. significant A 7.57 3.79 44.9 <. B.36.8 5.85 <. AB 3.53 4.88 94.7 <. Residual.5 7 9.367E-3 Lack of Fit. Pure Error.5 7 9.367E-3 Cor Total 3.7 35 The Model F-value of 79.57 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. Both frequency and environment, as well as their interaction are significant. The residual plots of the based on the transformed data look better. Residuals vs. Predicted Normal plot of residuals.6534 99.8783 Residuals.44 -.8988 Normal % probability 95 9 8 7 5 3 5 -.6484.65.7.5.93.36 -.6484 -.8988.44.8783.6534 Predicted Residual 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Environment Residuals vs. Frequency.6534.6534.8783.8783 Residuals.44 Residuals.44 -.8988 -.8988 -.6484 -.6484 3 3 Environment Frequency 5-3 An article in the IEEE Transactions on Electron Devices (Nov. 986, pp. 754) describes a study on polysilicon doping. The experiment shown below is a variation of their study. The response variable is base current. Polysilicon Anneal Temperature ( C) Doping (ions) 9 95 x 4.6.5. 4.4..58 x 3. 9.38.8 3.5..6 (a) Is there evidence (with =.5) indicating that either polysilicon doping level or anneal temperature affect base current? Design Expert Output Response: Base Current ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.74 5.55 35.9 <. significant A.98.98 5.6.79 B.9 55.59 865.6 <. AB.58.9 4.48.645 Residual.39 6.64 Lack of Fit. Pure Error.39 6.64 Cor Total 3.3 The Model F-value of 35.9 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Both factors, doping and anneal are significant. Their interaction is significant at the % level. 5-33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (b) Prepare graphical displays to assist in interpretation of this experiment..5 Interaction Graph Doping 9.888 Base Current 7.75 5.568 A- 3.3986 A+ 9 95 Anneal (c) Analyze the residuals and comment on model adequacy. Residuals vs. Predicted Normal plot of residuals.3 99 Residuals.6 8.8878E-6 -.6 Normal % probability 95 9 8 7 5 3 5 -.3 3.35 5. 7.7 8.93.8 -.3 -.6-8.8878E-6.6.3 Predicted Residual 5-34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.3 Residuals vs. Doping.3 Residuals vs. Anneal.6.6 Residuals 8.8878E-6 Residuals 8.8878E-6 -.6 -.6 -.3 -.3 3 Doping Anneal There is a funnel shape in the plot of residuals versus predicted, indicating some inequality of variance. (d) Is the model y x x x x x supported by this experiment (x = doping level, x = temperature)? Estimate the parameters in this model and plot the response surface. Design Expert Output Response: Base Current ANOVA for Response Surface Reduced Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.73 4 8.8 493.73 <. significant A.98.98 7.8.43 B 93.6 93.6 63.9 <. B 8.3 8.3 35.8 <. AB.56.56 9.84.64 Residual.4 7.57 Lack of Fit.4.4..6569 not significant Pure Error.39 6.64 Cor Total 3.3 The Model F-value of 493.73 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, B, AB are significant model terms. Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 9.94. 9.66. A-Doping -.9.69 -.45 -.. B-Anneal 3.4.84 3. 3.6. B -.6.5 -.95 -.5. AB.7.84.65.46. All of the coefficients in the assumed model are significant. The quadratic effect is easily observable in the response surface plot. 5-35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY. Base Current 975. Anneal 95. 95. 9 8 7 Base Current 9 8 7 6 5 4 3 6 5 4 9..E+.5E+.5E+.75E+.E+ Doping.E+.5E+.5E+ Doping.75E+.E+ 9.. 975. 95. 95. Anneal 5-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 6 The k Factorial Design Solutions 6- An engineer is interested in the effects of cutting speed (A), tool geometry (B), and cutting angle on the life (in hours) of a machine tool. Two levels of each factor are chosen, and three replicates of a 3 factorial design are run. The results follow: Treatment Replicate A B C Combination I II III - - - () 3 5 + - - a 3 43 9 - + - b 35 34 5 + + - ab 55 47 46 - - + c 44 45 38 + - + ac 4 37 36 - + + bc 6 5 54 + + + abc 39 4 47 (a) Estimate the factor effects. Which effects appear to be large? From the normal probability plot of effects below, factors B, C, and the AC interaction appear to be significant. DESIGN-EXPERT Plot Life Normal plot A: Cutting Speed B: Tool Geometry C: Cutting Angle Normal % probability 99 95 9 8 7 5 3 A C B 5 AC -8.8 3-3.7 9.5 6.9.33 Effect (b) Use the analysis of variance to confirm your conclusions for part (a). The analysis of variance confirms the significance of factors B, C, and the AC interaction. Design Expert Output Response: Life in hours 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.67 7 3.38 7.64.4 significant A.67.67..8837 B 77.67 77.67 5.55. C 8.7 8.7 9.9.77 AB 6.67 6.67.55.468 AC 468.7 468.7 5.5. BC 48.7 48.7.6.45 ABC 8.7 8.7.93.3483 Pure Error 48.67 6 3.7 Cor Total 95.33 3 The Model F-value of 7.64 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. The reduced model ANOVA is shown below. Factor A was included to maintain hierarchy. Design Expert Output Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59.67 4 379.9.54 <. significant A.67.67..8836 B 77.67 77.67 5.44 <. C 8.7 8.7 9.5.67 AC 468.7 468.7 5.45.9 Residual 575.67 9 3.3 Lack of Fit 93. 3 3..3.467 not significant Pure Error 48.67 6 3.7 Cor Total 95.33 3 The Model F-value of.54 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Effects B, C and AC are significant at %. (c) Write down a regression model for predicting tool life (in hours) based on the results of this experiment. y 4. 8333. 667x 5. 6667x 3. 467x 4. 467x ijk A B C A x C Design Expert Output Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 4.83. 38.48 43.9 A-Cutting Speed.7. -.9.5. B-Tool Geometry 5.67. 3.3 8.. C-Cutting Angle 3.4..6 5.77. AC -4.4. -6.77 -.6. Final Equation in Terms of Coded Factors: Life = +4.83 +.7 * A +5.67 * B +3.4 * C -4.4 * A * C Final Equation in Terms of Actual Factors: 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Life +4.83333 = +.6667 * Cutting Speed +5.66667 * Tool Geometry +3.4667 * Cutting Angle -4.4667 * Cutting Speed * Cutting Angle The equation in part (c) and in the given in the computer output form a hierarchial model, that is, if an interaction is included in the model, then all of the main effects referenced in the interaction are also included in the model. (d) Analyze the residuals. Are there any obvious problems? Normal plot of residuals Residuals vs. Predicted.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals 6.7967.8333 -.6 5-7.33333-7.33333 -.6 5.8333 6.7967.5 7.7 33.9 4.67 47.4 54.7 Residual Predicted There is nothing unusual about the residual plots. (e) Based on the analysis of main effects and interaction plots, what levels of A, B, and C would you recommend using? Since B has a positive effect, set B at the high level to increase life. The AC interaction plot reveals that life would be maximized with C at the high level and A at the low level. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Life 6 Interaction Graph Cutting Angle DESIGN-EXPERT Plot Life 6 One Factor Plot X = A: Cutting Speed Y = C: Cutting Angle 5.5 C- -. C+. Actual Factor B: Tool Geometry =. 4 Life X = B: Tool Geometry Actual Factors A: Cutting Speed =. C: Cutting Angle =. Life 5.5 4 3.5 3.5 -. -.5..5. -. -.5..5. Cutting Speed Tool Geometry 6- Reconsider part (c) of Problem 6-. Use the regression model to generate response surface and contour plots of the tool life response. Interpret these plots. Do they provide insight regarding the desirable operating conditions for this process? The response surface plot and the contour plot in terms of factors A and C with B at the high level are shown below. They show the curvature due to the AC interaction. These plots make it easy to see the region of greatest tool life. DESIGN-EXPERT Plot Life X = A: Cutting Speed Y = C: Cutting Angle. 45.8889 Life DESIGN-EXPERT Plot Life X = A: Cutting Speed Y = C: Cutting Angle Actual Factor B: Tool Geometry =..5 43.778 Actual Factor B: Tool Geometry =. 48.5 Cutting Angle. 4.6667 44.5833 4.6667 36.75 Life 3.8333 -.5 38.556 35.4444..5 -. -. -.5..5. Cutting Speed..5. -.5 Cutting Angle -.. Cutting Speed -.5 -. 6-3 Find the standard error of the factor effects and approximate 95 percent confidence limits for the factor effects in Problem 6-. Do the results of this analysis agree with the conclusions from the analysis of variance? 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY SE( effect ) S 3. 7 k 3 n 3 Variable Effect CI A.333 4.395 B.333 4.395 * AB -.667 4.395 C 6.833 4.395 * AC -8.833 4.395 * BC -.833 4.395 ABC -.67 4.395 The 95% confidence intervals for factors B, C and AC do not contain zero. This agrees with the analysis of variance approach.. 4 6-4 Plot the factor effects from Problem 6- on a graph relative to an appropriately scaled t distribution. Does this graphical display adequately identify the important factors? Compare the conclusions from this MSE 3. 7 plot with the results from the analysis of variance. S 3. 7 n 3 Scaled t Distribution AC C B -... Factor Effects This method identifies the same factors as the analysis of variance. 6-5 A router is used to cut locating notches on a printed circuit board. The vibration level at the surface of the board as it is cut is considered to be a major source of dimensional variation in the notches. Two factors are thought to influence vibration: bit size (A) and cutting speed (B). Two bit sizes (/6 and /8 inch) and two speeds (4 and 9 rpm) are selected, and four boards are cut at each set of conditions shown below. The response variable is vibration measured as a resultant vector of three accelerometers (x, y, and z) on each test circuit board. Treatment Replicate A B Combination I II III IV - - () 8. 8.9.9 4.4 + - a 7. 4..4.5 - + b 5.9 4.5 5. 4. 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY + + ab 4. 43.9 36.3 39.9 (a) Analyze the data from this experiment. Design Expert Output Response: Vibration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 638. 3 546.4 9.36 <. significant A 7.3 7.3 85.5 <. B 7.6 7.6 38. <. AB 33.63 33.63 5.8 <. Residual 7.7 5.98 Lack of Fit. Pure Error 7.7 5.98 Cor Total 79.83 5 The Model F-value of 9.36 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. (b) Construct a normal probability plot of the residuals, and plot the residuals versus the predicted vibration level. Interpret these plots. Normal plot of residuals Residuals vs. Predicted 3.65 99 Normal % probability 95 9 8 7 5 3 5 Residuals.75 -. 75 -. 75-3.9 75-3.9 75 -. 75 -. 75.75 3.65 4.9.6 7.6 33.94 4.7 Residual Predicted There is nothing unusual about the residual plots. (c) Draw the AB interaction plot. Interpret this plot. What levels of bit size and speed would you recommend for routine operation? To reduce the vibration, use the smaller bit. Once the small bit is specified, either speed will work equally well, because the slope of the curve relating vibration to speed for the small tip is approximately zero. The process is robust to speed changes if the small bit is used. 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Vibration 43.9 Interaction Graph Cutting Speed X = A: Bit Size Y = B: Cutting Speed Design Points 36.5 B- -. B+. Vibration 8.4.65.9 -. -.5..5. Bit Size 6-6 Reconsider the experiment described in Problem 6-. Suppose that the experimenter only performed the eight trials from replicate I. In addition, he ran four center points and obtained the following response values: 36, 4, 43, 45. (a) Estimate the factor effects. Which effects are large? DESIGN-EXPERT Plot Life Normal plot A: Cutting Speed B: Tool Geometry C: Cutting Angle Normal % probability 99 95 9 8 7 5 3 C B 5 AC - 3.7 5-7. 3 -.5 6..75 Effect Effects B, C, and AC appear to be large. (b) Perform an analysis of variance, including a check for pure quadratic curvature. What are your conclusions? SS PureQuadratic Design Expert Output Response: Life in hours n F n C n y y 8 4 4. 875 4. F F n C C 8 4. 47 6-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 48.88 7 49.84 9.77.439 significant A 3.3 3.3..683 B 35.3 35.3..93 C 9. 9..4.389 AB 6.3 6.3.4.57 AC 378. 378. 4.66.57 BC 55. 55. 3.6.54 ABC 9. 9. 5.94.97 Curvature.4.4.77E-3.967 not significant Pure Error 46. 3 5.33 Cor Total 94.9 The Model F-value of 9.77 implies the model is significant. There is only a 4.39% chance that a "Model F-Value" this large could occur due to noise. The "Curvature F-value" of. implies the curvature (as measured by difference between the average of the center points and the average of the factorial points) in the design space is not significant relative to the noise. There is a 96.7% chance that a "Curvature F-value" this large could occur due to noise. Design Expert Output Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 896.5 4 4.3 7.9.98 significant A 3.3 3.3..7496 B 35. 35..47.7 C 9. 9. 6.7.36 AC 378. 378. 3.34.8 Residual 98.4 7 8.35 Lack of Fit 5.4 4 38..49.4 not significant Pure Error 46. 3 5.33 Cor Total 94.9 The Model F-value of 7.9 implies the model is significant. There is only a.98% chance that a "Model F-Value" this large could occur due to noise. Effects B, C and AC are significant at 5%. There is no effect of curvature. (c) Write down an appropriate model for predicting tool life, based on the results of this experiment. Does this model differ in any substantial way from the model in Problem 7-, part (c)? Design Expert Output Final Equation in Terms of Coded Factors: Life = +4.88 +.6 * A +6.37 * B +4.87 * C -6.88 * A * C (d) Analyze the residuals. 6-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 3. 99 Normal % probability 95 9 8 7 5 3 5 Studentized Residuals.5. -.5-3. -. -. 5..5..3 3.9 4.5 49.3 58.38 Studentized Residuals Predicted (e) What conclusions would you draw about the appropriate operating conditions for this process? To maximize life run with B at the high level, A at the low level and C at the high level Cube Graph Life 58.38 45.88 B+ 34.88 49.88 B: Tool Geometry 45.63 33.3 C+ C: Cutting Angle B-.3 37.3 C- A- A+ A: Cutting Speed 6-7 An experiment was performed to improve the yield of a chemical process. Four factors were selected, and two replicates of a completely randomized experiment were run. The results are shown in the following table: Treatment Replicate Replicate Treatment Replicate Replicate Combination I II Combination I II () 9 93 d 98 95 a 74 78 ad 7 76 b 8 85 bd 87 83 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ab 83 8 abd 85 86 c 77 78 cd 99 9 ac 8 8 acd 79 75 bc 88 8 bcd 87 84 abc 73 7 abcd 8 8 (a) Estimate the factor effects. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -9.65 657.3 4.374 Error B -.35 3.78.84679 Error C -.6875 57.783 3.5538 Error D 3.9375 4.3 7.6 Error AB 4.65 3.3 8.67 Error AC.6875 3.785.3339 Error AD -.875 38.83.35 Error BC -.565.535.55533 Error BD -.875.85.784 Error CD.6875.78.3998 Error ABC -5.875 5.8 3.8 Error ABD 4.6875 75.78.89 Error ACD -.9375 7.35.4336 Error BCD -.9375 7.35.4336 Error ABCD.4375 47.533.956 (b) Prepare an analysis of variance table, and determine which factors are important in explaining yield. Design Expert Output Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 54.97 5.33 3. <. significant A 657.3 657.3 85.8 <. B 3.78 3.78.8.984 C 57.78 57.78 7.55.43 D 4.3 4.3 6.. AB 3.3 3.3 7.4.7 AC 3.78 3.78.49.493 AD 38.8 38.8 5..399 BC.53.53.33.5733 BD.8.8.37.854 CD.78.78.98.38 ABC 5.8 5.8 8. <. ABD 75.78 75.78.96. ACD 7.3 7.3.9.35 BCD 7.3 7.3.9.35 ABCD 47.53 47.53 6..4 Residual.5 6 7.66 Lack of Fit. Pure Error.5 6 7.66 Cor Total 67.47 3 The Model F-value of 3. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, D, AB, AD, ABC, ABD, ABCD are significant model terms. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY F 6.,, 853., and F 56.,, 6. therefore, factors A and D and interactions AB, ABC, and ABD are significant at %. Factor C and interactions AD and ABCD are significant at.5%. (b) Write down a regression model for predicting yield, assuming that all four factors were varied over the range from - to + (in coded units). Model with hierarchy maintained: Design Expert Output Final Equation in Terms of Coded Factors: yield = +8.78-4.53 * A -.66 * B -.34 * C +.97 * D +.3 * A * B +.34 * A * C -.9 * A * D -.8 * B * C -.94 * B * D +.84 * C * D -.59 * A * B * C +.34 * A * B * D -.47 * A * C * D -.47 * B * C * D +. * A * B * C * D Model without hierarchy terms: Design Expert Output Final Equation in Terms of Coded Factors: yield = +8.78-4.53 * A -.34 * C +.97 * D +.3 * A * B -.9 * A * D -.59 * A * B * C +.34 * A * B * D +. * A * B * C * D Confirmation runs might be run to see if the simpler model without hierarchy is satisfactory. (d) Plot the residuals versus the predicted yield and on a normal probability scale. Does the residual analysis appear satisfactory? There appears to be one large residual both in the normal probability plot and in the plot of residuals versus predicted. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 6.96875 99 Normal % probability 95 9 8 7 5 3 5 Residuals 3.96875.96875 -.35-5.35-5.35 -.35.96875 3.96875 6.96875 7.9 78.3 84.69 9.8 97.47 Residual Predicted (e) Two three-factor interactions, ABC and ABD, apparently have large effects. Draw a cube plot in the factors A, B, and C with the average yields shown at each corner. Repeat using the factors A, B, and D. Do these two plots aid in data interpretation? Where would you recommend that the process be run with respect to the four variables? Cube Graph yield Cube Graph yield 86.53 76.34 86. 83.5 B+ 84.3 84. B+ 84.56 77.6 B: B 85.4 77.47 C+ B: B 94.75 74.75 D+ C: C D: D B- 93.8 74.97 C- A- A+ A: A B- 83.94 77.69 D- A- A+ A: A Run the process at A low B low, C low and D high. 6-8 A bacteriologist is interested in the effects of two different culture media and two different times on the growth of a particular virus. She performs six replicates of a design, making the runs in random order. Analyze the bacterial growth data that follow and draw appropriate conclusions. Analyze the residuals and comment on the model s adequacy. Culture Medium Time 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 5 6 hr 3 8 4 5 6 9 7 37 39 3 34 8 hr 37 39 3 34 35 36 3 35 Design Expert Output Response: Virus growth ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 69.46 3 3.49 45. <. significant A 9.38 9.38.84.96 B 59.4 59.4 5.5 <. AB 9.4 9.4 8..4 Residual.7 5. Lack of Fit. Pure Error.7 5. Cor Total 793.63 3 The Model F-value of 45. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, AB are significant model terms. Normal plot of residuals Residuals vs. Predicted 4.66667 99 Norm al % probability 95 9 8 7 5 3 5 Residuals.66667.666667 -.33333-3.33333-3.33333 -.33333.666667.66667 4.66667 3.33 6.79 3.5 33.7 37.7 Residual Predicted Growth rate is affected by factor B (Time) and the AB interaction (Culture medium and Time). There is some very slight indication of inequality of variance shown by the small decreasing funnel shape in the plot of residuals versus predicted. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Virus growth 39 Interaction Graph Tim e X = A: Culture Medium Y = B: Time Design Points 34.5 B-. B+ 8. Virus growth 9.5 4.75 Culture Medium 6-9 An industrial engineer employed by a beverage bottler is interested in the effects of two different typed of 3-ounce bottles on the time to deliver -bottle cases of the product. The two bottle types are glass and plastic. Two workers are used to perform a task consisting of moving 4 cases of the product 5 feet on a standard type of hand truck and stacking the cases in a display. Four replicates of a factorial design are performed, and the times observed are listed in the following table. Analyze the data and draw the appropriate conclusions. Analyze the residuals and comment on the model s adequacy. Worker Bottle Type Glass 5. 4.89 6.65 6.4 4.98 5. 5.49 5.55 Plastic 4.95 4.43 5.8 4.9 4.7 4.5 4.75 4.7 Design Expert Output Response: Times ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4.86 3.6 3.4.4 significant A.. 6.8.7 B.54.54.4.7 AB.3.3.4.463 Residual.49. Lack of Fit. Pure Error.49. Cor Total 6.35 5 The Model F-value of 3.4 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. There is some indication of non-constant variance in this experiment. 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.6675 99 Norm al % probability 95 9 8 7 5 3 5 Residuals.3775.875 -. 5 -.4 9 5 -.4 9 5 -. 5.875.3775.6675 4.47 4.85 5.3 5.6 5.98 Residual Predicted Residuals vs. Worker.6675.3775 Residuals.875 -. 5 -.4 9 5 Worker 6- In problem 6-9, the engineer was also interested in potential fatigue differences resulting from the two types of bottles. As a measure of the amount of effort required, he measured the elevation of heart rate (pulse) induced by the task. The results follow. Analyze the data and draw conclusions. Analyze the residuals and comment on the model s adequacy. Worker Bottle Type Glass 39 45 3 58 35 6 Plastic 44 35 3 4 6 5 Design Expert Output 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Pulse ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 784.9 3 98.6 6.3. significant A 66.56 66.56 45.37 <. B 5.6 5.6.8.8 AB 5.56 5.56.9.3595 Residual 694.75 57.9 Lack of Fit. Pure Error 694.75 57.9 Cor Total 3478.94 5 The Model F-value of 6.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A are significant model terms. Normal plot of residuals Residuals vs. Predicted 3.75 99 Normal % probability 95 9 8 7 5 3 5 Residuals 6.6875 -.3 75-7.4 37 5-4.5-4.5-7.4 37 5 -.3 75 6.6875 3.75 3.5.9 8.88 36.56 44.5 Residual Predicted Residuals vs. Worker 3.75 6.6875 Residuals -.3 75-7.4 37 5-4.5 Worker There is an indication that one worker exhibits greater variability than the other. 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6- Calculate approximate 95 percent confidence limits for the factor effects in Problem 6-. Do the results of this analysis agree with the analysis of variance performed in Problem 6-? SE( effect ) S 57. 9 k n 4 3. 8 Variable Effect C.I. A -5.65 3.8(.96)= 7.448 B -5.5 3.8(.96)= 7.448 AB -7.5 3.8(.96)= 7.448 The 95% confidence intervals for factors A does not contain zero. This agrees with the analysis of variance approach. 6- An article in the AT&T Technical Journal (March/April 986, Vol. 65, pp. 39-5) describes the application of two-level factorial designs to integrated circuit manufacturing. A basic processing step is to grow an epitaxial layer on polished silicon wafers. The wafers mounted on a susceptor are positioned inside a bell jar, and chemical vapors are introduced. The susceptor is rotated and heat is applied until the epitaxial layer is thick enough. An experiment was run using two factors: arsenic flow rate (A) and deposition time (B). Four replicates were run, and the epitaxial layer thickness was measured (in mm). The data are shown below: Replicate Factor Levels A B I II III IV Low (-) High (+) - - 4.37 6.65 3.97 3.97 A 55% 59% + - 3.88 3.86 4.3 3.94 - + 4.8 4.757 4.843 4.878 B Short Long + + 4.888 4.9 4.45 4.93 ( min) (5 min) (a) Estimate the factor effects. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.375.459 6.79865 Error B.586.37358 3.96 Error AB.85.36969 5.3574 Error Lack Of Fit Error Pure Error 3.8848 64.655 (b) Conduct an analysis of variance. Which factors are important? Design Expert Output Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.9 3.7.9.45 not significant A.4.4.6.833 B.37.37 4.3.6 6-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY AB Residual.3 3.83.3.3.99.3386 Lack of Fit. Pure Error 3.83.3 Cor Total 5.9 5 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a 4.5 % chance that a "Model F-value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case there are no significant model terms. (c) Write down a regression equation that could be used to predict epitaxial layer thickness over the region of arsenic flow rate and deposition time used in this experiment. Design Expert Output Final Equation in Terms of Coded Factors: Thickness = +4.5 -.6 * A +.9 * B +.4 * A * B Final Equation in Terms of Actual Factors: Thickness = +37.6656 -.439 * Flow Rate -.48735 * Dep Time +.85 * Flow Rate * Dep Time (d) Analyze the residuals. Are there any residuals that should cause concern? Observation # falls outside the groupings in the normal probability plot and the plot of residual versus predicted. Normal plot of residuals Residuals vs. Predicted.64475 99 Norm al % probability 95 9 8 7 5 3 5 Residuals.85.5575 -.4875 -.635 -.635 -.4875.5575.85.64475 3.9 4.5 4.37 4.6 4.8 Residual Predicted (e) Discuss how you might deal with the potential outlier found in part (d). One approach would be to replace the observation with the average of the observations from that experimental cell. Another approach would be to identify if there was a recording issue in the original data. The first analysis below replaces the data point with the average of the other three. The second analysis assumes that the reading was incorrectly recorded and should have been 4.65. 6-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Analysis with the run associated with standard order replaced with the average of the remaining three runs in the cell, 3.97: Design Expert Output Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.97 3.99 53.57 <. significant A 7.439E-3 7.439E-3.4.5375 B.96.96 6.9 <. AB.76E-4.76E-4..953 Pure Error..8 Cor Total 3.9 5 The Model F-value of 53.57 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. Final Equation in Terms of Coded Factors: Thickness = +4.38 -. * A +.43 * B +3.688E-3 * A * B Final Equation in Terms of Actual Factors: Thickness = +3.3665 -. * Flow Rate +.999 * Dep Time +7.375E-4 * Flow Rate * Dep Time Normal plot of residuals Residuals vs. Predicted.43 99 Normal % probability 95 9 8 7 5 3 5 Residuals.375 -. 5 5 -.4475 -.3 7 4 -.374 -.4475 -.55.375.43 3.9 4.5 4.37 4.6 4.8 Residual Predicted A new outlier is present and should be investigated. Analysis with the run associated with standard order replaced with the value 4.65: 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 3.94 45.8 <. significant A.8.8.87.3693 B.8.8 34.47 <. AB 3.969E-3 3.969E-3.9.6699 Pure Error.5. Cor Total 3.7 5 The Model F-value of 45.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. Final Equation in Terms of Coded Factors: Thickness = +4.39 -.34 * A +.4 * B +.6 * A * B Final Equation in Terms of Actual Factors: Thickness = +5.556 -.5688 * Flow Rate -.35 * Dep Time +3.5E-3 * Flow Rate * Dep Time Normal plot of residuals Residuals vs. Predicted 3. 99 Normal % probability 95 9 8 7 5 3 5 Studentized Residuals.5. -.5-3. -3. -.9 6 -.9..6 3.9 4.5 4.37 4.6 4.8 Studentized Residuals Predicted Another outlier is present and should be investigated. 6-3 Continuation of Problem 6-. Use the regression model in part (c) of Problem 6- to generate a response surface contour plot for epitaxial layer thickness. Suppose it is critically important to obtain 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY layer thickness of 4.5 mm. What settings of arsenic flow rate and deposition time would you recommend? Arsenic flow rate may be set at any of the experimental levels, while the deposition time should be set at.4 minutes. DESIGN-EXPERT Plot 4 4 5. Thickness X = A: Flow Rate Y = B: Dep Time Design Points 3.75 Thickness 4.674 DESIGN-EXPERT Plot Thickness X = A: Flow Rate Y = B: Dep Time Design Points 6.65 5.4953 Interaction Graph Dep Time Dep Time.5 4.537 B-. B+ 5. Thickness 4.857.5 4.373 4.6 4.56 4.7 4 4. 55. 56. 57. 58. 59. 3.4864 55. 56. 57. 58. 59. Flow Rate Flow Rate 6-4 Continuation of Problem 6-3. How would your answer to Problem 6-3 change if arsenic flow rate was more difficult to control in the process than the deposition time? Running the process at a high level of Deposition Time there is no change in thickness as flow rate changes. 6-5 A nickel-titanium alloy is used to make components for jet turbine aircraft engines. Cracking is a potentially serious problem in the final part, as it can lead to non-recoverable failure. A test is run at the parts producer to determine the effects of four factors on cracks. The four factors are pouring temperature (A), titanium content (B), heat treatment method (C), and the amount of grain refiner used (D). Two replicated of a 4 design are run, and the length of crack (in m) induced in a sample coupon subjected to a standard test is measured. The data are shown below: Treatment Replicate Replicate A B C D Combination I II - - - - () 7.37 6.376 + - - - a 4.77 5.9 - + - - b.635.89 + + - - ab 7.73 7.85 - - + - c.43.5 + - + - ac 4.368 4.98 - + + - bc 9.36 9.53 + + + - abc 3.44.93 - - - + d 8.56 8.95 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY + - - + ad 6.867 7.5 - + - + bd 3.876 3.658 + + - + abd 9.84 9.639 - - + + cd.846.337 + - + + acd 6.5 5.94 - + + + bcd.9.935 + + + + abcd 5.653 5.53 (a) Estimate the factor effects. Which factors appear to be large? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 3.888 7.989.748 Model B 3.97588 6.46.99 Model C -3.5965 3.464 8.84 Model D.95775 3.663 5.3583 Model AB.934 9.967 5.969 Model AC -4.775 8.496.4548 Error AD.765.4688.8845 Error BC.96.7378.884 Error BD.475.7865.3 Error CD -.76875.4778.8685 Model ABC 3.375 78.75 3.768 Error ABD.98.7683.3464 Error ACD.95.963.534 Error BCD.3565.53.7746 Error ABCD.45.5963.7893 (b) Conduct an analysis of variance. Do any of the factors affect cracking? Use =.5. Design Expert Output Response: Crack Lengthin mm x ^- ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 57.95 5 38.6 468.99 <. significant A 7.9 7.9 898.34 <. B 6.46 6.46 558.7 <. C 3.46 3.46 74.8 <. D 3.66 3.66 377.8 <. AB 9.93 9.93 368.74 <. AC 8.5 8.5 583.6 <. AD.47.47.58.4586 BC.74.74.9.3547 BD.8.8..6453 CD.47.47.58.4564 ABC 78.75 78.75 97.33 <. ABD.77.77.95.345 ACD.96E-3.96E-3.36.858 BCD...3.78 ABCD.596E-3.596E-3..89 Residual.3 6.8 Lack of Fit. Pure Error.3 6.8 Cor Total 57.5 3 The Model F-value of 468.99 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, D, AB, AC, ABC are significant model terms. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Write down a regression model that can be used to predict crack length as a function of the significant main effects and interactions you have identified in part (b). Design Expert Output Final Equation in Terms of Coded Factors: Crack Length= +.99 +.5 *A +.99 *B -.8 *C +.98 *D +.97 *A*B -. *A*C +.57 * A * B * C (d) Analyze the residuals from this experiment. Normal plot of residuals Residuals vs. Predicted.454875 99 Normal % probability 95 9 8 7 5 3 5.3688 Residuals.5 -. 6 8 7 -.4 33 8 7 5 -.433875 -.687.5.3688.454875 4.9 8.6.93 5.8 9.66 Residual Predicted There is nothing unusual about the residuals. (e) Is there an indication that any of the factors affect the variability in cracking? By calculating the range of the two readings in each cell, we can also evaluate the effects of the factors on variation. The following is the normal probability plot of effects: 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Range Normal plot A: Pour Temp B: Titanium Content C: Heat Treat Method D: Grain Refiner Normal % probability 99 95 9 8 7 5 3 5 AB CD -. -..5.3. Effect It appears that the AB and CD interactions could be significant. The following is the ANOVA for the range data: Design Expert Output Response: Range ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.9.4.46.4 significant AB.3.3 9.98.75 CD.6.6.94.3 Residual.6 3.3 Cor Total.45 5 The Model F-value of.46 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case AB, CD are significant model terms. Final Equation in Terms of Coded Factors: Range = +.37 +.89 * A * B +. * C * D (f) What recommendations would you make regarding process operations? Use interaction and/or main effect plots to assist in drawing conclusions. From the interaction plots, choose A at the high level and B at the high level. In each of these plots, D can be at either level. From the main effects plot of C, choose C at the high level. Based on the range analysis, with C at the high level, D should be set at the low level. From the analysis of the crack length data: 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Crack Length 9.84 Interaction Graph B: Titanium Content DESIGN-EXPERT Plot Crack Length 9.84 Interaction Graph C: Heat Treat Method X = A: Pour Temp Y = B: Titanium Content X = A: Pour Temp Y = C: Heat Treat Method 5.895 B- -. B+. Actual Factors C: Heat Treat Method = D: Grain Refiner =..96 Crack Length 5.895 C - C Actual Factors B: Titanium Content =. D: Grain Refiner =..96 Crack Length 8.95 8.95 4.98 4.98 -. -.5..5. -. -.5..5. A: Pou r Tem p A: Pou r Tem p DESIGN-EXPERT Plot Crack Length 9.84 X = D: Grain Refiner One Factor Plot DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content Z = C: Heat Treat Method Cube Graph Crack Length.8 4.7 Actual Factors 5.895 A: Pour T emp =. B: Titanium Content =. C: Heat Treat Method = Crack Length.96 8.95 Actual Factor D: Grain Refiner =. B: Titanium Content B+.8.8 8.64 5. C+ C: Heat Treat Metho 4.98 -. -.5..5. B- 7.73 5.96 C- A- A+ A: Pour Tem p D: Grain Refiner From the analysis of the ranges: 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Range.66 Interaction Graph B: Titanium Content DESIGN-EXPERT Plot Range.66 Interaction Graph D: Grain Refiner X = A: Pour Temp Y = B: Titanium Content X = C: Heat Treat Method Y = D: Grain Refiner.55 B- -. B+. Actual Factors C: Heat Treat Method =. D: Grain Refiner =..384 Range.55 D- -. D+. Actual Factors A: Pour Temp =. B: Titanium Content =..384 Range.455.455.7.7 -. -.5..5. -. -.5..5. A: Pou r Tem p C: Heat Treat Method 6-6 Continuation of Problem 6-5. One of the variables in the experiment described in Problem 6-5, heat treatment method (c), is a categorical variable. Assume that the remaining factors are continuous. (a) Write two regression models for predicting crack length, one for each level of the heat treatment method variable. What differences, if any, do you notice in these two equations? Design Expert Output Final Equation in Terms of Coded Factors Heat Treat Method - Crack Length = +3.7869 +3.533 * Pour Temp +.93994 * Titanium Content +.97888 * Grain Refiner -.669 * Pour Temp * Titanium Content Heat Treat Method Crack Length = +.8994 -.49444 * Pour Temp +.3594 * Titanium Content +.97888 * Grain Refiner +.5358 * Pour Temp * Titanium Content (b) Generate appropriate response surface contour plots for the two regression models in part (a). 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content. Crack Length 8 DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content. Crack Length Actual Factors C: Heat Treat Method = -.5 D: Grain Refiner =. B: Titanium Content. 4 6 Actual Factors C: Heat T reat M ethod =.5 D: Grain Refiner =. B: Titanium Content. -.5 -.5 8 6 -. -. -.5..5. -. -. -.5..5. A: Pour Tem p A: Pour Temp (c) What set of conditions would you recommend for the factors A, B and D if you use heat treatment method C=+? High level of A, low level of B, and low level of D. (d) Repeat part (c) assuming that you wish to use heat treatment method C=-. Low level of A, low level of B, and low level of D. 6-7 An experimenter has run a single replicate of a 4 design. The following effect estimates have been calculated: A = 76.95 AB = -5.3 ABC = -.8 B = -67.5 AC =.69 ABD = -6.5 C = -7.84 AD = 9.78 ACD =. D = -8.73 BC =.78 BCD = -7.98 BD = 4.74 ABCD = -6.5 CD =.7 (a) Construct a normal probability plot of these effects. The plot from Minitab follows. 6-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot for Value ML Estimates 99 ML Estimates Percent 95 9 8 7 6 5 4 3 5 B AB A Mean -.57 StDev 3.96 Goodness of Fit AD*.465-5 Data 5 (b) Identify a tentative model, based on the plot of the effects in part (a). ŷ Intercept 38. 475x 33. 76x 5. 66x A B A x B 6-8 An article in Solid State Technology ( Orthogonal Design for Process Optimization and Its Application in Plasma Etching, May 987, pp. 7-3) describes the application of factorial designs in developing a nitride etch process on a single-wafer plasma etcher. The process uses C F 6 as the reactant gas. Four factors are of interest: anode-cathode gap (A), pressure in the reactor chamber (B), C F 6 gas flow (C), and power applied to the cathode (D). The response variable of interest is the etch rate for silicon nitride. A single replicate of a 4 design in run, and the data are shown below: Actual Etch Run Run Rate Factor Levels Number Order A B C D (A/min) Low (-) High (+) 3 - - - - 55 A (cm).8. 8 + - - - 669 B (mtorr) 4.5 55 3 - + - - 64 C (SCCM) 5 4 9 + + - - 65 D (W) 75 35 5 4 - - + - 633 6 5 + - + - 64 7 6 - + + - 6 8 3 + + + - 635 9 - - - + 37 4 + - - + 749 5 - + - + 5 + + - + 868 3 - - + + 75 4 + - + + 86 5 7 - + + + 63 6-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6 6 + + + + 79 (a) Estimate the factor effects. Construct a normal probability plot of the factor effects. Which effects appear large? DESIGN-EXPERT Plot Etch Rate Normal plot A: Gap B: Pressure C: Gas Flow D: Power Norm al % probability 99 95 9 8 7 5 3 5 AD A D - 5 3.6 3-3 8.69 76.5 9.9 36.3 Effect (b) Conduct an analysis of variance to confirm your findings for part (a). Design Expert Output Response: Etch Rate in A/min ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5.6E+5 3.7E+5 97.9 <. significant A 43.56 43.56 3.77.4 D 3.749E+5 3.749E+5 5.66 <. AD 944.56 944.56 54.3 <. Residual 857.75 738.5 Cor Total 5.34E+55 The Model F-value of 97.9 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, D, AD are significant model terms. (c) What is the regression model relating etch rate to the significant process variables? Design Expert Output Final Equation in Terms of Coded Factors: Etch Rate = +776.6-5.8 * A +53.6 * D -76.8 * A * D Final Equation in Terms of Actual Factors: Etch Rate = -545.375 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +4354.6875 +.485 * Gap * Power -5.365 * Gap * Power (d) Analyze the residuals from this experiment. Comment on the model s adequacy. Normal plot of residuals Residuals vs. Predicted 66.5 99 Norm al % probability 95 9 8 7 5 3 5 Residuals 3.75-3 -3 7.75-7.5-7.5-37.75-3 3.75 66.5 597. 7.94 86.88 94.8 56.75 Residual Predicted The residual versus predicted plot shows a slight football shape indicating very mild inequality of variance. (e) If not all the factors are important, project the 4 design into a k design with k<4 and conduct that analysis of variance. The analysis of variance table is the same as in part (b). Design Expert Output Response: Etch Rate in A/min ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5.6E+5 3.7E+5 97.9 <. significant A 43.56 43.56 3.77.4 B 3.749E+5 3.749E+5 5.66 <. AB 944.56 944.56 54.3 <. Residual 857.75 738.5 Lack of Fit. Pure Error 857.75 738.5 Cor Total 5.34E+55 The Model F-value of 97.9 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. (f) Draw graphs to interpret any significant interactions. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Etch Rate 88.87 Interaction Graph Power X = A: Gap Y = B: Power Design Points 954.49 B- 75. B+ 35. Etch Rate 89.433 684.76 55.8.9... Gap (g) Plot the residuals versus the actual run order. What problems might be revealed by this plot? Residuals vs. Run 66.5 3.75 Residuals -3-3 7.75-7.5 4 7 3 6 Run Number The plot of residuals versus run order can reveal trends in the process over time, inequality of variance with time, and possibly indicate that there may be factors that were not included in the original experiment. 6-9 Continuation of Problem 6-8. Consider the regression model obtained in part (c) of Problem 6-8. (a) Construct contour plots of the etch rate using this model. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot 35. Etch Rate X = A: Gap Y = D: Power Actual Factors B: Pressure = 5. C: Gas Flow = 6.5 3.5 98.5 Etch Rate 93.5 86.875 Power 3. 75.5 87.5 673.65 75..8.9... Gap (b) Suppose that it was necessary to operate this process at an etch rate of 8 Å/min. What settings of the process variables would you recommend? Run at the low level of anode-cathode gap (.8 cm) and at a cathode power level of about 86 watts. The curve is flatter (more robust) on the low end of the anode-cathode variable. 6- Consider the single replicate of the 4 design in Example 6-. Suppose we had arbitrarily decided to analyze the data assuming that all three- and four-factor interactions were negligible. Conduct this analysis and compare your results with those obtained in the example. Do you think that it is a good idea to arbitrarily assume interactions to be negligible even if they are relatively high-order ones? Design Expert Output Response: Etch Rate in A/min ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5.E+5 53.4 5.58. significant A 43.56 43.56.8.64 B.56.56 5.84E-3.9454 C 7.56 7.56..757 D 3.749E+5 3.749E+5 83.99 <. AB 48.6 48.6..744 AC 475.6 475.6..36 AD 944.56 944.56 46.34. BC 77.6 77.6 3.78.95 BD.56.56 7.669E-4.979 CD 8.6 8.6 8.866E-3.986 Residual 86.8 5 37.36 Cor Total 5.34E+55 The Model F-value of 5.58 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, D, AD are significant model terms. This analysis of variance identifies the same effects as the normal probability plot of effects approach used in Example 6-. In general, it is not a good idea to arbitrarily pool interactions. Use the normal 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY probability plot of effect estimates as a guide in the choice of which effects to tentatively include in the model. 6- An experiment was run in a semiconductor fabrication plant in an effort to increase yield. Five factors, each at two levels, were studied. The factors (and levels) were A = aperture setting (small, large), B = exposure time (% below nominal, % above nominal), C = development time (3 s, 45 s), D = mask dimension (small, large), and E = etch time (4.5 min, 5.5 min). The unreplicated 5 design shown below was run. () = 7 d = 8 e = 8 de = 6 a = 9 ad = ae = ade = b = 34 bd = 3 be = 35 bde = 3 ab = 55 abd = 5 abe = 5 abde = 53 c = 6 cd = 8 ce = 5 cde = 5 ac = acd = ace = acde = bc = 4 bcd = 44 bce = 45 bcde = 4 abc = 6 abcd = 6 abce = 65 abcde = 63 (a) Construct a normal probability plot of the effect estimates. Which effects appear to be large? DESIGN-EXPERT Plot Yield Normal plot A: Aperture B: Exposure Time C: Develop Tim e D: M ask Dim ension E: Etch Time Normal % probability 99 95 9 8 7 5 3 C AB A B 5 -. 9 7.59 6.38 5.6 33.94 Effect (b) Conduct an analysis of variance to confirm your findings for part (a). Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 585.3 4 896.8 99.83 <. significant A 6.8 6.8 38.7 <. B 94.3 94.3 355.34 <. C 75.78 75.78 57. <. AB 54.3 54.3 7.6 <. Residual 78.84 7.9 Cor Total 663.97 3 6-33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The Model F-value of 99.83 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. (c) Write down the regression model relating yield to the significant process variables. Design Expert Output Final Equation in Terms of Actual Factors: Aperture small Yield = +.465 +.65 * Exposure Time +.64583 * Develop Time Aperture large Yield = +.875 +.4688 * Exposure Time +.64583 * Develop Time (d) Plot the residuals on normal probability paper. Is the plot satisfactory? Normal plot of residuals 99 Norm al % probability 95 9 8 7 5 3 5 -.785 -.3 9 6 3-3.557E-5.396.785 There is nothing unusual about this plot. Residual (e) Plot the residuals versus the predicted yields and versus each of the five factors. Comment on the plots. 6-34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Aperture Residuals vs. Exposure Time.785.785.396 3.396 Residuals 3.557E-5 Residuals 3.557E-5 -.3963 3 -.3963 -.785 -.785 - - 3-7 7 3 Ape rture Expos ure Tim e Residuals vs. Develop Time Residuals vs. Mask Dimension.785.785.396 3.396 Residuals 3.557E-5 Residuals 3.557E-5 -.3963 3 -.3963 -.785 -.785 3 33 35 38 4 43 45 Develop Time Mas k Dim ens ion Residuals vs. Etch Time.785.396 Residuals 3.557E-5 -.3963 3 -.785 4.5 4.75 5. 5.5 5.5 Etch Time 6-35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The plot of residual versus exposure time shows some very slight inequality of variance. There is no strong evidence of a potential problem. (f) Interpret any significant interactions. DESIGN-EXPERT Plot Yield 65 Interaction Graph Ape rture X = B: Exposure Time Y = A: Aperture 5.5 A small A large Actual Factors C: Develop Tim e = 37.5 D: Mask Dimension = Small 35.5 E: Etch Time = 5. Yield.75 6 -. -.... Exposure Time Factor A does not have as large an effect when B is at its low level as it does when B is at its high level. (g) What are your recommendations regarding process operating conditions? For the highest yield, run with B at the high level, A at the high level and C at the high level. (h) Project the 5 design in this problem into a k design in the important factors. Sketch the design and show the average and range of yields at each run. Does this sketch aid in interpreting the results of this experiment? DESIGN-EASE Analysis Actual Yield 4.5 R=5 6.5 R=5 E x p o s u r e T i m e B+ 3.75 R=5 6. R=3 5.5 R=5 B- 7.5.5 C- A- R= R=3 A+ Aperture.75 R= D e v e l o p C+ T i m e 6-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY This cube plot aids in interpretation. The strong AB interaction and the large positive effect of C are clearly evident. 6- Continuation of Problem 6-. Suppose that the experimenter had run four runs at the center points in addition to the 3 trials in the original experiment. The yields obtained at the center point runs were 68, 74, 76, and 7. (a) Reanalyze the experiment, including a test for pure quadratic curvature. SS PureQuadratic n F n C n y y 343. 535 7 F F n C C 3 4 64. 337 Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 46.9 4 865.7 353.9 <. significant A 99.5 99.5.56 <. B 94.3 94.3 38. <. C 75.78 75.78 9.74 <. AB 54.3 54.3 6.6 <. Curvature 64.34 64.34 755.4 <. significant Residual 4.88 3 8. Cor Total 788.3 35 The Model F-value of 353.9 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. (b) Discuss what your next step would be. Add axial points and fit a second-order model. 6-3 In a process development study on yield, four factors were studied, each at two levels: time (A), concentration (B), pressure (C), and temperature (D). A single replicate of a 4 design was run, and the resulting data are shown in the following table: Actual Run Run Yield Factor Levels Number Order A B C D (lbs) Low (-) High (+) 5 - - - - A (h).5 3. 9 + - - - 8 B (%) 4 8 3 8 - + - - 3 C (psi) 6 8 4 3 + + - - 6 D (ºC) 5 5 5 3 - - + - 7 6 7 + - + - 5 7 4 - + + - 8 + + + - 5 9 6 - - - + 6-37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY + - - + 5 - + - + 3 5 + + - + 4 3 4 - - + + 9 4 6 + - + + 5 - + + + 7 6 + + + + 3 (a) Construct a normal probability plot of the effect estimates. Which factors appear to have large effects? DESIGN-EXPERT Plot Yield Normal plot A: Time B: Concentration C: Pressure D: Temperature Normal % probability 99 95 9 8 7 5 3 5 AC C AD D A -4. 5 -. 6.3.3 4.5 Effect A, C, D and the AC and AD interactions. (b) Conduct an analysis of variance using the normal probability plot in part (a) for guidance in forming an error term. What are your conclusions? Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 75.5 5 55. 33.9 <. significant A 8. 8. 49.85 <. C 6. 6. 9.85.5 D 4.5 4.5 6..5 AC 7.5 7.5 44.46 <. AD 64. 64. 39.38 <. Residual 6.5.6 Cor Total 9.75 5 The Model F-value of 33.9 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, D, AC, AD are significant model terms. (c) Write down a regression model relating yield to the important process variables. 6-38

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Final Equation in Terms of Coded Factors: Yield = +7.38 +.5 *A +. *C +.63 *D -.3 *A*C +. *A*D Final Equation in Terms of Actual Factors: Yield = +9.5-83.5 * Time +.4375 * Pressure -.63 * Temperature -.85 * Time * Pressure +.64 * Time * Temperature (d) Analyze the residuals from this experiment. Does your analysis indicate any potential problems? Normal plot of residuals Residuals vs. Predicted.375 99 Normal % probability 95 9 8 7 5 3 5 Residuals.65 -. 5 -.8 75 -.6 5 -.6 5 -.8 75 -. 5.65.375.63 4.8 8..9 4.38 Residual Predicted 6-39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.375 Residuals vs. Run.65 Residuals -. 5 -.8 75 -.6 5 4 7 3 6 Run Number There is nothing unusual about the residual plots. (e) Can this design be collapsed into a 3 design with two replicates? If so, sketch the design with the average and range of yield shown at each point in the cube. Interpret the results. DESIGN-EASE Analysis Actual yield 8. R=. R= t e m p e r a t u r e D+.5 R=3 8.5 R=3 4.5 R= D-.5 7. C- A- R= R= A+ time 5. C+ R= p r e s s u r e 6-4 Continuation of Problem 6-3. Use the regression model in part (c) of Problem 6-3 to generate a response surface contour plot of yield. Discuss the practical purpose of this response surface plot. 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The response surface contour plot shows the adjustments in the process variables that lead to an increasing or decreasing response. It also displays the curvature of the response in the design region, possibly indicating where robust operating conditions can be found. Two response surface contour plots for this process are shown below. These were formed from the model written in terms of the original design variables. DESIGN-EXPERT Plot Yield X = A: Time Y = C: Pressure 8. Yield DESIGN-EXPERT Plot 5. Yield X = A: Time Y = D: Temperature Yield Actual Factors B: Concentration = 6.75. D: Temperature = 37.5 7.8333 Actual Factors B: Concentration = 6. 43.75 C: Pressure = 7. 9.97 Pressure 7. 65. 4.967 6.375 9.97 Temperature 37.5 3.5 6.375 7.8333 3.4583 6..5.63.75.88 3. 5..5.63.75.88 3. Tim e Tim e 6-5 The scrumptious brownie experiment. The author is an engineer by training and a firm believer in learning by doing. I have taught experimental design for many years to a wide variety of audiences and have always assigned the planning, conduct, and analysis of an actual experiment to the class participants. The participants seem to enjoy this practical experience and always learn a great deal from it. This problem uses the results of an experiment performed by Gretchen Krueger at Arizona State University. There are many different ways to bake brownies. The purpose of this experiment was to determine how the pan material, the brand of brownie mix, and the stirring method affect the scrumptiousness of brownies. The factor levels were Factor Low (-) High (+) A = pan material Glass Aluminum B = stirring method Spoon Mixer C = brand of mix Expensive Cheap The response variable was scrumptiousness, a subjective measure derived from a questionnaire given to the subjects who sampled each batch of brownies. (The questionnaire dealt with such issues as taste, appearance, consistency, aroma, and so forth.) An eight-person test panel sampled each batch and filled out the questionnaire. The design matrix and the response data are shown below: Brownie Test Panel Results Batch A B C 3 4 5 6 7 8 - - - 9 8 9 + - - 5 6 4 9 6 5 3 - + - 9 4 + + - 6 7 5 3 3 5 - - + 5 8 6 8 9 4 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6 + - + 3 4 3 9 3 4 9 7 - + + 3 7 7 7 3 8 + + + 5 5 3 9 4 (a) Analyze the data from this experiment as if there were eight replicates of a 3 design. Comment on the results. Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 93.5 7 3.3..475 significant A 7.5 7.5.95. B 8.6 8.6.99.894 C.63.63..994 AB.6.6..994 AC.56.56.6.63 BC...7.6858 ABC.5.5.4.8396 Residual 338.5 56 6.4 Lack of Fit. Pure Error 338.5 56 6.4 Cor Total 43.75 63 The Model F-value of. implies the model is significant. There is only a 4.75% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A are significant model terms. In this analysis, A, the pan material and B, the stirring method, appear to be significant. There are 56 degrees of freedom for the error, yet only eight batches of brownies were cooked, one for each recipe. (b) Is the analysis in part (a) the correct approach? There are only eight batches; do we really have eight replicates of a 3 factorial design? The different rankings by the taste-test panel are not replicates, but repeat observations by differenttesters on the same batch of brownies. It is not a good idea to use the analysis in part (a) because the estimate of error may not reflect the batch-to-batch variation. (c) Analyze the average and standard deviation of the scrumptiousness ratings. Comment on the results. Is this analysis more appropriate than the one in part (a)? Why or why not? 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Average Normal plot DESIGN-EXPERT Plot Stdev Normal plot A: Pan M aterial B: Stirring Method C: Mix Brand 99 95 9 A A: Pan M aterial B: Stirring Method C: Mix Brand 99 95 9 C Normal % probability 8 7 5 3 5 B Normal % probability 8 7 5 3 5 AC A -.3.9.9.5.3 -.5 7 -. -.4 5..68 Effect Effect Design Expert Output Response: Average ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 5.64 76.3. significant A 9.3 9.3.93. B.5.5 3.34.7 Residual.37 5.74 Cor Total.65 7 The Model F-value of 76.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Design Expert Output Response: Stdev ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.5 3. 9.77.59 significant A.4.4.5.343 C.9.9 4.4.34 AC 4.9 4.9 3.75.8 Residual.8 4. Cor Total 6.87 7 The Model F-value of 9.77 implies the model is significant. There is only a.59% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case AC are significant model terms. Variables A and B affect the mean rank of the brownies. Note that the AC interaction affects the standard deviation of the ranks. This is an indication that both factors A and C have some effect on the variability in the ranks. It may also indicate that there is some inconsistency in the taste test panel members. For the analysis of both the average of the ranks and the standard deviation of the ranks, the mean square error is 6-43

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY now determined by pooling apparently unimportant effects. This is a more estimate of error than obtained assuming that all observations were replicates. 6-6 An experiment was conducted on a chemical process that produces a polymer. The four factors studied were temperature (A), catalyst concentration (B), time (C), and pressure (D). Two responses, molecular weight and viscosity, were observed. The design matrix and response data are shown below: Actual Run Run Molecular Factor Levels Number Order A B C D Weight Viscosity Low (-) High (+) 8 - - - - 4 4 A (ºC) 9 + - - - 4 5 B (%) 4 8 3 3 - + - - 35 5 C (min) 3 4 8 + + - - 5 63 D (psi) 6 75 5 3 - - + - 65 38 6 + - + - 65 55 7 4 - + + - 4 5 8 7 + + + - 75 6 9 6 - - - + 4 4 7 + - - + 39 55 - + - + 3 5 + + - + 5 5 3 4 - - + + 65 4 4 9 + - + + 63 49 5 5 - + + + 5 5 6 + + + + 7 6 7 55 5 8 5 5 46 9 6 4 55 475 5 (a) Consider only the molecular weight response. Plot the effect estimates on a normal probability scale. What effects appear important? 6-44

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Molecular Wt Half Normal plot A: Temperature B: Catalyst Con. C: Tim e D: Pressure Half Normal % probability 99 97 95 9 85 8 7 6 4 AB A C. 5.3.63 5.94.5 Effect A, C and the AB interaction. (b) Use an analysis of variance to confirm the results from part (a). Is there an indication of curvature? A,C and the AB interaction are significant. While the main effect of B is not significant, it could be included to preserve hierarchy in the model. There is no indication of quadratic curvature. Design Expert Output Response: Molecular Wt ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.89E+5 3 936.83 73. <. significant A 656.5 656.5 47.76 <. C.6E+5.6E+5 6.3 <. AB 576. 576. 44.9 <. Curvature 3645. 3645..84.5 not significant Residual 937.5 5 8.5 Lack of Fit 4.5 95.4.36.96 not significant Pure Error 785. 3 68.33 Cor Total 3.37E+59 The Model F-value of 73. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, AB are significant model terms. (c) Write down a regression model to predict molecular weight as a function of the important variables. Design Expert Output Final Equation in Terms of Coded Factors: Molecular Wt = +56.5 +6.87 * A +.63 * C +6. * A * B (d) Analyze the residuals and comment on model adequacy. 6-45

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 4.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals.65 -.5-5 3. 5-8 5-8 5-53.5 -.5.65 4.5 83.75 395. 56.5 67.5 78.75 Residual Predicted There are two residuals that appear to be large and should be investigated. (e) Repeat parts (a) - (d) using the viscosity response. DESIGN-EXPERT Plot Viscosity Normal plot A: Temperature B: Catalyst Con. C: Tim e D: Pressure Normal % probability 99 95 9 8 7 5 3 5 B A - 5. 5.3 35.6 65.94 96.5 Effect Design Expert Output Response: Viscosity ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 736.5 358.5 35.97 <. significant A 3756.5 3756.5 37.88 <. B 3336.5 3336.5 34.5 <. Curvature 6.5 6.5.63.856 not significant Residual 565. 6 978.3 Lack of Fit 348.5 3 37..43.498 not significant Pure Error 68.75 3 7.9 Cor Total 8673.75 9 6-46

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The Model F-value of 35.97 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Final Equation in Terms of Coded Factors: Viscosity = +5.6 +48.3 * A +45.63 * B Normal plot of residuals Residuals vs. Predicted 35.65 99 Norm al % probability 95 9 8 7 5 3 5 Residuals 3.5-9.37 5-6.87 5 3-9 4.37 5-94.375-6.875-9.375 3.5 35.65 46.87 453.75 5.6 547.5 594.37 Residual Predicted There is one large residual that should be investigated. 6-7 Continuation of Problem 6-6. Use the regression models for molecular weight and viscosity to answer the following questions. (a) Construct a response surface contour plot for molecular weight. In what direction would you adjust the process variables to increase molecular weight? Increase temperature, catalyst and time. 6-47

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Molecular Wt X = A: Temperature Y = B: Catalyst Con. Design Points Actual Factors C: T ime = 5. D: Pressure = 67.5 B: Catalys t Con. 8. 4 45 7. 6. Molecular W t 45 55 475 4 55 5 6 575 5. 4.. 5.. 5.. A: Temperature (a) Construct a response surface contour plot for viscosity. In what direction would you adjust the process variables to decrease viscosity? DESIGN-EXPERT Plot V i sco si ty X = A: Temperature Y = B: Catalyst Con. Design Points Actual Factors C: T ime = 5. D: Pressure = 67.5 B: Catalys t Con. 8. 7. 6. Viscosity 575 55 55 5 4 475 5. 45 45 4.. 5.. 5.. A: Temperature Decrease temperature and catalyst. (c) What operating conditions would you recommend if it was necessary to produce a product with a molecular weight between 4 and 5, and the lowest possible viscosity? 6-48

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Overlay Plot X = A: Temperature Y = B: Catalyst Con. 8. Overlay Plot Actual Factors C: T ime = 4.5 D: Pressure = 67.5 7. B: Catalys t Con. 6. 5. Molecular Wt: 5 Viscosity: 45 4.. 5.. 5.. A: Temperature Set the temperature between and 5, the catalyst between 4 and 5%, and the time at 4.5 minutes. The pressure was not significant and can be set at conditions that may improve other results of the process such as cost. 6-8 Consider the single replicate of the 4 design in Example 6-. Suppose that we ran five points at the center (,,,) and observed the following responses: 73, 75, 7, 69, and 76. Test for curvature in this experiment. Interpret the results. Design Expert Output Response: Filtration Rate ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5535.8 5 7.6 68. <. significant A 87.56 87.56 4.9 <. C 39.6 39.6 3.96. D 855.56 855.56 5.55 <. AC 34.6 34.6 8.7 <. AD 5.56 5.56 67.9 <. Curvature 8.55 8.55.75.66 not significant Residual 7.93 4 6.8 Lack of Fit 95.3 9.5.38.93 not significant Pure Error 3.8 4 8. Cor Total 579.9 The Model F-value of 68. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, D, AC, AD are significant model terms. The "Curvature F-value" of.75 implies the curvature (as measured by difference between the average of the center points and the average of the factorial points) in the design space is not significant relative to the noise. There is a.66% chance that a "Curvature F-value" this large could occur due to noise. There is no indication of curvature. 6-49

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6-9 A missing value in a k factorial. It is not unusual to find that one of the observations in a k design is missing due to faulty measuring equipment, a spoiled test, or some other reason. If the design is replicated n times (n>) some of the techniques discussed in Chapter 4 can be employed, including estimating the missing observations. However, for an unreplicated factorial (n-) some other method must be used. One logical approach is to estimate the missing value with a number that makes the highestorder interaction contrast zero. Apply this technique to the experiment in Example 6- assuming that run ab is missing. Compare the results with the results of Example 6-. Treatment Com bination R esponse R esponse * ABCD ABCD A B C D () 45 45 - - - - a 7-7 - - - - b 48-48 - - - - ab missing m issing * - - c 68-68 - - - - ac 6 6 - - bc 8 8 - - abc 65-65 - - d 43-43 - - - - ad - - bd 45 45 - - abd 4-4 - - cd 75 75 - - acd 86-86 - - bcd 7-7 - - abcd 96 96 Contrast(ABCD )= missing -54 = missing = 54 Substitute the value 54 for the missing run at ab. Term Effect SumSqr % Contribtn Model Intercept Model A.5 64.5 7.546 Model B.75.5.5684 Model C.5 56.5 8.59 Model D 6 4 7.935 Model AB -.5 6.5.494 Model AC -6.75.5 8.843 Model AD 8 96.765 Model BC 3.75 56.5.944465 Model BD 4.676 Model CD -.5 5.4976 Model ABC 3.5 4.5.79398 Model ABD 5.5.365 Model ACD -3 36.64458 Model BCD -4 64.7459 Model ABCD Lenth's ME.5676 Lenth's SME 3.4839 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Filtration Rate Normal plot A: Temperature B: Pressure C: Concentration D: Stirring Rate Normal % probability 99 95 9 8 7 5 3 5 AC C A AD D -6.75-7.5.75..5 Effect 6-3 An engineer has performed an experiment to study the effect of four factors on the surface roughness of a machined part. The factors (and their levels) are A = tool angle ( degrees, 5 degrees), B = cutting fluid viscosity (3, 4), C = feed rate ( in/min, 5 in/min), and D = cutting fluid cooler used (no, yes). The data from this experiment (with the factors coded to the usual -, + levels) are shown below. Run A B C D Surface Roughness - - - -.34 + - - -.36 3 - + - -.3 4 + + - -.8 5 - - + -.8 6 + - + -.9 7 - + + -.5 8 + + + -.6 9 - - - +.336 + - - +.344 - + - +.38 + + - +.84 3 - - + +.69 4 + - + +.84 5 - + + +.53 6 + + + +.63 (a) Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model. 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Surface Roughness Normal plot A: Tool Angle B: Viscosity C: Feed Rate D: Cutting Fluid 99 95 Normal % probability 9 8 7 5 3 5 B AB A C -. 9 -. 6 -. 4 -.. Effect (b) Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy? Design Expert Output Response: Surface Roughness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.46E-6 4.6E-6 4.97 <. significant A 8.556E-7 8.556E-7 6.43 <. B 3.8E-6 3.8E-6. <. C.3E-6.3E-6 73.96 <. AB.44E-6.44E-6 3.38 <. Residual.53E-7.393E-8 Cor Total 6.559E-6 5 The Model F-value of 4.97 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.665 99 Normal % probability 95 9 8 7 5 3 5 8.565E-5 Residuals 5E-6-7.5 6 5 E- 5 -.565 -.565-7.565E-5 5E-6 8.565E-5.665.5..5.3.35 Residual Predicted Residuals vs. Tool Angle.665 8.565E-5 Residuals 5E-6-7.565E-5 -.565 3 4 5 Tool Angle The plot of residuals versus predicted shows a slight u-shaped appearance in the residuals, and the plot of residuals versus tool angle shows an outward-opening funnel. (c) Repeat the analysis from parts (a) and (b) using /y as the response variable. Is there and indication that the transformation has been useful? The plots of the residuals are more representative of a model that does not violate the constant variance assumption. 6-53

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot./(Surface Roughness) Normal plot A: Tool Angle B: Viscosity C: Feed Rate D: Cutting Fluid Normal % probability 99 95 9 8 7 5 3 C AB A B 5-8.3 3.4 7.59.4 49.49 Effect Design Expert Output Response: Surface RoughnessTransform: Inverse ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.59E+5 4 547.8 455.7 <. significant A 46.9 46.9 5. <. B 89386.7 89386.7 57.99 <. C 876.9 876.9 53.63 <. AB 559.6 559.6 559.6 <. Residual 388.94 35.36 Cor Total.63E+55 The Model F-value of 455.7 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. DESIGN-EXPERT Plot./(Surface Roughness) Residuals vs. Predicted DESIGN-EXPERT Plot./(Surface Roughness) Residuals vs. Tool Angle 8.973 8.973 4.944 4.944 Residuals.85679 Residuals.85679-3.46-3. 4 6-7.577-7. 57 7 8.73 365.57 449.4 533.6 67. 3 4 5 Predicted Tool Angle 6-54

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (d) Fit a model in terms of the coded variables that can be used to predict the surface roughness. Convert this prediction equation into a model in the natural variables. Design Expert Output Final Equation in Terms of Coded Factors:./(Surface Roughness) = +397.8 +5.6 * A +74.74 * B +34.4 * C +58.7 * A * B 6-3 Resistivity on a silicon wafer is influenced by several factors. The results of a 4 factorial experiment performed during a critical process step is shown below. Run A B C D Resistivity - - - -.9 + - - -.8 3 - + - -.9 4 + + - - 5.75 5 - - + -.3 6 + - + - 9.53 7 - + + -.3 8 + + + - 5.35 9 - - - +.6 + - - +.73 - + - +.6 + + - + 4.68 3 - - + +.6 4 + - + + 9. 5 - + + +.7 6 + + + + 5.3 (a) Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model. 6-55

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Re si stivi ty Normal plot A: A B: B C: C D: D Normal % probability 99 95 9 8 7 5 3 5 B AB A -3. -.6 7.66 3.99 6.3 Effect (b) Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy? The normal probability plot of residuals is not satisfactory. The plots of residual versus predicted, residual versus factor A, and the residual versus factor B are funnel shaped indicating non-constant variance. Design Expert Output Response: Resistivity ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4. 3 7.4 48.8 <. significant A 59.83 59.83 333.9 <. B 36.9 36.9 75. <. AB 8.3 8.3 38.3 <. Residual 5.76.48 Cor Total 9.98 5 The Model F-value of 48.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. 6-56

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.375 99 Normal % probability 95 9 8 7 5 3 5 Residuals.665.75 -.6 47 5 -.3 5 -.3 5 -.6 47 5.75.665.375.9 3.4 5.75 8.8.4 Residual Predicted Residuals vs. A Residuals vs. B.375.375.665.665 Residuals.75 Residuals.75 -.6 47 5 -.6 47 5 -.3 5 -.3 5 - - A B (c) Repeat the analysis from parts (a) and (b) using ln(y) as the response variable. Is there any indication that the transformation has been useful? 6-57

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot L n (Re si stivi ty) Normal plot A: A B: B C: C D: D Normal % probability 99 95 9 8 7 5 3 5 B A -.6 3 -. 6.5.6.63 Effect Design Expert Output Response: Resistivity Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.5 6.8 553.44 <. significant A.57.57 96.95 <. B.58.58 43.94 <. Residual.4 3. Cor Total.3 5 The Model F-value of 553.44 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. The transformed data no longer indicates that the AB interaction is significant. A simpler model has resulted from the log transformation. Normal plot of residuals Residuals vs. Predicted.49585 99 Norm al % probability 95 9 8 7 5 3 5.579833 Residuals -. 33 6 8 -. 5 9 -. 6 8 -.68 -.59 -.3368.579833.49585.6.6.9.75.3 Residual Predicted 6-58

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. A Residuals vs. B.49585.49585.579833.579833 Residuals -.3368 Residuals -. 33 6 8 -.59 -. 5 9 -.68 -. 6 8 - - A B The residual plots are much improved. (d) Fit a model in terms of the coded variables that can be used to predict the resistivity. Design Expert Output Final Equation in Terms of Coded Factors: Ln(Resistivity) = +.9 +.8 * A -.3 * B 6.3 Continuation of Problem 6-3. Suppose that the experiment had also run four center points along with the 6 runs in Problem 6-3. The resistivity measurements at the center points are: 8.5, 7.63, 8.95, 6.48. Analyze the experiment again incorporating the center points. What conclusions can you draw now? DESIGN-EXPERT Plot Re si stivi ty Normal plot A: A B: B C: C D: D Norm al % probability 99 95 9 8 7 5 3 5 B AB A -3. -.6 7.66 3.99 6.3 Effect 6-59

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Resistivity ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4. 3 7.4 9.35 <. significant A 59.83 59.83 67.4 <. B 36.9 36.9 6.3 <. AB 8.3 8.3 3.58 <. Curvature 3.9 3.9 5.3 <. significant Residual 8.97 5.6 Lack of Fit 5.76.48.45.863 not significant Pure Error 3. 3.7 Cor Total 54.38 9 The Model F-value of 9.35 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. Normal plot of residuals Residuals vs. Predicted.375 99 Norm al % probability 95 9 8 7 5 3 5 Residuals.6575 -. 5 -.6 6 5 -.3 5 -.3 5 -.6 6 5 -. 5.6575.375.9 3.4 5.75 8.8.4 Residual Predicted Repeated analysis with the natural log transformation. 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot L n (Re si stivi ty) Normal plot A: A B: B C: C D: D Normal % probability 99 95 9 8 7 5 3 5 B A -.6 3 -. 6.5.6.63 Effect Design Expert Output Response: Resistivity Transform: Natural log Constant:. ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.5 6.8 49.37 <. significant A.57.57 853. <. B.58.58 7.54 <. Curvature.38.38 9.98 <. significant Residual. 6. Lack of Fit.4 3..59.78 not significant Pure Error.56 3.9 Cor Total 4.73 9 The Model F-value of 49.37 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. The "Curvature F-value" of 9.98 implies there is significant curvature (as measured by difference between the average of the center points and the average of the factorial points) in the design space. There is only a.% chance that a "Curvature F-value" this large could occur due to noise. The curvature test indicates that the model has significant pure quadratic curvature. 6.33 Often the fitted regression model from a k factorial design is used to make predictions at points of interest in the design space. (a) Find the variance of the predicted response ŷ at the point x, x,, x k in the design space. Hint: Remember that the x s are coded variables, and assume a k design with an equal number of replicates n at each design point so that the variance of a regression coefficient ˆ is covariance between any pair of regression coefficients is zero. Let s assume that the model can be written as follows: k n and that the 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY yˆ( x )= ˆ ˆ x ˆ x... ˆ pxp where x [ x, x,..., x k ] are the values of the original variables in the design at the point of interest where a prediction is required, and the variables in the model x, x,..., xp potentially include interaction terms among the original k variables. Now the variance of the predicted response is V[ yˆ ( x)] V( ˆ ˆ ˆ ˆ xx... pxp) V( ˆ ) V( ˆ x ) V( ˆ x )... V( ˆ x ) n p k xi i This result follows because the design is orthogonal and all model parameter estimates have the same variance. Remember that some of the x s involved in this equation are potentially interaction terms. (b) Use the result of part (a) to find an equation for a (-)% confidence interval on the true mean response at the point x, x,, x k in the design space. The confidence interval is /, dfe /, df E yˆ( x) t V[ yˆ( x)] y( x) yˆ( x) t V[ yˆ( x )] p p where df E is the number of degrees of freedom used to estimate and the estimate of has been used in computing the variance of the predicted value of the response at the point of interest. 6.34 Hierarchical Models. Several times we have utilized the hierarchy principal in selecting a model; that is, we have included non-significant terms in a model because they were factors involved in significant higher-order terms. Hierarchy is certainly not an absolute principle that must be followed in all cases. To illustrate, consider the model resulting from Problem 6-, which required that a nonsignificant main effect be included to achieve hierarchy. Using the data from Problem 6-: (a) Fit both the hierarchical model and the non-hierarchical model. Design Expert Output for Hierarchial Model Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59.67 4 379.9.54 <. significant A.67.67..8836 B 77.67 77.67 5.44 <. C 8.7 8.7 9.5.67 AC 468.7 468.7 5.45.9 Residual 575.67 9 3.3 Lack of Fit 93. 3 3..3.467 not significant Pure Error 48.67 6 3.7 Cor Total 95.33 3 The Model F-value of.54 implies the model is significant. There is only 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, C, AC are significant model terms. Std. Dev. 5.5 R-Squared.753 Mean 4.83 Adj R-Squared.6674 C.V. 3.48 Pred R-Squared.566 PRESS 98.5 Adeq Precision.747 The "Pred R-Squared" of.566 is in reasonable agreement with the "Adj R-Squared" of.6674. A difference greater than. between the "Pred R-Squared" and the "Adj R-Squared" indicates a possible problem with your model and/or data. Design Expert Output for Non-Hierarchical Model Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59. 3 56.33 7.57 <. significant B 77.67 77.67 6.74 <. C 8.7 8.7 9.7.54 AC 468.7 468.7 6.5.7 Residual 576.33 8.8 Lack of Fit 93.67 4 3.4.78.5566 not significant Pure Error 48.67 6 3.7 Cor Total 95.33 3 The Model F-value of 7.57 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, C, AC are significant model terms. The "Lack of Fit F-value" of.78 implies the Lack of Fit is not significant relative to the pure error. There is a 55.66% chance that a "Lack of Fit F-value" this large could occur due to noise. Non-significant lack of fit is good -- we want the model to fit. Std. Dev. 5.37 R-Squared.749 Mean 4.83 Adj R-Squared.6837 C.V. 3.5 Pred R-Squared.639 PRESS 89.9 Adeq Precision.3 The "Pred R-Squared" of.639 is in reasonable agreement with the "Adj R-Squared" of.6837. A difference greater than. between the "Pred R-Squared" and the "Adj R-Squared" indicates a possible problem with your model and/or data. (b) Calculate the PRESS statistic, the adjusted R and the mean square error for both models. The PRESS and R are in the Design Expert Output above. The PRESS is smaller for the nonhierarchical model than the hierarchical model suggesting that the non-hierarchical model is a better predictor. (c) Find a 95 percent confidence interval on the estimate of the mean response at a cube corner ( x = x = x 3 = ). Hint: Use the result of Problem 6-33. Design Expert Output Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Life 7.45.8.9 3.99 5.79 5.37 39.54 Life 36.7.9 3.6 4.74 5.8 4.7 48.6 Life 38.67.9 34. 43.4 5.8 6.57 5.76 Life 47.5.9 4.93 5.7 5.8 35.4 59.59 Life 43..9 38.43 47.57 5.8 3.9 55.9 Life 34.7.9 9.6 38.74 5.8.7 46.6 6-63

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Life 54.33.9 49.76 58.9 5.8 4.4 66.43 Life 45.5.9 4.93 5.7 5.8 33.4 57.59 (d) Based on the analyses you have conducted, which model would you prefer? Notice that PRESS is smaller and the adjusted R is larger for the non-hierarchical model. This is an indication that strict adherence to the hierarchy principle isn t always necessary. Note also that the confidence interval is shorter for the non-hierarchical model. 6-64

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 7 Blocking and Confounding in the k Factorial Design Solutions 7- Consider the experiment described in Problem 6-. Analyze this experiment assuming that each replicate represents a block of a single production shift. Source of Sum of Degrees of Mean Variation Squares Freedom Square F Cutting Speed (A).67.67 < Tool Geometry (B) 77.67 77.67.38* Cutting Angle (C) 8.7 8.7 8.4* AB 6.67 6.67 < AC 468.7 468.7 3.6* BC 48.7 48.7.4 ABC 8.7 8.7 < Blocks.58.9 Error 48.8 4 34.43 Total 95.33 3 Design Expert Output Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.58.9 Model 59.67 4 379.9.3. significant A.67.67..89 B 77.67 77.67.78. C 8.7 8.7 8.8.4 AC 468.7 468.7 3.84.7 Residual 575.8 7 33.83 Cor Total 95.33 3 The Model F-value of.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, C, AC are significant model terms. These results agree with the results from Problem 6-. Tool geometry, cutting angle and the cutting speed x cutting angle factors are significant at the 5% level. The Design Expert program also includes A, speed, in the model to preserve hierarchy. 7- Consider the experiment described in Problem 6-5. Analyze this experiment assuming that each one of the four replicates represents a block. Source of Sum of Degrees of Mean Variation Squares Freedom Square F Bit Size (A) 7.3 7.3 364.* 7-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Cutting Speed (B) 7.6 7.6 74.76* AB 33.63 33.63 99.88* Blocks 44.36 3 4.79 Error 7.36 9 3.4 Total 79.83 5 These results agree with those from Problem 6-5. Bit size, cutting speed and their interaction are significant at the % level. Design Expert Output Response: Vibration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 44.36 3 4.79 Model 638. 3 546.4 79.6 <. significant A 7.3 7.3 364. <. B 7.6 7.6 74.75 <. AB 33.63 33.63 99.88 <. Residual 7.36 9 3.4 Cor Total 79.83 5 The Model F-value of 79.6 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. 7-3 Consider the alloy cracking experiment described in Problem 6-5. Suppose that only 6 runs could be made on a single day, so each replicate was treated as a block. Analyze the experiment and draw conclusions. The analysis of variance for the full model is as follows: Design Expert Output Response: Crack Lengthin mm x ^- ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.6.6 Model 57.95 5 38.6 445. <. significant A 7.9 7.9 85.59 <. B 6.46 6.46 478.83 <. C 3.46 3.46 9.9 <. D 3.66 3.66 358.56 <. AB 9.93 9.93 349.96 <. AC 8.5 8.5 5.63 <. AD.47.47.55.478 BC.74.74.86.3678 BD.8.8..654 CD.47.47.55.4686 ABC 78.75 78.75 9.9 <. ABD.77.77.9.358 ACD.96E-3.96E-3.34.8557 BCD....735 ABCD.596E-3.596E-3.9.893 Residual.8 5.86 Cor Total 57.5 3 The Model F-value of 445. implies the model is significant. There is only 7-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, D, AB, AC, ABC are significant model terms. The analysis of variance for the reduced model based on the significant factors is shown below. The BC interaction was included to preserve hierarchy. Design Expert Output Response: Crack Lengthin mm x ^- ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.6.6 Model 57.74 8 7.34 56. <. significant A 7.9 7.9 79.8 <. B 6.46 6.46 87. <. C 3.46 3.46 53.59 <. D 3.66 3.66 453.9 <. AB 9.93 9.93 443. <. AC 8.5 8.5 9.5 <. BC.74.74.9.375 ABC 78.75 78.75 65.76 <. Residual.49.68 Cor Total 57.5 3 The Model F-value of 56. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, D, AB, AC, ABC are significant model terms. Blocking does not change the results of Problem 6-5. 7-4 Consider the data from the first replicate of Problem 6-. Suppose that these observations could not all be run using the same bar stock. Set up a design to run these observations in two blocks of four observations each with ABC confounded. Analyze the data. Block Block () a ab b ac c bc abc From the normal probability plot of effects, B, C, and the AC interaction are significant. Factor A was included in the analysis of variance to preserve hierarchy. 7-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Life Normal plot A: Cutting Speed B: Tool Geometry C: Cutting Angle 99 95 9 B Normal % probability 8 7 5 3 5 AC C -3.75-7. 3 -.5 6.3.75 Effect Design Expert Output Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 9.3 9.3 Model 896.5 4 4.3 7.3.38 not significant A 3.3 3.3..7797 B 35. 35..6.87 C 9. 9. 6..33 AC 378.3 378.3.35.73 Residual 6.5 3.6 Cor Total 48.88 7 The "Model F-value" of 7.3 implies the model is not significant relative to the noise. There is a.38 % chance that a "Model F-value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case there are no significant model terms. This design identifies the same significant factors as Problem 6-. 7-5 Consider the data from the first replicate of Problem 6-7. Construct a design with two blocks of eight observations each with ABCD confounded. Analyze the data. Block Block () a ab b ac c bc d ad abc bd abd cd acd abcd bcd 7-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The significant effects are identified in the normal probability plot of effects below: DESIGN-EXPERT Plot yield Normal plot A: A B: B C: C D: D 99 95 D Normal % probability 9 8 7 5 3 5 A AD ABC ABD AB -. -6. 5 -.5.5 5. Effect AC, BC, and BD were included in the model to preserve hierarchy. Design Expert Output Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 4.5 4.5 Model 89.5 8. 9.64.438 significant A 4. 4. 47.5.63 B.5.5.7.648 C.5.5.7.648 D...88.4 AB 8. 8. 9.6.53 AC....753 AD 56.5 56.5 6.68.84 BC 6.5 6.5.74.45 BD 9. 9..7.377 ABC 44. 44. 7..56 ABD 9.5 9.5.7.466 Residual 5.5 3 8.4 Cor Total 959.75 5 The Model F-value of 9.64 implies the model is significant. There is only a 4.38% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, D, ABC, ABD are significant model terms. 7-6 Repeat Problem 7-5 assuming that four blocks are required. Confound ABD and ABC (and consequently CD) with blocks. Block Block Block 3 Block 4 () ac c a ab bc abc b acd d ad cd 7-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY bcd abd bd abcd DESIGN-EXPERT Plot yield Normal plot A: A B: B C: C D: D Normal % probability 99 95 9 8 7 5 3 5 A D AB ABCD -. -6. 5 -.5.5 5. Effect Design Expert Output Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 43.5 3 8.8 Model 63.5 4 55.8 3.37.3 significant A 4. 4. 34.3.4 D.. 8.58.9 AB 8. 8. 6.95.99 ABCD 4.5 4.5 3.6.934 Residual 93.5 8.66 Cor Total 959.75 5 The Model F-value of 3.37 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, D, AB are significant model terms. 7-7 Using the data from the 5 design in Problem 6-, construct and analyze a design in two blocks with ABCDE confounded with blocks. Block Block Block Block () ae a e ab be b abe ac ce c ace bc abce abc bce ad de d ade bd abde abd bde cd acde acd cde abcd bcde bcd abcde 7-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The normal probability plot of effects identifies factors A, B, C, and the AB interaction as being significant. This is confirmed with the analysis of variance. DESIGN-EXPERT Plot Yield Normal plot A: Aperture B: Exposure Time C: Develop Time D: M ask Dim ension E: Etch Time Normal % probability 99 95 9 8 7 5 3 5 C AB A B -. 9 7.59 6.38 5.6 33.94 Effect Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.8.8 Model 585.3 4 896.8 958.5 <. significant A 6.8 6.8 369.43 <. B 94.3 94.3 349.35 <. C 75.78 75.78 48.47 <. AB 54.3 54.3 66.8 <. Residual 78.56 6 3. Cor Total 663.97 3 The Model F-value of 958.5 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. 7-8 Repeat Problem 7-7 assuming that four blocks are necessary. Suggest a reasonable confounding scheme. Use ABC, CDE, confounded with ABDE. The four blocks follow. Block Block Block 3 Block 4 () a ac c ab b bc abc acd cd d ad bcd abcd abd bd ace ce e ae bce abce abe be 7-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY de ade acde cde abde bde bcde abcde The normal probability plot of effects identifies the same significant effects as in Problem 7-7. DESIGN-EXPERT Plot Yield Normal plot A: Aperture B: Exposure Time C: Develop Time D: M ask Dim ension E: Etch Time Normal % probability 99 95 9 8 7 5 3 C AB A B 5 -. 9 7.59 6.38 5.6 33.94 Effect Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 3.84 3 4.6 Model 585.3 4 896.8 69.4 <. significant A 6.8 6.8 4.7 <. B 94.3 94.3 34. <. C 75.78 75.78 77. <. AB 54.3 54.3 86. <. Residual 65. 4.7 Cor Total 663.97 3 The Model F-value of 69.4 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. 7-9 Consider the data from the 5 design in Problem 6-. Suppose that it was necessary to run this design in four blocks with ACDE and BCD (and consequently ABE) confounded. Analyze the data from this design. Block Block Block 3 Block 4 () a b c ae e abe ace cd acd bcd d abc bc ac ab acde cde abcde ade 7-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY bce abce ce be abd bd ab abcd bde abde de bcde Even with four blocks, the same effects are identified as significant per the normal probability plot and analysis of variance below: DESIGN-EXPERT Plot Yield Normal plot A: Aperture B: Exposure Time C: Develop Time D: M ask Dim ension E: Etch Time Normal % probability 99 95 9 8 7 5 3 C AB A B 5 -. 9 7.59 6.37 5.6 33.94 Effect Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.59 3.86 Model 585.3 4 896.8 9.6 <. significant A 6.8 6.8 35.35 <. B 94.3 94.3 9.5 <. C 75.78 75.78 36.3 <. AB 54.3 54.3 58.65 <. Residual 76.5 4 3.8 Cor Total 663.97 3 The Model F-value of 9.6 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, AB are significant model terms. 7- Design an experiment for confounding a 6 factorial in four blocks. Suggest an appropriate confounding scheme, different from the one shown in Table 7-8. We choose ABCE and ABDF. Which also confounds with CDEF Block Block Block 3 Block 4 a c ac () b abc bc ab 7-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY cd ad d acd abcd bd abd bcd ace e ae ce bce abe be abce de acde cde ade abde bcde abcde bde cf af f acf abcf bf abf bcf adf cdf acdf df bdf abcdf bcdf abdf ef acef cef aef abef bcef abcef bef acdef def adef cdef bcdef abdef bdef abcdef 7- Consider the 6 design in eight blocks of eight runs each with ABCD, ACE, and ABEF as the independent effects chosen to be confounded with blocks. Generate the design. Find the other effects confound with blocks. Block Block Block 3 Block 4 Block 5 Block 6 Block 7 Block 8 b abc a c ac () bc ab acd d bcd abd bd abcd ad cd ce ae abce be abe bce e ace abde bcde de acde cde ade abcde bde abcf bf cf af f acf abf bcf de acdf abdf bcdf abcdf bdf cdf adf aef cef def abcef bcef abef acef ef bcdef abdef acdef def adef cdef bdef abcdef The factors that are confounded with blocks are ABCD, ABEF, ACE, BDE, CDEF, BCF, and ADF. 7- Consider the design in two blocks with AB confounded. Prove algebraically that SS AB = SS Blocks. If AB is confounded, the two blocks are: SS SS Blocks Blocks ab a b ab ab a Block Block () a ab b () + ab a + b ab a b 4 b ab 7-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY SS SS Blocks Blocks ab a b ab a b aab bab ab a b ab ab a b aab bab 4 ab a b SS AB 4 4 ab 7-3 Consider the data in Example 7-. Suppose that all the observations in block are increased by. Analyze the data that would result. Estimate the block effect. Can you explain its magnitude? Do blocks now appear to be an important factor? Are any other effect estimates impacted by the change you made in the data? 46 75 39 Block Effect yblock yblock 38. 65 8 8 8 This is the block effect estimated in Example 7- plus the additional units that were added to each observation in block. All other effects are the same. Source of Sum of Degrees of Mean Variation Squares Freedom Square F A 87.56 87.56 89.93 C 39.6 39.6 8.75 D 855.56 855.56 4.3 AC 34.6 34.6 63.8 AD 5.56 5.56 53.5 Blocks 5967.56 5967.56 Error 87.56 9.8 Total 69.93 5 Design Expert Output Response: Filtration in gal/hr ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 5967.56 5967.56 Model 5535.8 5 7.6 53.3 <. significant A 87.56 87.56 89.76 <. C 39.6 39.6 8.7.9 D 855.56 855.56 4.5. AC 34.6 34.6 63.5 <. AD 5.56 5.56 53.5 <. Residual 87.56 9.84 Cor Total 69.94 5 The Model F-value of 53.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, D, AC, AD are significant model terms. 7-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 7-4 Suppose that the data in Problem 7- we had confounded ABC in replicate I, AB in replicate II, and BC in replicate III. Construct the analysis of variance table. Replicate I Replicate II Replicate III (ABC Confounded) (AB Confounded) (BC Confounded) Block-> () a () a () b ab b ab b bc c ac c abc ac abc ab bc abc c bc a ac Source of Sum of Degrees of Mean Variation Squares Freedom Square F A.67.67 < B 77.67 77.67.77 C 8.7 8.7 7.55 AB (reps and III) 5. 5. < AC 468.7 468.7.6 BC (reps I and II).56.56 < ABC (reps II and III).6.6 < Blocks within replicates 9.83 3 5.87 Replicates.58 Error 48. 37. Total 95.33 3 7-5 Repeat Problem 7- assuming that ABC was confounded with blocks in each replicate. Replicate I, II, and III (ABC Confounded) Block-> () a ab b ac c bc abc Source of Sum of Degrees of Mean Variation Squares Freedom Square F A.67.67 < B 77.67 77.67.5 C 8.7 8.7 8.5 AB 6.67 6.67 < AC 468.7 468.7 3.46 BC 48.7 48.7.38 Blocks (or ABC) 9.83 9.83 Replicates/Lack of Fit 64.83 4 7-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error 47.5 34.79 Total 95.33 3 7-6 Suppose that in Problem 7-7 ABCD was confounded in replicate I and ABC was confounded in replicate II. Perform the statistical analysis of variance. Source of Sum of Degrees of Mean Variation Squares Freedom Square F A 657.3 657.3 84.89 B 3.78 3.78.78 C 57.78 57.78 7.46 D 4.3 4.3 6. AB 3.3 3.3 7.6 AC 3.78 3.78 < AD 38.8 38.8 4.95 BC.53.53 < BD.8.8 < CD.78.78.94 ABC 44. 44. 8.64 ABD 75.78 75.78.7 ACD 7.3 7.3 < BCD 7.3 7.3 < ABCD.56.56.36 Replicates.8.8 Blocks 8.8 5.35 Error.65 3 7.74 Total 67.47 3 7-7 Construct a 3 design with ABC confounded in the first two replicates and BC confounded in the third. Outline the analysis of variance and comment on the information obtained. Replicate I Replicate II Replicate III (ABC Confounded) (ABC Confounded) (BC Confounded) Block-> () a () a () b ab b ab b bc c ac c ac c abc ab bc abc bc abc a ac Source of Degrees of Variation Freedom A B C 7-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY AB AC BC ABC Replicates Blocks 3 Error Total 3 This design provides two-thirds information on BC and one-third information on ABC. 7-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 8 Two-Level Fractional Factorial Designs Solutions 8- Suppose that in the chemical process development experiment in Problem 6-7, it was only possible to run a one-half fraction of the 4 design. Construct the design and perform the statistical analysis, using the data from replicate. The required design is a 4- with I=ABCD. A B C D=ABC - - - - () 9 + - - + ad 7 - + - + bd 87 + + - - ab 83 - - + + cd 99 + - + - ac 8 - + + - bc 88 + + + + abcd 8 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A - 88 64.857 Model B -.44649 Model C 4 3 7.486 Model D -.44649 Model AB 6 7 6.74 Model AC -.44649 Model AD -5 5.67 Error BC Aliased Error BD Aliased Error CD Aliased Error ABC Aliased Error ABD Aliased Error ACD Aliased Error BCD Aliased Error ABCD Aliased Lenth's ME.5856 Lenth's SME 54.56 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 D B AD AB A. 3. 6. 9.. Effect The largest effect is A. The next largest effects are the AB and AD interactions. A plausible tentative model would be A, AB and AD, along with B and D to preserve hierarchy. Design Expert Output Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 44. 5 8.8 4.87.79 not significant A 88. 88. 6.94.543 B....7643 D....7643 AB 7. 7. 4.4.758 AD 5. 5..94.85 Residual 34. 7. Cor Total 448. 7 The "Model F-value" of 4.87 implies the model is not significant relative to the noise. There is a 7.9 % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 4. R-Squared.94 Mean 85. Adj R-Squared.7344 C.V. 4.85 Pred R-Squared -.43 PRESS 544. Adeq Precision 6.44 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 85..46 78.73 9.7 A-A -6..46 -.7.7. B-B -.5.46-6.77 5.77. D-D -.5.46-6.77 5.77. AB 3..46-3.7 9.7. AD -.5.46-8.77 3.77. Final Equation in Terms of Coded Factors: yield = +85. -6. * A -.5 * B -.5 * D +3. * A * B -.5 * A * D 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Final Equation in Terms of Actual Factors: yield = +85. -6. * A -.5 * B -.5 * D +3. * A * B -.5 * A * D The Design-Expert output indicates that we really only need the main effect of factor A. The updated analysis is shown below: Design Expert Output Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 88. 88..8.67 significant A 88. 88..8.67 Residual 6. 6 6.67 Cor Total 448. 7 The Model F-value of.8 implies the model is significant. There is only a.67% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 5.6 R-Squared.649 Mean 85. Adj R-Squared.5833 C.V. 6.8 Pred R-Squared.365 PRESS 84.44 Adeq Precision 4.648 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 85..83 8.53 89.47 A-A -6..83 -.47 -.53. Final Equation in Terms of Coded Factors: yield = +85. -6. * A Final Equation in Terms of Actual Factors: yield = +85. -6. * A 8- Suppose that in Problem 6-5, only a one-half fraction of the 4 design could be run. Construct the design and perform the analysis, using the data from replicate I. The required design is a 4- with I=ABCD. A B C D=ABC - - - - ().7 + - - + ad.86 - + - + bd.79 + + - - ab.67 - - + + cd.8 + - + - ac.5 - + + - bc.46 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY + + + + abcd.85 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A -.85.645 9.53 Error B -.5.945.884 Model C -.45.34445 4.55 Error D.55.65.767 Error AB -.8.8.5695 Model AC -.3.8.94 Error AD -.6.5 6.778 Error BC Aliased Error BD Aliased Error CD Aliased Error ABC Aliased Error ABD Aliased Error ACD Aliased Error BCD Aliased Error ABCD Aliased Lenth's ME.397 Lenth's SME.958 C, A and AC + BD are the largest three effects. Now because the main effects of A and C are large, the large effect estimate for the AC + BD alias chain probably indicates that the AC interaction is important. Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 AC A C....3.4 Effect Design Expert Output Response: Crack Lengthin mm x ^- ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.69 3.3 5.64.64 not significant A.6.6 4..6 C.34.34 8.48.436 AC.8.8 4.43.3 Residual.6 4.4 Cor Total.85 7 The Model F-value of 5.64 implies there is a 6.4% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.887 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Mean.55 Adj R-Squared.665 C.V. 3. Pred R-Squared.348 PRESS.65 Adeq Precision 5.7 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.55.7.35.75 A-Pour Temp -.4.7 -.34.55. C-Heat Tr Mtd -..7 -.4-9.648E-3. AC -.5.7 -.35.48. Final Equation in Terms of Coded Factors: Crack Length = +.55 -.4 * A -. * C -.5 * A * C Final Equation in Terms of Actual Factors: Crack Length = +.55 -.45 * Pour Temp -.75 * Heat Treat Method -.5 * Pour Temp * Heat Treat Method 8-3 Consider the plasma etch experiment described in Problem 6-8. Suppose that only a one-half fraction of the design could be run. Set up the design and analyze the data. Etch Rate Factor Levels A B C D=ABC (A/min) Low (-) High (+) - - - - 55 A (cm).8. + + - - 65 B (mtorr) 4.5 55 + - + - 64 C (SCCM) 5 - + + - 6 D (W) 75 35 + - - + 749 - + - + 5 - - + + 75 + + + + 79 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A 4 3.394 Error B.5 64.5.9479 Model C 9.5 6878 6.967 Model D -7 358.4859 Error AB -97.5 78.5 7.7775 Error AC -5.5 3.5.4636 Error AD -.79 Error BC Aliased Error BD Aliased Model CD Aliased Error ABC Aliased Error ABD Aliased Error ACD Aliased Error BCD Aliased Error ABCD Aliased Lenth's ME 6.6987 Lenth's SME 45.64 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 D AB C. 7.63 45.5 7.88 9.5 Effect The large AB + CD alias chain is most likely the CD interaction. Design Expert Output Response: Etch Rate in A/min ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.79E+5 3 937. 7.5 <. significant C.688E+5.688E+5 375.69 <. D 358. 358. 7.8. CD 78.5 78.5 73.65. Residual 797. 4 449.5 Cor Total.88E+5 7 The Model F-value of 7.5 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.9936 Mean 756. Adj R-Squared.9888 C.V..8 Pred R-Squared.9744 PRESS 788. Adeq Precision 3.56 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 756. 7.49 735.9 776.8 C-Gas Flow 45.5 7.49 4.44 66.6. D-Power -63.5 7.49-84.3-4.69. CD -98.75 7.49-9.56-77.94. Final Equation in Terms of Coded Factors: Etch Rate = +756. +45.5 * C -63.5 * D -98.75 * C * D Final Equation in Terms of Actual Factors: Etch Rate = -446.4667 +35.47333 * Gas Flow +4.57667 * Power -.533 * Gas Flow * Power 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 8-4 Problem 6-describes a process improvement study in the manufacturing process of an integrated circuit. Suppose that only eight runs could be made in this process. Set up an appropriate 5- design and find the alias structure. Use the appropriate observations from Problem 6- as the observations in this design and estimate the factor effects. What conclusions can you draw? I = ABD = ACE = BCDE A (ABD) =BD A (ACE) =CE A (BCDE) =ABCDE A=BD=CE=ABCDE B (ABD) =AD B (ACE) =ABCE B (BCDE) =CDE B=AD=ABCE=CDE C (ABD) =ABCD C (ACE) =AE C (BCDE) =BDE C=ABCD=AE=BDE D (ABD) =AB D (ACE) =ACDE D (BCDE) =BCE D=AB=ACDE=BCE E (ABD) =ABDE E (ACE) =AC E (BCDE) =BCD E=ABDE=AC=BCD BC (ABD) =ACD BC (ACE) =ABE BC (BCDE) =DE BC=ACD=ABE=DE BE (ABD) =ADE BE (ACE) =ABC BE (BCDE) =CD BE=ADE=ABC=CD A B C D=AB E=AC - - - + + de 6 + - - - - a 9 - + - - + be 35 + + - + - abd 5 - - + + - cd 8 + - + - + ace - + + - - bc 4 + + + + + abcde 63 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.5 53.5 8.9953 Model B 33.5.3 77.948 Model C.75 3.5 8.443 Model D 7.75.5 4.39 Error E.5.5.35678 Error BC -.75 6.5.583 Error BE.75 6.5.583 Lenth's ME 8.3 Lenth's SME 67.5646 Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 D C A B. 8.3 6.63 4.94 33.5 Effect 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The main A, B, C, and D are large. However, recall that you are really estimating A+BD+CE, B+AD, C+DE and D+AD. There are other possible interpretations of the experiment because of the aliasing. Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 85.5 4 73.88 94.37.7 significant A 53.3 53.3 33.94. B.. 96.46.4 C 3.3 3.3 3.99.4 D.3.3 6..78 Residual.38 3 7.46 Cor Total 837.88 7 The Model F-value of 94.37 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..73 R-Squared.99 Mean 3.38 Adj R-Squared.986 C.V. 8.99 Pred R-Squared.9439 PRESS 59. Adeq Precision 5.59 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 3.38.97 7.3 33.45 A-Aperture 5.63.97.55 8.7. B-Exposure Time 6.63.97 3.55 9.7. C-Develop Time 5.37.97.3 8.45. D-Mask Dimension 3.87.97.8 6.95. Final Equation in Terms of Coded Factors: Yield = +3.38 +5.63 * A +6.63 * B +5.37 * C +3.87 * D Final Equation in Terms of Actual Factors: Aperture small Mask Dimension Small Yield = -6. +.835 * Exposure Time +.7667 * Develop Time Aperture large Mask Dimension Small Yield = +5.5 +.835 * Exposure Time +.7667 * Develop Time Aperture small Mask Dimension Large Yield = +.75 +.835 * Exposure Time +.7667 * Develop Time Aperture large Mask Dimension Large Yield = +3. 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.835 * Exposure Time +.7667 * Develop Time 8-5 Continuation of Problem 8-4. Suppose you have made the eight runs in the 5- design in Problem 8-4. What additional runs would be required to identify the factor effects that are of interest? What are the alias relationships in the combined design? We could fold over the original design by changing the signs on the generators D = AB and E = AC to produce the following new experiment. A B C D=-AB E=-AC - - - - - () 7 + - - + + ade - + - + - bd 3 + + - - + abe 5 - - + - + ce 5 + - + + - acd - + + + + bcde 4 + + + - - abc 6 A (-ABD) =-BD A (-ACE) =-CE A (BCDE) =ABCDE A=-BD=-CE=ABCDE B (-ABD) =-AD B (-ACE) =-ABCE B (BCDE) =CDE B=-AD=-ABCE=CDE C (-ABD) =-ABCD C (-ACE) =-AE C (BCDE) =BDE C=-ABCD=-AE=BDE D (-ABD) =-AB D (-ACE) =-ACDE D (BCDE) =BCE D=-AB=-ACDE=BCE E (-ABD) =-ABDE E (-ACE) =-AC E (BCDE) =BCD E=-ABDE=-AC=BCD BC (-ABD) =-ACD BC (-ACE) =-ABE BC (BCDE) =DE BC=-ACD=-ABE=DE BE (-ABD) =-ADE BE (-ACE) =-ABC BE (BCDE) =CD BE=-ADE=-ABC=CD Assuming all three factor and higher interactions to be negligible, all main effects can be separated from their two-factor interaction aliases in the combined design. 8-6 R.D. Snee ( Experimenting with a Large Number of Variables, in Experiments in Industry: Design, Analysis and Interpretation of Results, by R.D. Snee, L.B. Hare, and J.B. Trout, Editors, ASQC, 985) describes an experiment in which a 5- design with I=ABCDE was used to investigate the effects of five factors on the color of a chemical product. The factors are A = solvent/reactant, B = catalyst/reactant, C = temperature, D = reactant purity, and E = reactant ph. The results obtained were as follows: e = -.63 d = 6.79 a =.5 ade = 5.47 b = -.68 bde = 3.45 abe =.66 abd = 5.68 c =.6 cde = 5. ace =. acd = 4.38 bce = -.9 bcd = 4.3 abc =.93 abcde = 4.5 (a) Prepare a normal probability plot of the effects. Which effects seem active? Factors A, B, D, and the AB, AD interactions appear to be active. 8-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Color Normal plot A: Solvent/Reactant B: Catalyst/Reactant C: Temperature D: Reactant Purity E: Reactant ph Normal % probability 99 95 9 8 7 5 3 5 B AD A AB D -.3 6.9.53.98 4.4 Effect Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.3 6.8644 5.98537 Model B -.34 7.84 6.665 Error C -.475.875.75889 Model D 4.4 78.456 68.386 Error E -.875.739.3888 Model AB.75 6.55 5.6698 Error AC -.7875.486.697 Model AD -.355 7.344 6.4364 Error AE.35.3665.3953 Error BC.675.5.978539 Error BD.45.4.9354 Error BE.875.3365.8886 Error CD -.75.363.7759 Error CE -.4.34.896 Error DE.875.365.6733 Lenth's ME.95686 Lenth's SME 3.977 Design Expert Output Response: Color ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.4 5. 4.53 <. significant A 6.86 6.86 7.94.8 B 7.8 7.8 8.3.63 D 78.5 78.5 9.37 <. AB 6.5 6.5 7.5.8 AD 7.34 7.34 8.49.55 Residual 8.65.86 Cor Total 4.69 5 The Model F-value of 4.53 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..93 R-Squared.946 Mean.7 Adj R-Squared.8869 C.V. 34.35 Pred R-Squared.87 PRESS.4 Adeq Precision 4.734 Coefficient Standard 95% CI 95% CI 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Estimate DF Error Low High VIF Intercept.7.3.9 3.3 A-Solvent/Reactant.66.3.4.7. B-Catalyst/Reactant -.67.3 -.9 -.5. D-Reactant Purity..3.69.73. AB.64.3..6. AD -.68.3 -. -.6. Final Equation in Terms of Coded Factors: Color = +.7 +.66 * A -.67 * B +. * D +.64 * A * B -.68 * A * D Final Equation in Terms of Actual Factors: Color = +.775 +.655 * Solvent/Reactant -.67 * Catalyst/Reactant +. * Reactant Purity +.6375 * Solvent/Reactant * Catalyst/Reactant -.6775 * Solvent/Reactant * Reactant Purity (b) Calculate the residuals. Construct a normal probability plot of the residuals and plot the residuals versus the fitted values. Comment on the plots. Design Expert Output Diagnostics Case Statistics Standard Actual Predicted Student Cook's Outlier Run Order Value Value Residual Leverage Residual Distance t Order -.63.47 -..375 -.5.5 -.66.5.86.65.375.88.78.87 6 3 -.68 -.4 -.54.375 -.73.53 -.73 4 4.66.8 -.4.375 -.87.3 -.78 5.6.47.59.375.59.466.84 8 6..86 -.64.375 -.874.76 -.863 5 7 -.9 -.4.53.375.7..68 8.93.8.3.375.8.3.7 3 9 6.79 6.5.54.375.738.54.7 4 5.47 4.93.54.375.738.54.7 5 3.45 3.63 -.8.375 -.48.6 -.36 6 5.68 4.86.8.375..4.7 3 5. 6.5 -.3.375 -.398.95 -.478 9 4 4.38 4.93 -.55.375 -.745.55 -.77 5 4.3 3.63.67.375.98.8.899 3 6 4.5 4.86 -.8.375 -.5. -.9 7 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.5875 99 Normal % probability 95 9 8 7 5 3 5 Residuals.95.45 -.4 3 -. 5 -. 5 -.4 3.45.95.5875 -. 4 -. 5.5 4.5 6.5 Residual Predicted The residual plots are satisfactory. (c) If any factors are negligible, collapse the 5- design into a full factorial in the active factors. Comment on the resulting design, and interpret the results. The design becomes two replicates of a 3 in the factors A, B and D. When re-analyzing the data in three factors, D becomes labeled as C. Design Expert Output Response: Color ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.5 7 5. 4.89.5 significant A 6.86 6.86 6.7.3 B 7.8 7.8 7.3.9 C 78.5 78.5 76.46 <. AB 6.5 6.5 6.36.357 AC 7.34 7.34 7.9.79 BC.4.4.3.649 ABC.3.3.3.6476 Residual 8.8 8. Lack of Fit. Pure Error 8.8 8. Cor Total 4.69 5 The Model F-value of 4.89 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.987 Mean.7 Adj R-Squared.8663 C.V. 37.34 Pred R-Squared.748 PRESS 3.7 Adeq Precision.736 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.7.5. 3.9 A-Solvent/Reactant.66.5.7.4. B-Catalyst/Reactant -.67.5 -.5 -.87. C-Reactant Purity..5.63.79. AB.64.5.55.. AC -.68.5 -.6 -.95. 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY BC..5 -.46.7. ABC -..5 -.7.46. Final Equation in Terms of Coded Factors: Color = +.7 +.66 * A -.67 * B +. * C +.64 * A * B -.68 * A * C +. * B * C -. * A * B * C Final Equation in Terms of Actual Factors: Color = +.775 +.655 * Solvent/Reactant -.67 * Catalyst/Reactant +. * Reactant Purity +.6375 * Solvent/Reactant * Catalyst/Reactant -.6775 * Solvent/Reactant * Reactant Purity +.5 * Catalyst/Reactant * Reactant Purity -. * Solvent/Reactant * Catalyst/Reactant * Reactant Purity 8-7 An article by J.J. Pignatiello, Jr. And J.S. Ramberg in the Journal of Quality Technology, (Vol. 7, 985, pp. 98-6) describes the use of a replicated fractional factorial to investigate the effects of five factors on the free height of leaf springs used in an automotive application. The factors are A = furnace temperature, B = heating time, C = transfer time, D = hold down time, and E = quench oil temperature. The data are shown below: A B C D E Free Height - - - - - 7.78 7.78 7.8 + - - + - 8.5 8.8 7.88 - + - + - 7.5 7.56 7.5 + + - - - 7.59 7.56 7.75 - - + + - 7.54 8. 7.88 + - + - - 7.69 8.9 8.6 - + + - - 7.56 7.5 7.44 + + + + - 7.56 7.8 7.69 - - - - + 7.5 7.5 7. + - - + + 7.88 7.88 7.44 - + - + + 7.5 7.56 7.5 + + - - + 7.63 7.75 7.56 - - + + + 7.3 7.44 7.44 + - + - + 7.56 7.69 7.6 - + + - + 7.8 7.8 7.5 + + + + + 7.8 7.5 7.59 (a) Write out the alias structure for this design. What is the resolution of this design? I=ABCD, Resolution IV A (ABCD)= BCD B (ABCD)= ACD C (ABCD)= ABD D (ABCD)= ABC E (ABCD)= ABCDE AB (ABCD)= CD AC (ABCD)= BD AD (ABCD)= BC 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY AE (ABCD)= BCDE BE (ABCD)= ACDE CE (ABCD)= ABDE DE (ABCD)= ABCE (b) Analyze the data. What factors influence the mean free height? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.483.735 4.374 Model B -.6375.3769.39 Model C -.495833.95.56 Model D.95.99988 3.45646 Model E -.3875.6849 3.66 Model AB -.95833.5.36396 Model AC.5.875E-5.64864 Model AD -.967.638.86 Model AE.6375.487687.6874 Error BC Aliased Error BD Aliased Model BE.597.86 9.7679 Error CD Aliased Model CE -.3967.3.449777 Model DE.395833.88.6545 Error Pure Error.6767.699 Lenth's ME.8857 Lenth's SME.35984 Normal plot 99 BE A 95 Normal % probability 9 8 7 5 3 5 E B -. 4 -....4 Effect Design Expert Output Response:Free Height ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.99 4.5 3.74 <. significant A.7.7 33.56 <. B.3.3 5.35.3 E.68.68 3.64 <. BE.8.8 3.39.7 Residual.9 43. Lack of Fit.7.5.7.844 not significant 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Pure Error.63 3. Cor Total.89 47 The Model F-value of 3.74 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..4 R-Squared.6883 Mean 7.63 Adj R-Squared.6593 C.V..9 Pred R-Squared.66 PRESS. Adeq Precision 3.796 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 7.63. 7.58 7.67 A-Furnace Temp...79.6. B-Heating Time -.8. -. -.4. E-Quench Temp -.. -.6 -.77. BE.76..34.. Final Equation in Terms of Coded Factors: Free Height = +7.63 +. * A -.8 * B -. * E +.76 * B * E Final Equation in Terms of Actual Factors: Free Height = +7.656 +.4 * Furnace Temp -.8875 * Heating Time -.937 * Quench Temp +.76458 * Heating Time * Quench Temp (c) Calculate the range and standard deviation of the free height for each run. Is there any indication that any of these factors affects variability in the free height? Design Expert Output (Range) Term Effect SumSqr % Contribtn Model Intercept Model A.375.57563 6.98 Model B -.65.637563 9.984 Model C.65.7565.863774 Error D.65.563 4.777 Model E -.375.7565.36999 Error AB.4375.76565.39937 Error AC -.3375.45565.4787 Error AD.365.5565.6474 Error AE -.375 5.65E-5.768 Model BC Aliased Error BD Aliased Model BE.65.565.336 Error CD Aliased Model CE -.365.7456 3.7 Error DE -.5.865.56656 Error ABC Aliased Error ABD Aliased Error ABE.35.3965.47 Error ACD Aliased Error ACE.4875.9565.9794 Error ADE.3875.776 4.38 Error BCD Aliased Model BCE Aliased Error BDE Aliased Error CDE Aliased Lenth's ME.336 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Lenth's SME.6494 Interaction ADE is aliased with BCE. Although the plot below identifies ADE, BCE was included in the analysis. DESIGN-EXPERT Plot Range Half Normal plot A: Furnace Temp B: Heating Time C: Transfer Time D: Hold Tim e E: Quench Temp Half Normal % probability 99 97 95 9 85 8 7 6 4 E C A B AD E CE..3.7..4 Effect Design Expert Output (Range) Response: Range ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 8.35 5.7.67 significant A.5.5 8.53.3 B.64.64.5.4 C.756E-3.756E-3.45.5 E 7.56E-4 7.56E-4..7345 BC 5.56E-3 5.56E-3.87.383 BE.56E-3.56E-3.7.689 CE.74.74.3. BCE.77.77.69.9 Residual.4 7 6.7E-3 Cor Total.3 5 The Model F-value of 5.7 implies the model is significant. There is only a.67% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..78 R-Squared.8668 Mean. Adj R-Squared.746 C.V. 35.5 Pred R-Squared.343 PRESS. Adeq Precision 7.66 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept..9.7.7 A-Furn Temp.57.9... B-Heat Time -.63.9 -. -.7. C-Transfer Time.3.9 -.33.59. E-Qnch Temp -6.875E-3.9 -.53.39. BC.8.9 -.8.64. BE 8.5E-3.9 -.38.54. CE -.68.9 -. -.. BCE.69.9.3.. Final Equation in Terms of Coded Factors: 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Range = +. +.57 * A -.63 * B +.3 * C -6.875E-3 * E +.8 * B * C +8.5E-3 * B * E -.68 * C * E +.69 * B * C * E Final Equation in Terms of Actual Factors: Range = +.937 +.56875 * Furnace Temp -.635 * Heating Time +.35 * Transfer Time -6.875E-3 * Quench Temp +.85 * Heating Time * Transfer Time +8.5E-3 * Heating Time * Quench Temp -.685 * Transfer Time * Quench Temp +.69375 * Heating Time * Transfer Time * Quench Temp Design Expert Output (StDev) Term Effect SumSqr % Contribtn Model Intercept Model A.65896.56698 6.873 Model B -.74887.445. Model C.567.446646.4894 Error D.35366.576 5.3858 Model E -.68434.876.53 Error AB.53974.94837.3 Error AC -.855.9978.564 Error AD.968.4536.56484 Error AE -.3935 4.3357E-5.46638 Model BC Aliased Error BD Aliased Model BE.87666.3743.337 Error CD Aliased Model CE -.7486.4385.78 Error DE -.46779 8.7537E-5.9455 Error ABC Aliased Error ABD Aliased Error ABE.55599.968437.48 Error ACD Aliased Error ACE.9974.59587.784 Error ADE Aliased Error BCD Aliased Model BCE.764346.3369 5.633 Error BDE Aliased Error CDE Aliased Lenth's ME.596836 Lenth's SME.66 Interaction ADE is aliased with BCE. Although the plot below identifies ADE, BCE was included in the analysis. 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot StDev Half Normal plot A: Furnace Temp B: Heating Time C: Transfer Time D: Hold Tim e E: Quench Temp Half Normal % probability 99 97 95 9 85 8 7 6 4 C BE A B CE AD E...4.6.8 Effect Design Expert Output (StDev) Response: StDev ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 8. 6.8. significant A.6.6.39.46 B.. 3.56.78 C 4.466E-4 4.466E-4.3.63 E.87E-4.87E-4..735 BC.453E-3.453E-3.96.3589 BE 3.74E-4 3.74E-4..6653 CE.. 3.55.78 BCE.3.3 5.5.56 Residual. 7.58E-3 Cor Total.93 5 The Model F-value of 6.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..39 R-Squared.8863 Mean. Adj R-Squared.7565 C.V. 33.7 Pred R-Squared.46 PRESS.55 Adeq Precision 7.86 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept. 9.78E-3.94.4 A-Furnace Temp.3 9.78E-3 8.34E-3.54. B-Heating Time -.36 9.78E-3 -.59 -.3. C-Transfer Time 5.83E-3 9.78E-3 -.8.8. E-Quench Temp -3.4E-3 9.78E-3 -.6.. BC 9.53E-3 9.78E-3 -.3.3. BE 4.383E-3 9.78E-3 -.9.7. CE -.36 9.78E-3 -.59 -.3. BCE.38 9.78E-3.5.6. Final Equation in Terms of Coded Factors: StDev = +. +.3 * A -.36 * B +5.83E-3 * C 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3.4E-3 * E +9.53E-3 * B * C +4.383E-3 * B * E -.36 * C * E +.38 * B * C * E Final Equation in Terms of Actual Factors: StDev = +.744 +.395 * Furnace Temp -.35744 * Heating Time +5.835E-3 * Transfer Time -3.47E-3 * Quench Temp +9.534E-3 * Heating Time * Transfer Time +4.3833E-3 * Heating Time * Quench Temp -.3574 * Transfer Time * Quench Temp +.387 * Heating Time * Transfer Time * Quench Temp (d) Analyze the residuals from this experiment, and comment on your findings. The residual plot follows. All plots are satisfactory. Normal plot of residuals Residuals vs. Predicted.479 99 Norm al % probability 95 9 8 7 5 3 5.875 Residuals -.43547 -. 88 9 5 8 -.3 34 3 7 5 -.334375 -.88958 -.43547.875.479 7.38 7.54 7.7 7.86 8. Residual Predicted 8-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Furnace Temp Residuals vs. Heating Time.479.479.875.875 Residuals -.43547 Residuals -.43547 -.88958 -. 88 9 5 8 -.334375 -.3 34 3 7 5 - - Furnace Temp Heating Time Residuals vs. Transfer Time Residuals vs. Hold Time.479.479.875.875 Residuals -.43547 Residuals -.43547 -.88958 -. 88 9 5 8 -.334375 -.3 34 3 7 5 - - Transfer Tim e Hold Time Residuals vs. Quench Temp.479.875 Residuals -.43547 -.88958 -.334375 - Quench Temp 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (e) Is this the best possible design for five factors in 6 runs? Specifically, can you find a fractional design for five factors in 6 runs with a higher resolution than this one? This was not the best design. A resolution V design is possible by setting the generator equal to the highest order interaction, ABCDE. 8-8 An article in Industrial and Engineering Chemistry ( More on Planning Experiments to Increase Research Efficiency, 97, pp. 6-65) uses a 5- design to investigate the effect of A = condensation, B = amount of material, C = solvent volume, D = condensation time, and E = amount of material on yield. The results obtained are as follows: e = 3. ad = 6.9 cd = 3.8 bde = 6.8 ab = 5.5 bc = 6. ace = 3.4 abcde = 8. (a) Verify that the design generators used were I = ACE and I = BDE. A B C D=BE E=AC - - - - + e + - - + - ad - + - + + bde + + - - - ab - - + + - cd + - + - + ace - + + - - bc + + + + + abcde (b) Write down the complete defining relation and the aliases for this design. I=BDE=ACE=ABCD. A (BDE) =ABDE A (ACE) =CE A (ABCD) =BCD A=ABDE=CE=BCD B (BDE) =DE B (ACE) =ABCE B (ABCD) =ACD B=DE=ABCE=ACD C (BDE) =BCDE C (ACE) =AE C (ABCD) =ABD C=BCDE=AE=ABD D (BDE) =BE D (ACE) =ACDE D (ABCD) =ABC D=BE=ACDE=ABC E (BDE) =BD E (ACE) =AC E (ABCD) =ABCDE E=BD=AC=ABCDE AB (BDE) =ADE AB (ACE) =BCE AB (ABCD) =CD AB=ADE=BCE=CD AD (BDE) =ABE AD (ACE) =CDE AD (ABCD) =BC AD=ABE=CDE=BC (c) Estimate the main effects. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A -.55 4.655 5.83 Model B -5.75 53.563 59.6858 Model C.75.35.5349 Model D -.675.95.545 Model E.75.353.5349 (d) Prepare an analysis of variance table. Verify that the AB and AD interactions are available to use as error. The analysis of variance table is shown below. Part (b) shows that AB and AD are aliased with other factors. If all two-factor and three factor interactions are negligible, then AB and AD could be pooled as an estimate of error. 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 79.83 5 5.97 3..537 not significant A 4.65 4.65.94.4349 B 53.56 53.56.8.84 C.35.35.9.853 D.9.9.8.798 E.35.35.9.853 Residual 9.9 4.96 Cor Total 89.74 7 The "Model F-value" of 3. implies the model is not significant relative to the noise. There is a 5.37 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..3 R-Squared.8895 Mean 9.4 Adj R-Squared.634 C.V..57 Pred R-Squared -.7674 PRESS 58.6 Adeq Precision 5.44 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 9.4.79 5.85.6 A-Condensation -.76.79-4.5.6. B-Material -.59.79-5.97.8. C-Solvent.4.79 -.5 4.5. D-Time -.34.79-3.7 3.5. E-Material.4.79 -.5 4.5. Final Equation in Terms of Coded Factors: Yield = +9.4 -.76 * A -.59 * B +.4 * C -.34 * D +.4 * E Final Equation in Terms of Actual Factors: Yield = +9.375 -.765 * Condensation -.5875 * Material +.375 * Solvent -.3375 * Time +.375 * Material (e) Plot the residuals versus the fitted values. Also construct a normal probability plot of the residuals. Comment on the results. The residual plots are satisfactory. 8-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals.55 99 Residuals.775.77636E-5 -.7 75 Normal % probability 95 9 8 7 5 3 5 -.5 5 3.95 6.38 8.8.4 3.68 -.5 5 -.7 75 -.77636E-5.775.55 Predicted Residual 8-9 Consider the leaf spring experiment in Problem 8-7. Suppose that factor E (quench oil temperature) is very difficult to control during manufacturing. Where would you set factors A, B, C and D to reduce variability in the free height as much as possible regardless of the quench oil temperature used? DESIGN-EXPERT Plot Free Height 8.8 Interaction Graph Heating Time X = E: Quench Temp Y = B: Heating Time 7.95 B- -. B+. Actual Factors A: Furnace Temp =. C: Transfer Time =. 7.65 D: Hold Time =. Free Height 7.385 7. -. -.5..5. Quench Tem p Run the process with A at the high level, B at the low level, C at the low level and D at either level (the low level of D may give a faster process). 8- Construct a 7- design by choosing two four-factor interactions as the independent generators. Write down the complete alias structure for this design. Outline the analysis of variance table. What is the resolution of this design? I=CDEF=ABCG=ABDEFG, Resolution IV A B C D E F=CDE G=ABC 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - - - - - - - () + - - - - - + ag 3 - + - - - - + bg 4 + + - - - - - ab 5 - - + - - + + cfg 6 + - + - - + - acf 7 - + + - - + - bcf 8 + + + - - + + abcfg 9 - - - + - + - df + - - + - + + adfg - + - + - + + bdfg + + - + - + - abdf 3 - - + + - - + cdg 4 + - + + - - - acd 5 - + + + - - - bcd 6 + + + + - - + abcdg 7 - - - - + + - ef 8 + - - - + + + aefg 9 - + - - + + + befg + + - - + + - abef - - + - + - + ceg + - + - + - - ace 3 - + + - + - - bce 4 + + + - + - + abceg 5 - - - + + - - de 6 + - - + + - + adeg 7 - + - + + - + bdeg 8 + + - + + - - abde 9 - - + + + + + cdefg 3 + - + + + + - acdef 3 - + + + + + - bcdef 3 + + + + + + + abcdefg Alias Structure A (CDEF)= ACDEF A(ABCG)= BCG A (ABDEFG)= BDEFG A=ACDEF=BCG=BDEFG B (CDEF)= BCDEF B(ABCG)= ACG B (ABDEFG)= ADEFG B=BCDEF=ACG=ADEFG C (CDEF)= DEF C(ABCG)= ABG C (ABDEFG)= ABCDEFG C=DEF=ABG=ABCDEFG D (CDEF)= CEF D(ABCG)= ABCDG D (ABDEFG)= ABEFG D=CEF=ABCDG=ABEFG E (CDEF)= CDF E(ABCG)= ABCEG E (ABDEFG)= ABDFG E=CDF=ABCEG=ABDFG F (CDEF)= CDE F(ABCG)= ABCFG F (ABDEFG)= ABDEG F=CDE=ABCFG=ABDEG G (CDEF)= CDEFG G(ABCG)= ABC G (ABDEFG)= ABDEF G=CDEFG=ABC=ABDEF AB (CDEF)= ABCDEF AB(ABCG)= CG AB (ABDEFG)= DEFG AB=ABCDEF=CG=DEFG AC (CDEF)= ADEF AC(ABCG)= BG AC (ABDEFG)= BCDEFG AC=ADEF=BG=BCDEFG AD (CDEF)= ACEF AD(ABCG)= BCDG AD (ABDEFG)= BEFG AD=ACEF=BCDG=BEFG AE (CDEF)= ACDF AE(ABCG)= BCEG AE (ABDEFG)= BDFG AE=ACDF=BCEG=BDFG AF (CDEF)= ACDE AF(ABCG)= BCFG AF (ABDEFG)= BDEG AF=ACDE=BCFG=BDEG AG (CDEF)= ACDEFG AG(ABCG)= BC AG (ABDEFG)= BDEF AG=ACDEFG=BC=BDEF BD (CDEF)= BCEF BD(ABCG)= ACDG BD (ABDEFG)= AEFG BD=BCEF=ACDG=AEFG BE (CDEF)= BCDF BE(ABCG)= ACEG BE (ABDEFG)= ADFG BE=BCDF=ACEG=ADFG BF (CDEF)= BCDE BF(ABCG)= ACFG BF (ABDEFG)= ADEG BF=BCDE=ACFG=ADEG CD (CDEF)= EF CD(ABCG)= ABDG CD (ABDEFG)= ABCEFG CD=EF=ABDG=ABCEFG CE (CDEF)= DF CE(ABCG)= ABEG CE (ABDEFG)= ABCDFG CE=DF=ABEG=ABCDFG CF (CDEF)= DE CF(ABCG)= ABFG CF (ABDEFG)= ABCDEG CF=DE=ABFG=ABCDEG DG (CDEF)= CEFG DG(ABCG)= ABCD DG (ABDEFG)= ABEF DG=CEFG=ABCD=ABEF EG (CDEF)= CDFG EG(ABCG)= ABCE EG (ABDEFG)= ABDF EG=CDFG=ABCE=ABDF FG (CDEF)= CDEG FG(ABCG)= ABCF FG (ABDEFG)= ABDE FG=CDEG=ABCF=ABDE Analysis of Variance Table Source Degrees of Freedom A B C D E F G AB=CG 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY AC=BG AD AE AF AG=BC BD BE CD=EF CE=DF CF=DE DG EG FG Error 9 Total 3 8- Consider the 5 design in Problem 6-. Suppose that only a one-half fraction could be run. Furthermore, two days were required to take the 6 observations, and it was necessary to confound the 5- design in two blocks. Construct the design and analyze the data. A B C D E=ABCD Data Blocks = AB Block - - - - + e 8 + + - - - - a 9 - - + - - - b 34 - + + - - + abe 5 + - - + - - c 6 + + - + - + ace - - + + - + bce 45 - + + + - - abc 6 + - - - + - d 8 + + - - + + ade - - + - + + bde 3 - + + - + - abd 5 + - - + + + cde 5 + + - + + - acd - - + + + - bcd 44 - + + + + + abcde 63 + Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.875 473.63 8.6343 Model B 33.65 45.56 8.5455 Model C.65 45.56 8.488 Error D -.65.565.8586 Error E.375.565.667 Error AB Aliased Error AC.65.565.8586 Error AD.875 3.65.558965 Error AE.375 7.565.383 Error BC.875 3.65.558965 Error BD -.375.565.667 Error BE.5.65.475 Error CD.65.565.8586 Error CE.65.565.8586 Error DE -.65.565.9786 Lenth's ME.4663 Lenth's SME 5.57 The AB interaction in the above table is aliased with the three-factor interaction BCD, and is also confounded with blocks. 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 A C B. 8.4 6.8 5. 33.63 Effect Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 3.6 3.6 Model 5447.9 3 85.73 63.3 <. significant A 473.6 473.6 64. <. B 45.56 45.56 569.96 <. C 45.56 45.56 56.76 <. Residual 3.69.88 Cor Total 568.94 5 The Model F-value of 63.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..7 R-Squared.994 Mean 3.44 Adj R-Squared.996 C.V. 5.58 Pred R-Squared.9878 PRESS 67.4 Adeq Precision 58. Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 3.44.4 9.5 3.37 Block 3.56 Block -3.56 A-Aperture 5.44.4 4.5 6.37. B-Exposure Time 6.8.4 5.88 7.75. C-Develop Time 5.3.4 4.38 6.5. Final Equation in Terms of Coded Factors: Yield = +3.44 +5.44 * A +6.8 * B +5.3 * C Final Equation in Terms of Actual Factors: Aperture small Yield = -.565 +.8463 * Exposure Time 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.7833 * Develop Time Aperture large Yield = +9.35 +.8463 * Exposure Time +.7833 * Develop Time 8- Analyze the data in Problem 6-3 as if it came from a IV design with I = ABCD. Project the design into a full factorial in the subset of the original four factors that appear to be significant. 4 Run Yield Factor Levels Number A B C D=ABC (lbs) Low (-) High (+) - - - - () A (h).5 3. + - - + ad 5 B (%) 4 8 3 - + - + bd 3 C (psi) 6 8 4 + + - - ab 6 D (ºC) 5 5 5 - - + + cd 9 6 + - + - ac 5 7 - + + - bc 8 + + + + abcd 3 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 3.75 8.5 8.3974 Error B.5.5.8766 Model C.75 5.5 9.8937 Model D 4.5 36.5 3.634 Error AB -.75.5.735895 Model AC -4.5 36.5 3.634 Model AD 4.5 36.5 3.634 Lenth's ME.74 Lenth's SME 5.6734 Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 C A AD D AC..6.3 3.9 4.5 Effect Design Expert Output Response: Yield in lbs ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Sum of Mean F Source Squares DF Square Value Prob > F Model 5.63 5 3.3 48.5.3 significant A 8.3 8.3 45..5 C 5.3 5.3 4..389 D 36. 36. 57.8.69 AC 36. 36. 57.8.69 AD 36.3 36.3 57.8.69 Residual.5.6 Cor Total 5.88 7 The Model F-value of 48.5 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..79 R-Squared.998 Mean 7.88 Adj R-Squared.974 C.V. 4.4 Pred R-Squared.869 PRESS. Adeq Precision 7.89 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 7.88.8 6.67 9.8 A-Time.87.8.67 3.8. C-Pressure.37.8.7.58. D-Temperature.3.8.9 3.33. AC -.3.8-3.33 -.9. AD.3.8.9 3.33. Final Equation in Terms of Coded Factors: Yield = +7.88 +.87 * A +.37 * C +.3 * D -.3 * A * C +.3 * A * D Final Equation in Terms of Actual Factors: Yield = +7.75-94.5 * Time +.475 * Pressure -.7 * Temperature -.85 * Time * Pressure +.68 * Time * Temperature 8-3 Repeat Problem 8- using I = -ABCD. Does use of the alternate fraction change your interpretation of the data? Run Yield Factor Levels Number A B C D=ABC (lbs) Low (-) High (+) - - - + d A (h).5 3. + - - - a 8 B (%) 4 8 3 - + - - b 3 C (psi) 6 8 4 + + - + abd 4 D (ºC) 5 5 5 - - + - c 7 6 + - + + acd 7 - + + + bcd 7 8 + + + - abc 5 Design Expert Output Term Effect SumSqr % Contribtn Model Intercept 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Model A 5.5 55.5 4.87 Error B.75.5.8346 Model C.5 3.5.3696 Model D.5.5 7.5695 Error AB -.75.5.8346 Model AC -4.5 36.5 6.784 Model AD 3.75 8.5.856 Lenth's ME.744 Lenth's SME 3.44 Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 C D AD AC A..3.63 3.94 5.5 Effect Design Expert Output Response: Yield in lbs ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.63 5 6.5 3.58.4 significant A 55.3 55.3 49..98 C 3.3 3.3.78.375 D.3.3 9..955 AC 36.3 36.3 3..98 AD 8.3 8.3 5..377 Residual.5. Cor Total 34.88 7 The Model F-value of 3.58 implies the model is significant. There is only a 4.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..6 R-Squared.9833 Mean 6.88 Adj R-Squared.946 C.V. 6.9 Pred R-Squared.733 PRESS 36. Adeq Precision 4.45 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 6.88.37 5.6 8.49 A-Time.63.37. 4.4. C-Pressure.63.37 -.99.4. D-Temperature.3.37 -.49.74. AC -.3.37-3.74 -.5. AD.88.37.6 3.49. Final Equation in Terms of Coded Factors: Yield = 8-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +6.88 +.63 * A +.63 * C +.3 * D -.3 * A * C +.88 * A * D Final Equation in Terms of Actual Factors: Yield = +9.5-7.5 * Time +.4 * Pressure -.56 * Temperature -.85 * Time * Pressure +.6 * Time * Temperature 4 8-4 Project the IV design in Example 8- into two replicates of a design in the factors A and B. Analyze the data and draw conclusions. Design Expert Output Response: Filtration Rate ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 78.5 3 4.83.4.753 not significant A 7. 7..3.39 B 4.5 4.5 7.68E-3.9344 AB.. 3.44E-3.956 Residual 343. 4 585.75 Lack of Fit. Pure Error 343. 4 585.75 Cor Total 37.5 7 The "Model F-value" of.4 implies the model is not significant relative to the noise. There is a 75.3 % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 4. R-Squared.37 Mean 7.75 Adj R-Squared -.3349 C.V. 34. Pred R-Squared -.53 PRESS 937. Adeq Precision.98 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 7.75 8.56 46.99 94.5 A-Temperature 9.5 8.56-4.6 33.6. B-Pressure.75 8.56-3. 4.5. AB -.5 8.56-4.6 3.6. Final Equation in Terms of Coded Factors: Filtration Rate = +7.75 +9.5 * A +.75 * B -.5 * A * B Final Equation in Terms of Actual Factors: Filtration Rate = +7.75 +9.5 * Temperature +.75 * Pressure -.5 * Temperature * Pressure 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6 3 8-5 Construct a III design. Determine the effects that may be estimated if a second fraction of this design is run with all signs reversed. A B C D=AB E=AC F=BC - - - + + + def + - - - - + af - + - - + - be + + - + - - abd - - + + - - cd + - + - + - ace - + + - - + bcf + + + + + + abcdef Principal Fraction l A =A+BD+CE l B =B+AD+CF l C =C+AE+BF l D =D+AB+EF l E =E+AC+DF l F =F+BC+DE l BE =BE+CD+AF Second Fraction l* A =A-BD-CE l* B =B-AD-CF l* C =C-AE-BF l* D =D-AB-EF l* E =E-AC-DF l* F =F-BC-DE l* BE =BE+CD+AF By combining the two fractions we can estimate the following: ( l i +l* I )/ ( l i -l* I )/ A BD+CE B AD+CF C AE+BF D AB+EF E AC+DF F BC+DE BE+CD+AF 6 3 8-6 Consider the III design in Problem 8-5. Determine the effects that may be estimated if a second fraction of this design is run with the signs for factor A reversed. Principal Fraction l A =A+BD+CE l B =B+AD+CF l C =C+AE+BF l D =D+AB+EF l E =E+AC+DF l F =F+BC+DE l BE =BE+CD+AF Second Fraction l* A =-A+BD+CE l* B =B-AD+CF l* C =C-AE+BF l* D =D-AB+EF l* E =E-AC+DF l* F =F+BC+DE l* BE =BE+CD-AF By combining the two fractions we can estimate the following: ( l i -l* I )/ ( l i +l* I )/ A BD+CE AD B+CF 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY AE AB AC AF C+BF D+EF E+DF F+BC+DE 8-7 Fold over the design is a 8 4 7 4 III design in Table 8-9 to produce a eight-factor design. Verify that the resulting IV design. Is this a minimal design? H A B C D=AB E=AC F=BC G=ABC + - - - + + + - Original + + - - - - + + Design + - + - - + - + + + + - + - - - + - - + + - - + + + - + - + - - + - + + - - + - + + + + + + + + - + + + - - - + Second - - + + + + - - Set of - + - + + - + - Runs w/ - - - + - + + + all Signs - + + - - + + - Switched - - + - + - + + - + - - + + - + - - - - - - - - After folding the original design over, we add a new factor H, and we have a design with generators 8 4 D=ABH, E=ACH, F=BCH, and G=ABC. This is a IV design. It is a minimal design, since it contains k=(8)=6 runs. 8-8 Fold over a design. Compare this 5 III design to produce a six-factor design. Verify that the resulting design is a 6 IV design to the in Table 8-. 6 IV F A B C D=AB E=BC + - - - + + Original + + - - - + Design + - + - - - + + + - + - + - - + + - + + - + - - + - + + - + + + + + + + - + + + - - Second - - + + + - Set of - + - + + + Runs w/ - - - + - + all Signs - + + - - + Switched - - + - + + - + - - + - - - - - - - If we relabel the factors from left to right as A, B, C, D, E, F, then this design becomes IV with generators I=ABDF and I=BCEF. It is not a minimal design, since k=(6)= runs, and the design contains 6 runs. 6 8-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 8-9 An industrial engineer is conducting an experiment using a Monte Carlo simulation model of an inventory system. The independent variables in her model are the order quantity (A), the reorder point (B), the setup cost (C), the backorder cost (D), and the carrying cost rate (E). The response variable is 5 average annual cost. To conserve computer time, she decides to investigate these factors using a III design with I = ABD and I = BCE. The results she obtains are de = 95, ae = 34, b = 58, abd = 9, cd = 9, ac = 87, bce = 55, and abcde = 85. (a) Verify that the treatment combinations given are correct. Estimate the effects, assuming three-factor and higher interactions are negligible. A B C D=AB E=BC - - - + + de + - - - + ae - + - - - b + + - + - abd - - + + - cd + - + - - ac - + + - + bce + + + + + abcde Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 49 48 43.95 Model B 45 45 37.675 Error C.5.5.8 Error D -8 648 5.938 Error E -4.5 4.5 3.8486 Error AC 3.5 364.5 3.3368 Error AE -4.5 4.5 3.8486 Lenth's ME 8.877 Lenth's SME 95.937 DESIGN-EXPERT Plot Avg Annual Cost Normal plot Half Normal plot A: Order Quantity B: Re -o rd er P o in t C: Setup Cost D: Backorder Cost E: Ca rryin g Co st Normal % probability 99 95 9 8 7 5 3 5 B A Half Normal % probability 99 97 95 9 85 8 7 6 4 B A -8. -. 5 5.5 3.5 49...5 4.5 36.75 49. Effect Effect (b) Suppose that a second fraction is added to the first, for example ade = 36, e = 93, ab = 87, bd = 53, acd = 39, c = 99, abce - 9, and bcde = 5. How was this second fraction obtained? Add this data to the original fraction, and estimate the effects. 8-33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY This second fraction is formed by reversing the signs of factor A. A B C D=AB E=BC + - - + + ade - - - - + e + + - - - ab - + - + - bd + - + + - acd - - + - - c + + + - + abce - + + + + bcde Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 44.5 783.5 39.589 Model B 49.5 97.5 48.9666 Error C 6.5 69.8593 Error D -8 56.9 Error E -8.5 7.5.3743 Error AB - 4.877 Error AC 7.5.5.6 Error AD -4.5 7.5.36464 Error AE -6 44.76759 Error BD 4.75 9.5.455486 Error CD -8.5 89.45856 Error DE 6.5 56.5.788584 Error ACD -6.5 56.5.788584 Error ADE 4 64.334 Lenth's ME 5.88 Lenth's SME 5.573 DESIGN-EXPERT Plot Avg Annual Cost Normal plot Half Normal plot A: Order Quantity B: Re -o rd er P o in t C: Setup Cost D: Backorder Cost E: Ca rryin g Co st Normal % probability 99 95 9 8 7 5 3 5 A B Half Normal % probability 99 97 95 9 85 8 7 6 4 A B -9. 5 5.63.5 35.38 5.5..3 4.63 36.94 49.5 Effect Effect (c) Suppose that the fraction abc = 89, ce = 96, bcd = 54, acde = 35, abe = 93, bde = 5, ad = 37, and () = 98 was run. How was this fraction obtained? Add this data to the original fraction and estimate the effects. This second fraction is formed by reversing the signs of all factors. A B C D=AB E=BC + + + - - abc 8-34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - + + + - bcd + - + + + acde - - + - + ce + + - - + abe - + - + + bde + - - + - ad - - - - - () Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 43.75 7656.5 38.563 Model B 5.5.3 5.3364 Error C 4.5 8.43678 Error D -8.75 36.5.565 Error E -7.5 5.33 Error AB -9.5 34.5.7566 Error AC 6 44.7765 Error AD -5.5.5.54945 Error AE -6.5 69.844 Error BC -7 96.9768 Error BD 5.5.5.54945 Error BE 6 44.7765 Error ABC -8 56.758 Error ABE 7.5 5.33 Lenth's ME 6.5964 Lenth's SME 54.5583 DESIGN-EXPERT Plot Avg Annual Cost Normal plot Half Normal plot A: Order Quantity B: Re -o rd er P o in t C: Setup Cost D: Backorder Cost E: Ca rryin g Co st Normal % probability 99 95 9 8 7 5 3 5 A B Half Normal % probability 99 97 95 9 85 8 7 6 4 A B -9. 5 5.63.5 35.38 5.5..56 5.3 37.69 5.5 Effect Effect 8- Construct a 5 design. Show how the design may be run in two blocks of eight observations each. Are any main effects or two-factor interactions confounded with blocks? A B C D E=ABCD Blocks = AB Block - - - - + e + + - - - - a - - + - - - b - + + - - + abe + - - + - - c + + - + - + ace - 8-35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - + + - + bce - + + + - - abc + - - - + - d + + - - + + ade - - + - + + bde - + + - + - abd + - - + + + cde + + - + + - acd - - + + + - bcd - + + + + + abcde + Blocks are confounded with AB and CDE. 8- Construct a 7 design. Show how the design may be run in four blocks of eight observations each. Are any main effects or two-factor interactions confounded with blocks? A B C D E F=CDE G=ABC Block=ACE Block=BFG Block assignment - - - - - - - () - - + - - - - - + ag + + 4 3 - + - - - - + bg - - 4 + + - - - - - ab + + 4 5 - - + - - + + cfg + - 3 6 + - + - - + - acf - + 7 - + + - - + - bcf + - 3 8 + + + - - + + abcfg - + 9 - - - + - + - df - + + - - + - + + adfg + - 3 - + - + - + + bdfg - + + + - + - + - abdf + - 3 3 - - + + - - + cdg + + 4 4 + - + + - - - acd - - 5 - + + + - - - bcd + + 4 6 + + + + - - + abcdg - - 7 - - - - + + - ef + + 4 8 + - - - + + + aefg - - 9 - + - - + + + befg + + 4 + + - - + + - abef - - - - + - + - + ceg - + + - + - + - - ace + - 3 3 - + + - + - - bce - + 4 + + + - + - + abceg + - 3 5 - - - + + - - de + - 3 6 + - - + + - + adeg - + 7 - + - + + - + bdeg + - 3 8 + + - + + - - abde - + 9 - - + + + + + cdefg - - 3 + - + + + + - acdef + + 4 3 - + + + + + - bcdef - - 3 + + + + + + + abcdefg + + 4 Blocks are confounded with ACE, BFG, and ABCEFG. 8- Irregular fractions of the k [John (97)]. Consider a 4 design. We must estimate the four main effects and the six two-factor interactions, but the full 4 factorial cannot be run. The largest possible block contains runs. These runs can be obtained from the four one-quarter fractions defined by I = AB = ACD = BCD by omitting the principal fraction. Show how the remaining three 4- fractions can be combined to estimate the required effects, assuming that three-factor and higher interactions are negligible. This design could be thought of as a three-quarter fraction. The four 4- fractions are as follows: 8-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY () I=+AB=+ACD=+BCD Runs: c,d,ab,abcd () I=+AB=-ACD=-BCD Runs: (), cd, abc, abd (3) I=-AB=+ACD=-BCD Runs: a, bc, bd, acd (4) I=-AB=-ACD=+BCD Runs: b, ac, ad, bcd If we do not run the principal fraction (), then we can combine the remaining 3 fractions to from 3 onehalf fractions of the 4 as follows: Fraction : () + (3) implies I=-BCD. This fraction estimates: A, AB, AC, and AD Fraction : () + (4) implies I=-ACD. This fraction estimates: B, BC, BD, and AB Fraction 3: (3) + (4) implies I=-AB. This fraction estimates: C, D, and CD In estimating these effects we assume that all three-factor and higher interactions are negligible. Note that AB is estimated in two of the one-half fractions: and. We would average these quantities and obtain a single estimate of AB. John (97, pp. 6-63) discusses this design and shows that the estimates obtained above are also the least squares estimates. John also derives the variances and covariances of these estimators. 8-3 Carbon anodes used in a smelting process are baked in a ring furnace. An experiment is run in the furnace to determine which factors influence the weight of packing material that is stuck to the anodes after baking. Six variables are of interest, each at two levels: A = pitch/fines ratio (.45,.55); B = packing material type (, ); C = packing material temperature (ambient, 35 C); D = flue location (inside, outside); E = pit temperature (ambient, 95 C); and F = delay time before packing (zero, 4 hours). A 6-3 design is run, and three replicates are obtained at each of the design points. The weight of packing material stuck to the anodes is measured in grams. The data in run order are as follows: abd = (984, 86, 936); abcdef = (75, 976, 457); be = (7,, 89); af = (474, 64, 54); def = (3, 56, 93); cd = (765, 75, 8); ace = (338, 54, 94); and bcf = (35, 99, 53). We wish to minimize the amount stuck packing material. (a) Verify that the eight runs correspond to a 6 3 III design. What is the alias structure? A B C D=AB E=AC F=BC - - - + + + def + - - - - + af - + - - + - be + + - + - - abd - - + + - - cd + - + - + - ace - + + - - + bcf + + + + + + abcdef I=ABD=ACE=BCF=BCDE=ACDF=ABEF=DEF, Resolution III A=BD=CE=CDF=BEF B=AD=CF=CDE=AEF C=AE=BF=BDE=ADF 8-37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY D=AB=EF=BCE=ACF E=AC=DF=BCD=ABF F=BC=DE=ACD=ABE CD=BE=AF=ABC=ADE=BDF=CEF (b) Use the average weight as a response. What factors appear to be influential? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 37.9 37996..947 Error B -8.9 56.56.49675 Error C.8 94..666559 Model D -59.6 3668 43.3443 Model E 99.7667 746.7 8.6735 Model F 43.567 7863 34.3345 Error BC -38.36 69.69.8377 Lenth's ME 563.3 Lenth's SME 348.4 Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 E A F D. 64.9 9.8 94.7 59.6 Effect Factors A, D, E and F (and their aliases) are apparently important. (c) Use the range of the weights as a response. What factors appear to be influential? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A 44.5 396.5.33 Error B 3.5 364.5.9639 Model C -9 338 7.956 Error D 75.5 4.5 6.48 Model E 44 447.3367 Model F 63 5338 8.6 Model AF 45 45.648 Lenth's ME 78.384 Lenth's SME 743.7 Factors C, E, F and the AF interaction (and their aliases) appear to be large. 8-38

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Half Normal plot 99 Half Normal % probability 97 95 9 85 8 7 6 4 C AF E F. 4.75 8.5.5 63. Effect (d) What recommendations would you make to the process engineers? It is not known exactly what to do here, since A, D, E and F are large effects, and because the design is resolution III, the main effects are aliased with two-factor interactions. Note, for example, that D is aliased with EF and the main effect could really be a EF interaction. If the main effects are really important, then setting all factors at the low level would minimize the amount of material stuck to the anodes. It would be necessary to run additional experiments to confirm these findings. 8-4 A 6-run experiment was performed in a semiconductor manufacturing plant to study the effects of six factors on the curvature or camber of the substrate devices produced. The six variables and their levels are shown below: Lamination Lamination Lamination Firing Firing Firing Temperature Time Pressure Temperature Cycle Time Dew Point Run (c) (s) (tn) (c) (h) (c) 55 5 58 7.5 75 5 58 9 6 3 55 5 5 58 9 4 75 5 5 58 7.5 6 5 55 58 9 6 6 75 58 7.5 7 55 5 58 7.5 6 8 75 5 58 9 9 55 5 6 7.5 6 75 5 6 9 55 5 5 6 9 6 75 5 5 6 7.5 3 55 6 9 4 75 6 7.5 6 5 55 5 6 7.5 6 75 5 6 9 6 Each run was replicated four times, and a camber measurement was taken on the substrate. The data are shown below: Camber for Replicate (in/in) Total Mean Standard Run 3 4 ( -4 in/in) ( -4 in/in) Deviation.67.8.49.85 69 57.5 4.48.6.66.44. 9 48..976 8-39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3.4.43.4.5 76 44. 4.83 4.73.8.39.3 3 55.75 5.5 5.47.47.4.89 3 55.75.4 6.9.58.47.96 9 3. 63.639 7..9.9.86 389 97.5 6.9 8.55.5.6.69 9 5. 39.4 9.3.3.77.69 5.5 6.75.78.58.6.45 34 85.5 5.34.43.7.8.8 6 3.5 7.68.86.37.58.59 64 6..83 3..86..58 455 3.75 3. 4.65.9.6.7 37 9.75 9.5 5.55.58.45.45 63 5.75 6.75 6.93.4..33 46 5. 7.45 (a) What type of design did the experimenters use? The 6 IV, a 6-run design. (b) What are the alias relationships in this design? The defining relation is I=ABCE=ACDF=BDEF A (ABCE)= BCE A (ACDF)= CDF A (BDEF)= ABCDEF A=BCE=CDF=ABDEF B (ABCE)= ACE B (ACDF)= ABCDF B (BDEF)= DEF B=ACE=ABCDF=DEF C (ABCE)= ABE C (ACDF)= ADF C (BDEF)= BCDEF C=ABE=ADF=BCDEF D (ABCE)= ABCDE D (ACDF)= ACF D (BDEF)= BEF D=ABCDE=ACF=BEF E (ABCE)= ABC E (ACDF)= ACDEF E (BDEF)= BDF E=ABC=ABDEF=BDF F (ABCE)= ABCEF F (ACDF)= ACD F (BDEF)= BDE F=ABCEF=ACD=BDE AB (ABCE)= CE AB (ACDF)= BCDF AB (BDEF)= ADEF AB=CE=BCDF=ADEF AC (ABCE)= BE AC (ACDF)= DF AC (BDEF)= ABCDEF AC=BE=DF=ABCDEF AD (ABCE)= BCDE AD (ACDF)= CF AD (BDEF)= ABEF AD=BCDE=CF=ABEF AE (ABCE)= BC AE (ACDF)= CDEF AE (BDEF)= ABDF AE=BC=CDEF=ABDF AF (ABCE)= BCEF AF (ACDF)= CD AF (BDEF)= ABDE AF=BCEF=CD=ABDE BD (ABCE)= ACDE BD (ACDF)= ABCF BD (BDEF)= EF BD=ACDE=ABCF=EF BF (ABCE)= ACEF BF (ACDF)= ABCD BF (BDEF)= DE BF=ACEF=ABCD=DE (c) Do any of the process variables affect average camber? Yes, per the analysis below, variables A, C, D, and F affect average camber. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 38.963 654.79.96 Error B 5.785 33.69.7344 Model C 56.33 558.355 Error D -4.88 88.69.3759 Model E -34.4687 475.38 8.848 Model F -77.4688 45.6 4.89 Error AB 9.563 467.85.4969 Error AC.463 8.6 3.449 Error AD -.88 597.9.553 Error AE 8.563 38.6.49 Error AF -9.787 555.3.64483 Error BC Aliased Error BD 3.33.75 3.687 Error BE Aliased Error BF 7.465 9.4.373 Error CD Aliased Error CE Aliased Error CF Aliased Error DE Aliased Error DF Aliased Error EF Aliased Error ABC Aliased Error ABD.535.89.997 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error ABE Aliased Error ABF -7.3438 3..469 Lenth's ME 7.936 Lenth's SME 46.4 DESIGN-EXPERT Plot Cam ber Avg Normal plot A: Lam Temp B: Lam Time C: Lam Pres D: Fire Temp E: Fire Time F: Fire DP Norm al % probability 99 95 9 8 7 5 3 5 F E A C -7 7.47-4 4.9 -.7.66 56.3 Effect Design Expert Output Response: Camber Avg in in/in ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4737.8 4 84.7.39.7 significant A 654.79 654.79 5.8.344 C 558. 558..8.5 E 475.38 475.38 4.57.558 F 45.63 45.63 3.9.5 Residual 435. 39.55 Cor Total 5885.8 5 The Model F-value of.39 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3.4 R-Squared.855 Mean 7. Adj R-Squared.7348 C.V. 3.3 Pred R-Squared.5886 PRESS 493.8 Adeq Precision.478 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 7. 8.6 89.7 4.76 A-Lam Temp 9.45 8.6.7 37.9. C-Lam Pres 8. 8.6.7 45.76. E-Fire Time -7.3 8.6-34.98.5. F-Fire DP -38.73 8.6-56.48 -.99. Final Equation in Terms of Coded Factors: Camber Avg = +7. +9.45 * A +8. * C -7.3 * E -38.73 * F Final Equation in Terms of Actual Factors: 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Camber Avg = +63.738 +.9453 * Lam Temp +.65 * Lam Pres -.9978 * Fire Time -.946 * Fire DP (d) Do any of the process variables affect the variability in camber measurements? Yes, A, B, F, and AF interaction affect the variability in camber measurements. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 5.935.69 7.663 Model B -6.5773 99. 3.558 Error C 5.8745 38.39 3.77437 Error D -3.95 43.36.8564 Error E -.3375.85.597466 Model F -9.56 34.694 9.37 Error AB.9555 3.65.9984 Error AC.54 5.483.696757 Error AD -4.665 85.68.343 Error AE -.85.996.355347 Model AF -.8745 473.9.9337 Error BC Aliased Error BD -4.85575 94.33.57879 Error BE Aliased Error BF 8.85 7.59 7.38689 Error CD Aliased Error CE Aliased Error CF Aliased Error DE Aliased Error DF Aliased Error EF Aliased Error ABC Aliased Error ABD -.685.8564.57593 Error ABE Aliased Error ABF 3.3985 46.94.633 Lenth's ME 7.839 Lenth's SME 36.6 DESIGN-EXPERT Plot Cam ber StDev Normal plot A: Lam Temp B: Lam Time C: Lam Pres D: Fire Temp E: Fire Time F: Fire DP Norm al % probability 99 95 9 8 7 5 3 5 B F AF A - 6.58-8.4 6 -.3 4 7.78 5.9 Effect 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Camber StDev ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 96.6 4 73.65..8 significant A.69.69 5.3.5 B 99. 99. 6.55.9 F 34.69 34.69 5.6.44 AF 473. 473. 7..8 Residual 73.65 66.4 Cor Total 3657.7 5 The Model F-value of. implies the model is significant. There is only a.8% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 8.5 R-Squared.8 Mean 5.35 Adj R-Squared.776 C.V. 3.5 Pred R-Squared.5773 PRESS 545.84 Adeq Precision 9.56 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 5.35.4.87 9.84 A-Lam Temp 7.95.4 3.47.44. B-Lam Time -8.9.4 -.77-3.8. F-Fire DP -4.63.4-9. -.4. AF -5.44.4-9.9 -.95. Final Equation in Terms of Coded Factors: Camber StDev = +5.35 +7.95 * A -8.9 * B -4.63 * F -5.44 * A * F Final Equation in Terms of Actual Factors: Camber StDev = -4.46746 +4.96373 * Lam Temp -.55 * Lam Time +.384 * Fire DP -.84 * Lam Temp * Fire DP (e) If it is important to reduce camber as much as possible, what recommendations would you make? 8-43

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY One Factor Plot One Factor Plot 3 3 8.375 8.375 Camber Avg 3.75 Camber Avg 3.75 8.5 8.5 3.5 3.5 55. 6. 65. 7. 75. 5. 6.5 7.5 8.75. Lam Temp Lam Pres One Factor Plot One Factor Plot 3 3 8.375 8.375 Camber Avg 3.75 Camber Avg 3.75 8.5 8.5 3.5 3.5 7.5.38 3.5 6.3 9...5 3. 4.5 6. Fire Tim e Fire DP 8-44

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Cam ber StDev 63.639 Interaction Graph Fire DP DESIGN-EXPERT Plot Cam ber StDev 63.639 One Factor Plot X = A: Lam Temp Y = F: Fire DP 48.75 F-. F+ 6. Actual Factors B: Lam Time = 7.5 C: Lam Pres = 7.5 33.86 D: Fire Temp = 6. E: Fire Time = 3.5 Camber StDev X = B: Lam Time Actual Factors 48.75 A: Lam Temp = 65. C: Lam Pres = 7.5 D: Fire Temp = 6. E: Fire Tim e = 3.5 F: Fire DP = 3. 33.86 Camber StDev 8.97 8.97 4.83 4.83 55. 6. 65. 7. 75.. 3.75 7.5.5 5. Lam Temp Lam Time Run A and C at the low level and E and F at the high level. B at the low level enables a lower variation without affecting the average camber. 8-5 A spin coater is used to apply photoresist to a bare silicon wafer. This operation usually occurs early in the semiconductor manufacturing process, and the average coating thickness and the variability in the coating thickness has an important impact on downstream manufacturing steps. Six variables are used in the experiment. The variables and their high and low levels are as follows: Factor Low Level High Level Final Spin Speed 735 rpm 665 rpm Acceleration Rate 5 Volume of Resist Applied 3 cc 5 cc Time of Spin 4 s 6 s Resist Batch Variation Batch Batch Exhaust Pressure Cover Off Cover On The experimenter decides to use a 6- design and to make three readings on resist thickness on each test wafer. The data are shown in table 8-9. Table 8-9 A B C D E F Resist Thick ness Run Volume Batch Time Speed Acc. Cover Left Center Right Avg. Range 5 4 735 5 Off 453 453 455 455.7 6 5 6 735 5 Off 4446 4464 448 4446 36 3 3 6 665 5 Off 445 449 445 4464.7 38 4 3 4 735 Off 436 438 438 437.3 5 3 4 735 5 Off 437 495 489 497 8 6 5 6 665 Off 447 449 4495 4485.7 5 7 3 6 735 5 On 4496 45 448 4493.3 8 5 4 665 Off 454 4547 4538 454.3 9 9 5 4 665 5 Off 46 4643 463 465.7 3 3 4 665 5 On 4653 467 4645 4656 5 3 4 665 On 448 4486 447 4478.7 6 3 6 735 Off 4 433 47 43.7 6 3 5 6 665 5 On 46 464 469 466.7 4 3 6 665 On 4455 448 4466 4467 5 5 5 4 735 On 455 488 443 46 45 6 5 6 735 5 On 449 4534 453 455.7 44 7 3 4 735 5 On 454 455 454 4535 37 8 3 4 665 Off 4494 453 4496 4497.7 9 8-45

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 9 5 6 735 Off 493 436 43 43.3 3 3 6 735 5 Off 4534 4545 45 453.3 33 5 4 665 On 446 4457 4436 445 4 3 6 665 5 On 465 4688 4656 4664.7 38 3 5 4 735 Off 43 444 43 435 4 4 3 6 735 On 45 48 48 4.3 5 5 4 735 5 On 438 439 4376 438.7 5 6 3 6 665 Off 4533 45 45 45.7 7 3 4 735 On 494 43 47 498.7 58 8 5 6 665 5 Off 4666 4695 467 4677.7 9 9 5 6 735 On 48 43 497 496.7 33 3 5 6 665 On 4465 4496 4463 4474.7 33 3 5 4 665 5 On 4653 4685 4665 4667.7 3 3 3 4 665 5 Off 4683 47 4677 469.7 35 (a) Verify that this is a 6- design. Discuss the alias relationships in this design. I=ABCDEF. This is a resolution VI design where main effects are aliased with five-factor interactions and two-factor interactions are aliased with four-factor interactions. (b) What factors appear to affect average resist thickness? Factors B, D, and E appear to affect the average resist thickness. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A 9.95 788.45.7795 Model B 73.575 4336. 5.9378 Error C 3.375 9.5.4648 Model D -7.6 34999 46.98 Model E -8.95 6769 36.67 Error F -5.665 56.5.35877 Error AB -9 648.886387 Error AC -7.3 46.3.58355 Error AD -3.865 9.35.6358 Error AE -7. 43.8.55639 Error AF -6.9875 586.6.797 Error BC.875 946.5.949 Error BD 8.5 64.5.359 Error BE -8.35 649.78.87958 Error BF -3.375 734.45.53 Error CD -4.9875 4995.68357 Error CE 8. 537.9.7358 Error CF -6.7875 368.56.5448 Error DE -38.5375 88..65 Error DF -3. 8.9.57 Error EF -4.65 3554.8.8544 Error ABC.375.5.53887 Error ABD Aliased Error ABE 6.5 78.9795 Error ABF 3.45 7893.96.798 Error ACD 5.5875 943.76.65883 Error ACE Aliased Error ACF Aliased Error ADE 9.5375 77.7.99543 Error ADF Aliased Error AEF Aliased Error BCD 9.875 6768.66.95873 Error BCE -.65.5.88965 Error BCF Aliased Error BDE -.8875 8.53.389863 Error BDF 3.95 4.8.7739 Error BEF Aliased Error CDE Aliased Error CDF Aliased Error CEF 3.375 78.75.77 8-46

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error DEF Aliased Lenth's ME 8.678 Lenth's SME 54.48 DESIGN-EXPERT Plot Thick Avg Normal plot A: Volume B: Batch C: Tim e D: Speed E: Acc F: Cover Norm al % probability 99 95 9 8 7 5 3 5 D E B - 7. 6-3 6.9-6 6.74 3.4 73.57 Effect Design Expert Output Response: Thick Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.54E+5 3.8E+5 79. <. significant B 4336.4 4336.4 5.74.5 D 3.43E+5 3.43E+5 4.63 <. E.677E+5.677E+5 97.7 <. Residual 7759.83 8 75.4 Cor Total 7.3E+53 The Model F-value of 79. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 5.46 R-Squared.8946 Mean 4458.5 Adj R-Squared.8833 C.V..8 Pred R-Squared.863 PRESS.6E+5 Adeq Precision 4.993 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 4458.5 9.7 4439.5 4477.5 B-Batch 36.79 9.7 7.79 55.78. D-Speed -3.53 9.7 -.53-84.53. E-Acc -9.46 9.7 -.46-7.47. Final Equation in Terms of Coded Factors: Thick Avg = +4458.5 +36.79 * B -3.53 * D -9.46 * E Final Equation in Terms of Actual Factors: Batch Batch Thick Avg = +6644.7875 8-47

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -.958 * Speed -.95 * Acc Batch Batch Thick Avg = +678.365 -.958 * Speed -.95 * Acc (c) Since the volume of resist applied has little effect on average thickness, does this have any important practical implications for the process engineers? Yes, less material could be used. (d) Project this design into a smaller design involving only the significant factors. Graphically display the results. Does this aid in interpretation? Cube Graph Thick Avg 46.73 43.3 D+ 449.66 4483.3 Speed 4433.79 457.37 E+ Acc D- 466.7 469.9 E- B- B+ Batch The cube plot usually assists the experimenter in drawing conclusions. (e) Use the range of resist thickness as a response variable. Is there any indication that any of these factors affect the variability in resist thickness? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A -.65 3.5.777387 Model B.5 36.5.89866 Error C -.75 6.5.55 Error D.65.5.5554 Model E -5.375 3.5 5.74956 Model F 7.75 48.5.953 Model AB.65 3.5.777387 Error AC -3.5 98.43789 Error AD -.5.5.3955 Error AE.875 8.5.699649 Model AF.75 4.5.6947 Error BC Error BD.5.5.3955 Error BE -5.375 3.5 5.74956 Model BF 3.5 84.5.6 Error CD 3.75.5.79859 8-48

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error CE 3.75.5.79859 Error CF 4.875 9.5 4.796 Error DE 5.375 3.5 5.74956 Error DF 5.5 4 6.9 Model EF 8 5.7367 Error ABC Aliased Error ABD Aliased Error ABE 3.65 5.5.653 Model ABF 9 648 6.99 Error ACD -6.5 338 8.48 Error ACE Aliased Error ACF Aliased Error ADE -3.375 9.5.6686 Error ADF -.5.49758 Error AEF 8.99 Error BCD Aliased Error BCE Aliased Error BCF Aliased Error BDE -.65 55.5.373 Error BDF -.5.49758 Error BEF Aliased Error CDE Aliased Error CDF Aliased Error CEF.5 36.5.89866 Error DEF 3.79645 Lenth's ME 9.54 Lenth's SME 7.399 DESIGN-EXPERT Plot Thick StDev Normal plot A: Volume B: Batch C: Tim e D: Speed E: Acc F: Cover Normal % probability 99 95 9 8 7 5 3 B BF AB AF A EF F ABF 5 E -6.5 -.6.5 5.3 9. Effect Design Expert Output Response: Thick StDev ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3. 9 4.78.48.4 significant A 3.3 3.3.34.8545 B 36.3 36.3.4.5346 E 3. 3..55.48 F 48.5 48.5 5.9.33 AB 3. 3..34.8545 AF 4.5 4.5.7.686 BF 84.5 84.5.93.345 EF 5. 5. 5.64.67 ABF 648. 648. 7.4.39 8-49

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residual 996.88 9.77 Cor Total 49.88 3 The Model F-value of.48 implies the model is significant. There is only a 4.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 9.53 R-Squared.53 Mean 6.56 Adj R-Squared.3 C.V. 35.87 Pred R-Squared -.5 PRESS 44.79 Adeq Precision 5.586 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 6.56.68 3.7 3.6 A-Volume -.3.68-3.8 3.8. B-Batch.6.68 -.43 4.56. E-Acc -.69.68-6.8.8. F-Cover 3.88.68.38 7.37. AB.3.68-3.8 3.8. AF.88.68 -.6 4.37. BF.63.68 -.87 5.. EF 4..68.5 7.49. ABF 4.5.68. 7.99. Final Equation in Terms of Coded Factors: Thick StDev = +6.56 -.3 * A +.6 * B -.69 * E +3.88 * F +.3 * A * B +.88 * A * F +.63 * B * F +4. * E * F +4.5 * A * B * F Final Equation in Terms of Actual Factors: Batch Batch Cover Off Thick StDev = +.39583 +3. * Volume -.8967 * Acc Batch Batch Cover Off Thick StDev = +54.7783-5.375 * Volume -.8967 * Acc Batch Batch Cover On Thick StDev = +4.565-4.5 * Volume +.75 * Acc Batch Batch Cover On Thick StDev = +9.4375 +5.375 * Volume +.75 * Acc 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The model here for variability isn t very strong. Notice the small value of R, and in particular, the adjusted R. Often we find that obtaining a good model for a response that expresses variability isn t as easy as finding a satisfactory model for a response that essentially measures the mean. (f) Where would you recommend that the process engineers run the process? Considering only the average thickness results, the engineers could use factors B, D and E to put the process mean at target. Then the engineer could consider the other factors on the range model to try to set the factors to reduce the variation in thickness at that mean. 8-6 Harry and Judy Peterson-Nedry (two friends of the author) own a vineyard in Oregon. They grow several varieties of grapes and manufacture wine. Harry and Judy have used factorial designs for process and product development in the winemaking segment of their business. This problem describes the experiment conducted for their 985 Pinot Noir. Eight variables, shown below, were originally studied in this experiment: Variable Low Level High Level A Pinot Noir Clone Pommard Wadenswil B Oak Type Allier Troncais C Age of Barrel Old New D Yeast/Skin Contact Champagne Montrachet E Stems None All F Barrel Toast Light Medium G Whole Cluster None % H Fermentation Temperature Low (75 F Max) High (9 F Max) Harry and Judy decided to use a 84 IV design with 6 runs. The wine was taste-tested by a panel of experts on 8 March 986. Each expert ranked the 6 samples of wine tasted, with rank being the best. The design and taste-test panel results are shown in Table 8-3. Table 8-3 Run A B C D E F G H HPN JPN CAL DCM RGB y bar s - - - - - - - - 6 3 7 9.6 3.5 + - - - - + + + 7 4 4 9.8 3. 3 - + - - + - + + 4 3 5.6.7 4 + + - - + + - - 9 9 7 9 9..79 5 - - + - + + + - 8 8 8 9..4 6 + - + - + - - + 6 5 6 6 5..73 7 - + + - - + - + 6 5 6 5 3 5.. 8 + + + - - - + - 5 6 6 5 4 5..84 9 - - - + + + - + 3 3..84 + - - + + - + - 7 4 7 6 7..55 - + - + - + + - 3 3 8 8 8.8 3.96 + + - + - - - + 3 5 4.8.79 3 - - + + - - + + 4 5 9.6 3.9 4 + - + + - + - - 4 4.4.5 5 - + + + + - - - 5 5 9 6 9. 4. 6 + + + + + + + + 4 3 3.6.4 (a) What are the alias relationships in the design selected by Harry and Judy? E = BCD, F = ACD, G = ABC, H = ABD Defining Contrast : I = BCDE = ACDF = ABEF = ABCG = ADEG = BDFG = CEFG = ABDH = ACEH = BCFH = DEFH = CDGH = BEGH = AFGH = ABCDEFGH Aliases: A = BCG = BDH = BEF = CDF = CEH = DEG = FGH 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B = ACG = ADH = AEF = CDE = CFH = DFG = EGH C = ABG = ADF = AEH = BDE = BFH = DGH = EFG D = ABH = ACF = AEG = BCE = BFG = CGH = EFH E = ABF = ACH = ADG = BCD = BGH = CFG = DFH F = ABE = ACD = AGH = BCH = BDG = CEG = DEH G = ABC = ADE = AFH = BDF = BEH = CDH = CEF H = ABD = ACE = AFG = BCF = BEG = CDG = DEF AB = CG = DH = EF AC = BG = DF = EH AD = BH = CF = EG AE = BF = CH = DG AF = BE = CD = GH AG = BC = DE = FH AH = BD = CE = FG (b) Use the average ranks ( y ) as a response variable. Analyze the data and draw conclusions. You will find it helpful to examine a normal probability plot of effect estimates. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A.5 5.65.799 Error B.5 6.5.3883 Error C.875 4.65 5.57776 Model D -3.975 63.5 5.687 Error E.575 9.95 3.93566 Model F -.65 7.565.934 Model G 3.775 57.5.695 Error H.5.5.996 Error AB -.75.5.8944 Error AC.975 5.65 6.8858 Model AD -.375.565 8.949 Error AE.575 9.95 3.93566 Error AF.375 7.565.99959 Error AG.75.35.9984 Error AH.85 3.35 5.844 Lenth's ME 6.73 Lenth's SME.39 DESIGN-EXPERT Plot Taste Avg Normal plot A: A B: B C: C D: D E: E F: F G: G H: H Normal % probability 99 95 9 8 7 5 3 5 D AD F G -3.9 8 -. 4 -..84 3.78 Effect 8-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Taste Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 75.39 5 35.8 4.57.98 significant A 5.6 5.6.66.4355 D 63. 63. 8.4.67 F 7.56 7.56 3.59.873 G 57. 57. 7.43.4 AD.56.56.94.7 Residual 76.7 7.67 Cor Total 5. 5 The Model F-value of 4.57 implies the model is significant. There is only a.98% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..77 R-Squared.6957 Mean 8.8 Adj R-Squared.5435 C.V. 3.43 Pred R-Squared.9 PRESS 96.4 Adeq Precision 7.57 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 8.8.69 7.7.36 A-A.56.69 -.98.. D-D -.99.69-3.53 -.44. F-F -.3.69 -.86.3. G-G.89.69.34 3.43. AD -.9.69 -.73.36. Final Equation in Terms of Coded Factors: Taste Avg = +8.8 +.56 * A -.99 * D -.3 * F +.89 * G -.9 * A * D Factors D, F, G and the AD interaction are important. Factor A is added to the model to preserve hierarchy. Notice that the AD interaction is aliased with other two-factor interactions that could also be important. So the interpretation of the two-factor interaction is somewhat uncertain. Normally, we would add runs to the design to isolate the significant interactions, but that won t work very well here because each experiment requires a full growing season. In other words, it would require a very long time to add runs to dealias the alias chain of interest. 8-53

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Taste Avg 5. Interaction Graph D X = A: A Y = D: D Design Points.95 D D D D Actual Factors B: B = B C: C = C E: E = E F: F = F G: G = G H: H = H Tas te Avg 8.7 5.45. A A A One Factor Plot One Factor Plot 5. 5..95.95 Tas te Avg 8.7 Tas te Avg 8.7 5.45 5.45.. F F G G F G (c) Use the standard deviation of the ranks (or some appropriate transformation such as log s) as a response variable. What conclusions can you draw about the effects of the eight variables on variability in wine quality? 8-54

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Ln(Taste StDev) Normal plot A: A B: B C: C D: D E: E F: F G: G H: H Normal % probability 99 95 9 8 7 5 3 5 Warning! No term s are s ele -.3 9 -. -..9.38 Effect There do not appear to be any significant factors. (d) After looking at the results, Harry and Judy decide that one of the panel members (DCM) knows more about beer than he does about wine, so they decide to delete his ranking. What affect would this have on the results and on conclusions from parts (b) and (c)? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.65.565 4.957 Error B.65 7.56 6.494 Error C.5 9 3.43348 Model D -4.5 8 3.93 Error E.4375 3.7656 9.665 Model F -.375.565 8.6753 Model G.9375 34.556 3.676 Error H -.6875.8963.768 Error AB -.565.656.48833 Error AC.375.565 8.6753 Model AD -.5 9 3.43348 Error AE.6875.896.768 Error AF.875 3.65.6834 Error AG.85.646.739 Error AH.35.396 8.647 Lenth's ME 6.6579 Lenth's SME.75 8-55

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Taste Avg Normal plot A: A B: B C: C D: D E: E F: F G: G H: H Normal % probability 99 95 9 8 7 5 3 5 D F AD A G -4.5 -.6 4 -.7 8.8.94 Effect Design Expert Output Response: Taste Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 57.64 5 3.53 3..646 not significant A.56.56..3384 D 8. 8. 7.75.93 F.56.56.6.74 G 34.5 34.5 3.3.99 AD 9. 9..86.375 Residual 4.48.45 Cor Total 6.3 5 The Model F-value of 3. implies there is a 6.46% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3.3 R-Squared.64 Mean 8.5 Adj R-Squared.4 C.V. 38.3 Pred R-Squared -.4 PRESS 67.48 Adeq Precision 5.778 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 8.5.8 6.7.3 A-A.8.8 -.99.6. D-D -.5.8-4.5 -.45. F-F -.9.8 -.99.6. G-G.47.8 -.33 3.7. AD -.75.8 -.55.5. Final Equation in Terms of Coded Factors: Taste Avg = +8.5 +.8 * A -.5 * D -.9 * F +.47 * G -.75 * A * D The results are the same for average taste without DCM as they were with DCM. 8-56

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Taste StDev Normal plot A: A B: B C: C D: D E: E F: F G: G H: H Normal % probability 99 95 9 8 7 5 3 5 Warning! No term s are s ele -.7 4 -.3 5.3.4.79 Effect The standard deviation response is much the same with or without DCM s responses. Again, there are no significant factors. (e) Suppose that just before the start of the experiment, Harry and Judy discovered that the eight new barrels they ordered from France for use in the experiment would not arrive in time, and all 6 runs would have to be made with old barrels. If Harry and Judy just drop column C from their design, what does this do to the alias relationships? Do they need to start over and construct a new design? The resulting design is a 7 3 IV BEGH = AFGH. with defining relations: I = ABEF = ADEG = BDFG = ABDH = DEFH = (f) Harry and Judy know from experience that some treatment combinations are unlikely to produce good results. For example, the run with all eight variables at the high level generally results in a poorly rated wine. This was confirmed in the 8 March 986 taste test. They want to set up a new design to make the run with all eight factors at the high level. What design would you suggest? By changing the sign of any of the design generators, a design that does not include the principal fraction will be generated. This will give a design without an experimental run combination with all of the variables at the high level. 8-7 In an article in Quality Engineering ( An Application of Fractional Factorial Experimental Designs, 988, Vol. pp. 9-3) M.B. Kilgo describes an experiment to determine the effect of CO pressure (A), CO temperature (B), peanut moisture (C), CO flow rate (D), and peanut particle size (E) on the total yield of oil per batch of peanuts (y). The levels she used for these factors are as follows: A B C D E Coded Pressure Temp Moisture Flow Particle Size Level (bar) (C) (% by weight) (liters/min) (mm) - 45 5 5 4.8 55 95 5 6 4.5 She conducted the 6-run fractional factorial experiment shown below: A B C D E y 45 5 5 4.8 63 8-57

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 55 5 5 4 4.5 3 45 95 5 4 4.5 36 4 55 95 5 4.8 99 5 45 5 5 4 4.5 4 6 55 5 5 4.8 66 7 45 95 5 4.8 7 8 55 95 5 4 4.5 54 9 45 5 5 6 4.5 3 55 5 5 6.8 74 45 95 5 6.8 8 55 95 5 6 4.5 33 3 45 5 5 6.8 63 4 55 5 5 6 4.5 5 45 95 5 6 4.5 44 6 55 95 5 6.8 96 (a) What type of design has been used? Identify the defining relation and the alias relationships. A 5 V, 6-run design, with I= -ABCDE. A(-ABCDE)= -BCDE B(-ABCDE)= -ACDE C(-ABCDE)= -ABDE D(-ABCDE)= -ABCE E(-ABCDE)= -ABCD AB(-ABCDE)= -CDE AC(-ABCDE)= -BDE AD(-ABCDE)= -BCE AE(-ABCDE)= -BCD BC(-ABCDE)= -ADE BD(-ABCDE)= -ACE BE(-ABCDE)= -ACD CD(-ABCDE)= -ABE CE(-ABCDE)= -ABD DE(-ABCDE)= -ABC A = -BCDE B = -ACDE C = -ABDE D = -ABCE E = -ABCD AB = -CDE AC = -BDE AD = -BCE AE = -BCD BC = -ADE BD = -ACE BE = -ACD CD = -ABE CE = -ABD DE = -ABC (b) Estimate the factor effects and use a normal probability plot to tentatively identify the important factors. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A 7.5 5.79 Model B 9.75 56.5 5.56 Error C.5 6.5.637 Error D Model E 44.5 79 76.4354 Error AB 5.5.5.6388 Error AC.5 6.5.637 Error AD -4 64.6758 Error AE 7 96.8934 Error BC 3 36.34739 Error BD -.75.5.89 Error BE.5.5.443 Error CD.5.5.9547 Error CE -6.5 56.5.5777 Error DE 3.5 49.47836 Lenth's ME.5676 Lenth's SME 3.4839 8-58

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Yield Half Normal plot A: Pressure B: Temperature C: M oisture D: Flow E: Particle Size Half Normal % probability 99 97 95 9 85 8 7 6 4 B E..3.5 33.38 44.5 Effect (c) Perform an appropriate statistical analysis to test the hypothesis that the factors identified in part above have a significant effect on the yield of peanut oil. Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 948.5 474.63 69.89 <. significant B 56.5 56.5 3..3 E 79. 79. 6.78 <. Residual 88.75 3 67.83 Cor Total 363. 5 The Model F-value of 69.89 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 8.4 R-Squared.949 Mean 54.5 Adj R-Squared.98 C.V. 5.8 Pred R-Squared.87 PRESS 335.67 Adeq Precision 8.7 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 54.5.6 49.8 58.7 B-Temperature 9.88.6 5.43 4.3. E-Particle Size.5.6 7.8 6.7. (d) Fit a model that could be used to predict peanut oil yield in terms of the factors that you have identified as important. Design Expert Output Final Equation in Terms of Coded Factors: Yield = +54.5 +9.88 * B +.5 * E Final Equation in Terms of Actual Factors: Yield = 8-59

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -5.4975 +.84 * Temperature +6.6498 * Particle Size (e) Analyze the residuals from this experiment and comment on model adequacy. The residual plots are satisfactory. There is a slight tendency for the variability of the residuals to increase with the predicted value of y. Normal plot of residuals Residuals vs. Predicted.65 99 Norm al % probability 95 9 8 7 5 3 5 Residuals 5.65 -.3 75-8.3 75-5.37 5-5.375-8.3 75 -.3 75 5.65.65.3 38.9 54.5 7.3 86.38 Residual Predicted Residuals vs. Temperature Residuals vs. Particle Size.65.65 5.65 5.65 Residuals -.3 75 Residuals -.3 75-8.3 75-8.3 75-5.375-5.37 5 5 37 48 6 7 83 95.8.97.67 3.36 4.5 Tem perature Particle Size 8-8 A 6-run fractional factorial experiment in factors on sand-casting of engine manifolds was conducted by engineers at the Essex Aluminum Plant of the Ford Motor Company and described in the article Evaporative Cast Process 3. Liter Intake Manifold Poor Sandfill Study, by D. Becknell (Fourth Symposium on Taguchi Methods, American Supplier Institute, Dearborn, MI, 986, pp. -3). The purpose was to determine which of factors has an effect on the proportion of defective castings. The design and the resulting proportion of nondefective castings p observed on each run are shown below. 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY This is a resolution III fraction with generators E=CD, F=BD, G=BC, H=AC, J=AB, and K=ABC. Assume that the number of castings made at each run in the design is. F&T s Run A B C D E F G H J K p arcsin Modification - - - - + + + + + -.958.364.363 + - - - + + + - - +..57.555 3 - + - - + - - + - +.977.49.47 4 + + - - + - - - + -.775.77.76 5 - - + - - + - - + +.958.364.363 6 + - + - - + - + - -.958.364.363 7 - + + - - - + - - -.83.4.3 8 + + + - - - + + + +.96.59.59 9 - - - + - - + + + -.679.969.968 + - - + - - + - - +.78.8.83 - + - + - + - + - +..57.556 + + - + - + - - + -.896.4.4 3 - - + + + - - - + +.958.364.363 4 + - + + + - - + - -.88.3.3 5 - + + + + + + - - -.84.6.6 6 + + + + + + + + + +.955.357.356 (a) Find the defining relation and the alias relationships in this design. I=CDE=BDF=BCG=ACH=ABJ=ABCK=BCEF=BDEG=ADEH=ABCDEJ=ABDEK=CDFG=ABCDFH = ADFJ=ACDFK=ABGH=ACGJ=AGK=BCHJ=BHK=CKJ (b) Estimate the factor effects and use a normal probability plot to tentatively identify the important factors. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.875.56463.497 Error B.665.7556.7353 Error C.765.456.9355 Error D -.55.868 7.88369 Error E.36375.5956 3.8393 Model F.7375.4676 33.4537 Error G -.5875.353 7.5 Error H.865.37756.37754 Error J -.875.6636.48986 Model K.9965.3976 8.7988 Error AB Aliased Error AC Aliased Error AD.4875 9.565E-5.689584 Error AE -.3465.479556 3.4787 Error AF.4875.4756.7954 Error BE -.535.89 8.899 Error DK.5375.945563.6859 Lenth's ME.345 Lenth's SME.9399 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot p Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J K: K Normal % probability 99 95 9 8 7 5 3 5 K F -. 5 -..3.7. Effect (c) Fit an appropriate model using the factors identified in part (b) above. Design Expert Output Response: p ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.86.43.7.8 significant F.46.46.5.48 K.4.4 9.9.77 Residual.5 3 4.3E-3 Cor Total.4 5 The Model F-value of.7 implies the model is significant. There is only a.8% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..63 R-Squared.65 Mean.89 Adj R-Squared.5645 C.V. 7.9 Pred R-Squared.48 PRESS.79 Adeq Precision 7.556 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.89.6.86.93 F-F.54.6..88. K-K.5.6.6.84. Final Equation in Terms of Coded Factors: p = +.89 +.54 * F +.5 * K Final Equation in Terms of Actual Factors: p = +.896 +.53688 * F +.498 * K (d) Plot the residuals from this model versus the predicted proportion of nondefective castings. Also prepare a normal probability plot of the residuals. Comment on the adequacy of these plots. 8-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The residual versus predicted plot identifies an inequality of variances. This is likely caused by the response variable being a proportion. A transformation could be used to correct this. Normal plot of residuals Residuals vs. Predicted.8885 99 Normal % probability 95 9 8 7 5 3 5.3987 Residuals -. 3 7 5 -.599688 -. 9 5 6 3 -.9563 -.599688 -.375.3987.8885.79.84.89.94. Residual Predicted (e) In part (d) you should have noticed an indication that the variance of the response is not constant (considering that the response is a proportion, you should have expected this). The previous table also shows a transformation on P, the arcsin square root, that is a widely used variance stabilizing transformation for proportion data (refer to the discussion of variance stabilizing transformations is Chapter 3). Repeat parts (a) through (d) above using the transformed response and comment on your results. Specifically, are the residuals plots improved? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.3.496.88453 Error B.5.5E-7 5.39875E-5 Error C -.5.865.396 Error D -.835.7889 6.63 Error E.5875.386.9846 Model F.965.5456 33.685 Error G -.85.59 5.59764 Error H.565.656.733 Error J -.535.34.44936 Model K.945.53 3.6778 Error AD -.3.496.88453 Error AF.55.3.85 Error BE -.4.4364 9.3486 Error DH -.5.565.935 Error DK.35.9.47734 Lenth's ME.535 Lenth's SME.4684 As with the original analysis, factors F and K remain significant with a slight increase with the R. 8-63

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot arcsin Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J K: K Normal % probability 99 95 9 8 7 5 3 5 F K -. -. 3.5.. Effect Design Expert Output Response: arcsin ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3.5.59.9 significant F.5.5.7.35 K.5.5.47.37 Residual.6 3. Cor Total.46 5 The Model F-value of.59 implies the model is significant. There is only a.9% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.6595 Mean.8 Adj R-Squared.67 C.V. 8.63 Pred R-Squared.484 PRESS.4 Adeq Precision 8.93 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.8.8..34 F-F.98.8.39.6. K-K.97.8.38.6. Final Equation in Terms of Coded Factors: arcsin = +.8 +.98 * F +.97 * K Final Equation in Terms of Actual Factors: arcsin = +.76 +.985 * F +.975 * K The inequality of variance has improved; however, there remain hints of inequality in the residuals versus predicted plot and the normal probability plot now appears to be irregular. 8-64

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.43875 Normal % probability 99 95 9 8 7 5 3 5.59375 Residuals -. 5 5 -. 9 6 5 -. 94 5 -.945 -.965 -.55.59375.43875.8.8.8.37.47 Residual Predicted (f) There is a modification to the arcsin square root transformation, proposed by Freeman and Tukey ( Transformations Related to the Angular and the Square Root, Annals of Mathematical Statistics, Vol., 95, pp. 67-6) that improves its performance in the tails. F&T s modification is: arcsin npˆ arcsin n npˆ n Rework parts (a) through (d) using this transformation and comment on the results. (For an interesting discussion and analysis of this experiment, refer to Analysis of Factorial Experiments with Defects or Defectives as the Response, by S. Bisgaard and H.T. Fuller, Quality Engineering, Vol. 7, 994-5, pp. 49-443.) Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.35.38756.87894 Error B.5 6.5E-8.466E-5 Error C -.7875.786.87566 Error D -.865.7376 6.444 Error E.57875.3398 3.458 Model F.9375.4833 33.375 Error G -.8375.5846 5.846 Error H.55875.488.8983 Error J -.4965.98556.639 Model K.9875.45733 3.79 Error AD -.7875.386.69938 Error AF.4965.98556.639 Error BE -.65.456 9.9 Error DH -.5375.945563.753 Error DK.365.356.539 Lenth's ME.9348 Lenth's SME.388464 As with the prior analysis, factors F and K remain significant. 8-65

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot F&T Transform Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J K: K Normal % probability 99 95 9 8 7 5 3 5 F K -. -. 3.5..9 Effect Design Expert Output Response: F&T Transform ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.9.5.67.9 significant F.5.5.77.34 K.5.5.57.36 Residual.5 3. Cor Total.44 5 The Model F-value of.67 implies the model is significant. There is only a.9% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.66 Mean.7 Adj R-Squared.688 C.V. 8.45 Pred R-Squared.4864 PRESS.3 Adeq Precision 8. Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.7.7..33 F-F.96.7.38.5. K-K.95.7.37.5. Final Equation in Terms of Coded Factors: F&T Transform = +.7 +.96 * F +.95 * K Final Equation in Terms of Actual Factors: F&T Transform = +.7356 +.9688 * F +.95437 * K The residual plots appears as they did with the arcsin square root transformation. 8-66

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.4487 Normal % probability 99 95 9 8 7 5 3 5.66875 Residuals -.85 -. 6 3 3 -. 89 8 3 -.8983 -.633 -.85.66875.4487.8.8.7.37.47 Residual Predicted 8-9 A 6-run fractional factorial experiment in 9 factors was conducted by Chrysler Motors Engineering and described in the article Sheet Molded Compound Process Improvement, by P.I. Hsieh and D.E. Goodwin (Fourth Symposium on Taguchi Methods, American Supplier Institute, Dearborn, MI, 986, pp. 3-). The purpose was to reduce the number of defects in the finish of sheet-molded grill opening panels. The design, and the resulting number of defects, c, observed on each run, is shown below. This is a resolution III fraction with generators E=BD, F=BCD, G=AC, H=ACD, and J=AB. F&T s Run A B C D E F G H J c c Modification - - - - + - + - + 56 7.48 7.5 + - - - + - - + - 7 4. 4.8 3 - + - - - + + - -.4.57 4 + + - - - + - + + 4.. 5 - - + - + + - + + 3.73.87 6 + - + - + + + - - 4.. 7 - + + - - - - + - 5 7.7 7. 8 + + + - - - + - +.4.57 9 - - - + - + + + +.. + - - + - + - - -..5 - + - + + - + + - 3.73.87 + + - + + - - - + 3.46 3.54 3 - - + + - - - - + 3.73.87 4 + - + + - - + + - 4.. 5 - + + + + + - - -..5 6 + + + + + + + + +..5 (a) Find the defining relation and the alias relationships in this design. I=BDE=BCDF=CEF=ACG=ABCDEG=ABDEG=AEFG=ACDH=ABCEH=ABFH=ADEFH=DGH= BEGH=BCRG=CDEFGH=ABJ=ADEJ=ACDFJ=ABCEFJ=BCGJ=CDEGJ=DEGJ=BEFGJ=BCDHJ= CEHJ=FHJ=BDEFHJ=ABDGHJ=AEGHJ=ACEGJ=ABCDEFGHJ (b) Estimate the factor effects and use a normal probability plot to tentatively identify the important factors. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept 8-67

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Model A -9.375 35.56 7.75573 Model B -.875 4.65.39 Model C -3.65 5.565.5957 Model D -4.375 86.56 8.346 Error E 3.65 5.565.5957 Model F -6.65 5.56 4.3895 Model G -.5 8.65.39847 Error H.375.565.49 Error J.5.65.3788 Model AD.65 54.563.95 Error AE.5 8.65.39847 Model AF 9.875 39.63 8.657 Error AH.375 7.565.66834 Model BC.375 57.563.478 Model BG -.65 637.56 4.65 Lenth's ME 3.9775 Lenth's SME 8.3764 DESIGN-EXPERT Plot c Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J Normal % probability 99 95 9 8 7 5 3 5 F BG D A AD BC AF - 6.6-9.5 6 -.5 4.56.6 Effect (c) Fit an appropriate model using the factors identified in part (b) above. Design Expert Output Response: c ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4454.3 445.4 8.6.9 significant A 35.56 35.56.3.5 B 4.6 4.6.89.3883 C 5.56 5.56 3.33.74 D 86.56 86.56 5.44.8 F 5.56 5.56 7.4.4 G 8.6 8.6.5.3333 AD 54.56 54.56 34.9. AF 39.6 39.6 4.75.4 BC 57.56 57.56 3.84.3 BG 637.56 637.56 4.45.4 Residual 78.8 5 5.76 Cor Total 453.94 5 The Model F-value of 8.6 implies the model is significant. There is only a.9% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3.97 R-Squared.986 8-68

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Mean.6 Adj R-Squared.9478 C.V. 39.46 Pred R-Squared.8 PRESS 87.4 Adeq Precision 7.77 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.6.99 7.5.6 A-A -4.69.99-7.4 -.4. B-B -.94.99-3.49.6. C-C -.8.99-4.36.74. D-D -7.9.99-9.74-4.64. F-F -8.3.99 -.86-5.76. G-G -.6.99-3.6.49. AD 5.8.99 3.6 8.36. AF 4.94.99.39 7.49. BC 5.69.99 3.4 8.4. BG -6.3.99-8.86-3.76. Final Equation in Terms of Coded Factors: c = +.6-4.69 * A -.94 * B -.8 * C -7.9 * D -8.3 * F -.6 * G +5.8 * A * D +4.94 * A * F +5.69 * B * C -6.3 * B * G Final Equation in Terms of Actual Factors: c = +.65-4.6875 * A -.9375 * B -.85 * C -7.875 * D -8.35 * F -.65 * G +5.85 * A * D +4.9375 * A * F +5.6875 * B * C -6.35 * B * G (d) Plot the residuals from this model versus the predicted number of defects. Also, prepare a normal probability plot of the residuals. Comment on the adequacy of these plots. 8-69

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 3.85 99 Normal % probability 95 9 8 7 5 3 5 Residuals.965 -.965-3.8 5-3.8 5 -.965.965 3.85-3.8.8 5.44 4.6 54.69 Residual Predicted There is a significant problem with inequality of variance. This is likely caused by the response variable being a count. A transformation may be appropriate. (e) In part (d) you should have noticed an indication that the variance of the response is not constant (considering that the response is a count, you should have expected this). The previous table also shows a transformation on c, the square root, that is a widely used variance stabilizing transformation for count data (refer to the discussion of variance stabilizing transformations in Chapter 3). Repeat parts (a) through (d) using the transformed response and comment on your results. Specifically, are the residual plots improved? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.895 3.4 4.936 Model B -.375.5555.74375 Error C -.6575.79.37 Model D -.65 8.756 5.66 Error E.4875.9565.7387 Model F -.675 7.96 36.4439 Model G -.385.599.79456 Error H.7.96.39754 Error J.6.44.9965 Error AD.45 5.44 7.77 Error AE.555.3.656 Error AF.86.9584 3.96436 Error AH.45.75.96875 Error BC.675.575.59 Model BG -.6.3684 3.894 Lenth's ME.7978 Lenth's SME 4.689 The analysis of the data with the square root transformation identifies only D, F, the BG interaction as being significant. The original analysis identified factor A and several two factor interactions as being significant. 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot sq rt Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J Normal % probability 99 95 9 8 7 5 3 5 F D BG -.6 -.6 7 -.7 3..4 Effect Design Expert Output Response: sqrt ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 57.4 5.48 6.67.56 significant B.56.56.3.586 D 8.7 8.7.87.8 F 7. 7. 5.8.6 G.59.59.34.57 BG.37.37 6.3.34 Residual 7..7 Cor Total 74.6 5 The Model F-value of 6.67 implies the model is significant. There is only a.56% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..3 R-Squared.7694 Mean.3 Adj R-Squared.654 C.V. 56.5 Pred R-Squared.497 PRESS 44.5 Adeq Precision 8.4 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.3.33.59 3.5 B-B -.9.33 -.9.54. D-D -.8.33 -.8 -.35. F-F -.3.33 -.3 -.57. G-G -.9.33 -.9.54. BG -.8.33 -.54 -.74. Final Equation in Terms of Coded Factors: sqrt = +.3 -.9 * B -.8 * D -.3 * F -.9 * G -.8 * B * G Final Equation in Terms of Actual Factors: sqrt = 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.35 -.865 * B -.85 * D -.3375 * F -.95 * G -.85 * B * G The residual plots are acceptable; although, there appears to be a slight u shape to the residuals versus predicted plot. Normal plot of residuals Residuals vs. Predicted.975 99 Norm al % probability 95 9 8 7 5 3 5.9535 Residuals -.6875 -.96 -. 5 -. 5 -.96 -.6875.9535.975 -. 5.44.3 3.83 5.5 Residual Predicted (f) There is a modification to the square root transformation proposed by Freeman and Tukey ( Transformations Related to the Angular and the Square Root, Annals of Mathematical Statistics, Vol., 95, pp. 67-6) that improves its performance. F&T s modification to the square root transformation is: c c Rework parts (a) through (d) using this transformation and comment on the results. (For an interesting discussion and analysis of this experiment, refer to Analysis of Factorial Experiments with Defects or Defectives as the Response, by S. Bisgaard and H.T. Fuller, Quality Engineering, Vol. 7, 994-5, pp. 49-443.) Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Error A -.86.9584 4.385 Model B -.35.45.6655 Error C -.65.464.78 Model D -.995 5.9 3.5977 Error E.55..497 Model F -.45 3.55 34.8664 Model G -.45.6485.9654 Error H.5.5.358 Error J.75.35.448383 Error AD.65 5.456 8.54 Error AE.55..55 Error AF.885 3.53 4.6757 Error AH.75.5.3645 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error BC.755.653 3.35735 Model BG -.54 9.4864 4.63 Lenth's ME.4 Lenth's SME 4.34453 As with the square root transformation, factors D, F, and the BG interaction remain significant. DESIGN-EXPERT Plot F&T Normal plot A: A B: B C: C D: D E: E F: F G: G H: H J: J Norm al % probability 99 95 9 8 7 5 3 5 F D BG -.4 -.5 3 -.6 3.7.6 Effect Design Expert Output Response: F&T ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 5. 5. 5.73.95 significant B.4.4.4.6334 D 5.9 5.9 9..9 F 3.5 3.5 3.47.43 G.65.65.37.556 BG 9.49 9.49 5.43.4 Residual 7.47.75 Cor Total 67.46 5 The Model F-value of 5.73 implies the model is significant. There is only a.95% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..3 R-Squared.74 Mean.5 Adj R-Squared.67 C.V. 5.63 Pred R-Squared.3373 PRESS 44.7 Adeq Precision 7.86 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.5.33.78 3.5 B-B -.6.33 -.9.57. D-D -..33 -.73 -.6. F-F -..33 -.95 -.48. G-G -..33 -.94.53. BG -.77.33 -.5 -.34. Final Equation in Terms of Coded Factors: F&T = +.5 -.6 * B 8-73

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -. * D -. * F -. * G -.77 * B * G Final Equation in Terms of Actual Factors: F&T = +.55 -.65 * B -.9975 * D -.5 * F -.5 * G -.77 * B * G The following interaction plots appear as they did with the square root transformation; a slight u shape is observed in the residuals versus predicted plot. Normal plot of residuals Residuals vs. Predicted.675 99 Normal % probability 95 9 8 7 5 3 5 Residuals.465.5 -.9965 -. 7 5 -. 7 5 -.9965.5.465.675 -.8 3.76.35 3.94 5.53 Residual Predicted 8-3 An experiment is run in a semiconductor factory to investigate the effect of six factors on transistor 6 gain. The design selected is the IV shown below. Standard Run Order Order A B C D E F Gain - - - - - - 455 8 + - - - + - 5 3 5 - + - - + + 487 4 9 + + - - - + 596 5 3 - - + - + + 43 6 4 + - + - - + 48 7 - + + - - - 458 8 + + + - + - 549 9 5 - - - + - + 454 3 + - - + + + 57 - + - + + - 487 6 + + - + - - 596 3 - - + + + - 446 4 4 + - + + - - 473 5 7 - + + + - + 46 6 6 + + + + + + 563 8-74

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Use a normal plot of the effects to identify the significant factors. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 76 34 55.74 Model B 53.75 556. 7.6459 Model C -3.5 366.5 8.75637 Error D 3.75 56.5.34566 Error E 6.38766 Error F.75.5.9355 Model AB 6.75 86.5 6.8473 Model AC -8.5 7.5.653 Error AD -.75.5.53865 Error AE -3.5 49.7 Error AF 5.5.5.6375 Error BD.5.399 Error BF.5 5.5987 Error ABD 3.5 49.7 Error ABF -.5 5.5987 Lenth's ME 9.63968 Lenth's SME 9.57 DESIGN-EXPERT Plot Gain Normal plot A: A B: B C: C D: D E: E F: F Norm al % probability 99 95 9 8 7 5 3 5 C AC AB B A -3.5-3.6 9.87 49.44 76. Effect (b) Conduct appropriate statistical tests for the model identified in part (a). Design Expert Output Response: Gain ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4455. 5 89. 39.6 <. significant A 34. 34. 667.75 <. B 556.5 556.5 334. <. C 366.5 366.5 5.79 <. AB 86.5 86.5 8.7 <. AC 7.5 7.5 7.87.86 Residual 346. 34.6 Cor Total 48. 5 The Model F-value of 39.6 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. 8-75

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Std. Dev. 5.88 R-Squared.997 Mean 497.75 Adj R-Squared.9876 C.V..39 Pred R-Squared.9788 PRESS 885.76 Adeq Precision 44.49 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 497.75.47 494.47 5.3 A-A 38..47 34.7 4.8. B-B 6.87.47 3.6 3.5. C-C -5.3.47-8.4 -.85. AB 3.38.47. 6.65. AC -4..47-7.4 -.85. Final Equation in Terms of Coded Factors: Gain = +497.75 +38. * A +6.87 * B -5.3 * C +3.38 * A * B -4. * A * C Final Equation in Terms of Actual Factors: Gain = +497.75 +38. * A +6.875 * B -5.5 * C +3.375 * A * B -4.5 * A * C (c) Analyze the residuals and comment on your findings. The residual plots are acceptable. The normality and equality of variance assumptions are verified. There does not appear to be any trends or interruptions in the residuals versus run order plot. Normal plot of residuals Residuals vs. Predicted.75 99 Normal % probability 95 9 8 7 5 3 5 Residuals 6.5.5-3. 5-7.7 5-7.7 5-3. 5.5 6.5.75 435.5 475.5 55.5 555.5 595.5 Residual Predicted 8-76

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Run Residuals vs. A.75.75 6.5 6.5 Residuals.5 Residuals.5-3. 5-3. 5-7.7 5-7.7 5 4 7 3 6 - Run Number A Residuals vs. B Residuals vs. C.75.75 6.5 6.5 Residuals.5 Residuals.5-3. 5-3. 5-7.7 5-7.7 5 - - B C Residuals vs. D Residuals vs. E.75.75 6.5 6.5 Residuals.5 Residuals.5-3. 5-3. 5-7.7 5-7.7 5 - - D E 8-77

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. F.75 6.5 Residuals.5-3. 5-7.7 5 - F (d) Can you find a set of operating conditions that produce gain of 5 5? Yes, see the graphs below. DESIGN-EXPERT Plot Gain 596 Interaction Graph B DESIGN-EXPERT Plot Gain 596 Interaction Graph C X = A: A Y = B: B X = A: A Y = C: C B- -. B+. Actual Factors C: C =. D: D =. E: E =. F: F =. 554.5 Gain 53 C- -. C+. Actual Factors B: B =. D: D =. E: E =. F: F =. 554.5 Gain 53 47.5 47.5 43 43 -. -.5..5. -. -.5..5. A A 8-78

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Overlay Plot X = A: A Y = B: B. Overlay Plot Actual Factors C: C = -. D: D =. E: E =. F: F =..5 B. Gain: 55 Gain: 475 -.5 -. -. -.5..5. A 8-3 Heat treating is often used to carbonize metal parts, such as gears. The thickness of the carbonized layer is a critical output variable from this process, and it is usually measured by performing a carbon 6 analysis on the gear pitch (top of the gear tooth). Six factors were studied on a IV design: A = furnace temperature, B = cycle time, C = carbon concentration, D = duration of the carbonizing cycle, E = carbon concentration of the diffuse cycle, and F = duration of the diffuse cycle. The experiment is shown below: Standard Run Order Order A B C D E F Pitch 5 - - - - + - 74 7 + - - - - - 9 3 8 - + - - - + 33 4 + + - - + + 7 5 - - + - - + 5 6 + - + - + + 7 6 - + + - + - 54 8 + + + - - - 44 9 6 - - - + + + 9 + - - + - + 88 4 - + - + - - 35 3 + + - + + - 7 3 - - + + - - 6 4 3 + - + + + - 75 5 5 - + + + + + 6 6 4 + + + + - + 93 (a) Estimate the factor effects and plot them on a normal probability plot. Select a tentative model. Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A 5.5 4.8777 Error B - 4.64 Model C -3 676.7755 Model D 37 5476.484 Model E 34.5 476 9.545 Error F 4.5 8.3356 Error AB -4 64.6737 Error AC -.5 5.63 Error AD 4 64.6737 Error AE 4.64 8-79

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error BD 4.5 8.3356 Model CD 4.5 84 3.455 Model DE - 936 7.94778 Error ABD.5.456 Error ABF 6 44.5957 Lenth's ME 5.435 Lenth's SME 3.39 Factors A, C, D, E and the two factor interactions CD and DE appear to be significant. The model can be found in the Design Expert Output below. DESIGN-EXPERT Plot Pitch Normal plot A: A B: B C: C D: D E: E F: F Norm al % probability 99 95 9 8 7 5 3 5 DE C CD D E A -. -3.8 8 4.5 3.38 5.5 Effect (b) Perform appropriate statistical tests on the model. Design Expert Output Response: Pitch ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 389. 6 398.83 76.57 <. significant A.. 96.7 <. C 676. 676. 3..57 D 5476. 5476. 5.3 <. E 476. 476. 9.56 <. CD 84. 84. 6.7.3 DE 936. 936. 37.3. Residual 468. 9 5. Cor Total 4359. 5 The Model F-value of 76.57 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 7. R-Squared.988 Mean 35.75 Adj R-Squared.968 C.V. 5.3 Pred R-Squared.9393 PRESS 479. Adeq Precision 8.68 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 35.75.8 3.67 39.83 A-A 5.5.8.7 9.33. C-C -6.5.8 -.58 -.4. D-D 8.5.8 4.4.58. 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY E-E 7.5.8 3.7.33. CD 7.5.8 3.7.33. DE -..8-5.8-6.9. Final Equation in Terms of Coded Factors: Pitch = +35.75 +5.5 * A -6.5 * C +8.5 * D +7.5 * E +7.5 * C * D -. * D * E Final Equation in Terms of Actual Factors: Pitch = +35.75 +5.5 * A -6.5 * C +8.5 * D +7.5 * E +7.5 * C * D -. * D * E (c) Analyze the residuals and comment on model adequacy. The residual plots are acceptable. The normality and equality of variance assumptions are verified. There does not appear to be any trends or interruptions in the residuals versus run order plot. The plots of the residuals versus factors C and E identify reduced variation at the lower level of both variables while the plot of residuals versus factor F identifies reduced variation at the upper level. Because C and E are significant factors in the model, this might not affect the decision on the optimum solution for the process. However, factor F is not included in the model and may be set at the upper level to reduce variation. Normal plot of residuals Residuals vs. Predicted 8.5 99 Norm al % probability 95 9 8 7 5 3 5 Residuals 3.5 -. 5-7.6 5-3 - 3-7.6 5 -. 5 3.5 8.5 5. 84.3 8.5 5.38 86.5 Residual Predicted 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Run Residuals vs. A 8.5 8.5 3.5 3.5 Residuals -. 5 Residuals -. 5-7.6 5-7.6 5-3 - 3 4 7 3 6 - Run Number A Residuals vs. B Residuals vs. C 8.5 8.5 3.5 3.5 Residuals -. 5 Residuals -. 5 3-7.6 5-7.6 5-3 - 3 - - B C Residuals vs. D Residuals vs. E 8.5 8.5 3.5 3.5 Residuals -. 5 Residuals -. 5-7.6 5-7.6 5-3 - 3 - - D E 8-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. F 8.5 3.5 Residuals -. 5 3-7.6 5-3 - F (d) Interpret the results of this experiment. Assume that a layer thickness of between 4 and 6 is desirable. The graphs below identify a region that is acceptable between 4 and 6. DESIGN-EXPERT Plot Pitch 93 Interaction Graph D DESIGN-EXPERT Plot Pitch 93 Interaction Graph E X = C: C Y = D: D X = D: D Y = E: E D- -. D+. Actual Factors A: A =. B: B =. E: E =. F: F =. 58.5 Pitch 3.5 E- -. E+. Actual Factors A: A =. B: B =. C: C =. F: F =. 58.5 Pitch 3.5 88.75 88.75 54 54 -. -.5..5. -. -.5..5. C D 8-83

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Overlay Plot X = C: C Y = D: D Coded Factors A: A =.5 B: B =. E: E =. F: F =...5 Overlay Plot Pitch: 6 D. Pitch: 4 -.5 -. -. -.5..5. C 8-3 Five factors are studied in the irregular fractional factorial design of resolution V shown below. Standard Run Order Order A B C D E Gain - - - - - 6.33 - + - - - 8.43 3 5 + + - - - 7.7 4 4 - - + - - 6.95 5 5 + - + - - 4.58 6 9 - + + - - 9. 7 6 - - - + - 8.96 8 7 + - - + - 3.56 9 8 + + - + - 9.5 3 + - + + - 5.74 3 - + + + -.73 + + + + -.5 3 - - - - + 5.58 4 + - - - +.3 5 9 + + - - + 6.78 6 + - + - + 3.39 7 - + + - + 8.63 8 6 + + + - + 9. 9 3 - - - + + 7.96 8 - + - + +.49 4 + + - + + 9.3 7 - - + + + 7.6 3 + - + + + 6.3 4 4 - + + + +.4 (a) Analyze the data from this experiment. What factors influence the response y? Design Expert Output Term Effect SumSqr % Contribtn Model Intercept Model A.95 5.8959.736 Model B 5.375 7.94 37.77 Model C -4.597 3.79.993 Model D.35 7.853 6.438 Error E -.475.996338.693 Model AB.4548.6896.88 Model AC -3.7585 8.845 8.355 Error AD -.8843.48.63 8-84

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error AE.337.768.75 Error BC.4887 7.5E-5.668E-5 Error BD.337.74.7494 Error BE.8664.774.5744 Error CD -.8458.9983.937 Error CE.9468.5833.5367 Error DE.9898.55.338 Error ABC -.3639.6433.584844 Error ABD.6736.75.6344 Error ABE Aliased Error ACD.89835.65.478947 Error ACE Aliased Error ADE.6.65.3844 Error BCD.5534.444.3985 Error BCE.89846.484.78 Error BDE.4848..54 Error CDE.573.3785.837783 Lenth's ME.45535 Lenth's SME.88839 Factors A, B, C, D, and the AB and AC interactions appear to be significant. DESIGN-EXPERT Plot y Normal plot A: A B: B C: C D: D E: E Normal % probability 99 95 9 8 7 5 3 5 AC D AB A B C -4. 6 -.7 9.58.96 5.33 Effect Design Expert Output Response: y ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 447.8 6 74.63 346.86 <. significant A 5.9 5.9 36.54 <. B 85.9 85.9 399.3 <. C 7.86 7.86 39.3 <. D 7.9 7.9 6.8 <. AB.69.69 58.98 <. AC 8.85 8.85 385. <. Residual 3.66 7. Cor Total 45.46 3 The Model F-value of 346.86 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..46 R-Squared.999 Mean 9.97 Adj R-Squared.989 C.V..3 Pred R-Squared.983 8-85

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY PRESS 7.6 Adeq Precision 6.974 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 9.97.95 9.77.7 A-A.46.95.6.66. B-B...79..3 C-C -.8. -.3 -.6.3 D-D.7.95.87.7. AB.77..56.98. AC -.97. -.8 -.76. Final Equation in Terms of Coded Factors: y = +9.97 +.46 * A +. * B -.8 * C +.7 * D +.77 * A * B -.97 * A * C Final Equation in Terms of Actual Factors: y = +9.97458 +.4565 * A +.687 * B -.85 * C +.665 * D +.775 * A * B -.976 * A * C DESIGN-EXPERT Plot y 9.3 Interaction Graph B DESIGN-EXPERT Plot y 9.3 Interaction Graph C X = A: A Y = B: B X = A: A Y = C: C B- -. B+. Actual Factors C: C =. D: D =. E: E =. y 5.33.35 C- -. C+. Actual Factors B: B =. D: D =. E: E =. y 5.33.35 7.37 7.37 3.39 3.39 -. -.5..5. -. -.5..5. A A 8-86

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot y 9.3 One Factor Plot X = D: D Actual Factors A: A =. B: B =. C: C =. E: E =. y 5.33.35 7.37 3.39 -. -.5..5. D (b) Analyze the residuals. Comment on model adequacy. The residual plots are acceptable. The normality and equality of variance assumptions are verified. There does not appear to be any trends or interruptions in the residuals versus run order plot. Normal plot of residuals Residuals vs. Predicted.786667 99 Norm al % probability 95 9 8 7 5 3 5.3779 Residuals -.4583 -.4 6 4 6 -.8 77 8 3 -.87783 -.4646 -.4583.3779.786667 3.79 7.6.43 5.5 9.7 Residual Predicted 8-87

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals vs. Run Residuals vs. A.786667.786667.3779.3779 Residuals -.4583 Residuals -.4583 -.4646 -.4 6 4 6 -.87783 -.8 77 8 3 4 7 3 6 9 - Run Number A Residuals vs. B Residuals vs. C.786667.786667.3779.3779 Residuals -.4583 Residuals -.4583 -.4646 -.4 6 4 6 -.87783 -.8 77 8 3 - - B C Residuals vs. D Residuals vs. E.786667.786667.3779.3779 Residuals -.4583 Residuals -.4583 -.4646 -.4 6 4 6 -.87783 -.8 77 8 3 - - D E 8-88

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 9 Three-Level and Mixed-Level Factorial and Fractional Factorial Design Solutions 9- The effects of developer concentration (A) and developer time (B) on the density of photographic plate film are being studied. Three strengths and three times are used, and four replicates of a 3 factorial experiment are run. The data from this experiment follow. Analyze the data using the standard methods for factorial experiments. Development Time (minutes) Developer Concentration 4 8 % 3 5 5 4 4 4 6 % 4 6 6 8 9 7 5 7 7 8 5 4% 7 8 7 8 7 9 8 Design Expert Output Response: Data ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4. 8 8.3.66 <. significant A 98. 99. 37.69 <. B.7.36 4.3.36 AB 3.8 4.8.3.8677 Residual 7. 7.63 Lack of Fit. Pure Error 7. 7.63 Cor Total 95. 35 The Model F-value of.66 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Concentration and time are significant. The interaction is not significant. By letting both A and B be treated as numerical factors, the analysis can be performed as follows: Design Expert Output Response: Data ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model. 5 44. 7.87 <. significant A 9.67 9.67 77.88 <. B.4.4 8.9.56 A 5.56 5.56.5.444 B.68.68.8.638 AB.6.6.5.8748 Residual 74. 3.47 Lack of Fit 3. 3.7.4.7488 not significant Pure Error 7. 7.63 Cor Total 95. 35 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The Model F-value of 7.87 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. 9- Compute the I and J components of the two-factor interaction in Problem 9-. B 7 A 8 3 3 35 39 77 78 7 6 36 AB Totals = 77, 78, 7; SS AB. 39 IAB AB Totals = 78, 74, 74; SS AB. 89 J AB SS AB I 78 74 AB J AB 74 3. 8 6 36 9-3 An experiment was performed to study the effect of three different types of 3-ounce bottles (A) and three different shelf types (B) -- smooth permanent shelves, end-aisle displays with grilled shelves, and beverage coolers -- on the time it takes to stock ten -bottle cases on the shelves. Three workers (factor C) were employed in this experiment, and two replicates of a 3 3 factorial design were run. The observed time data are shown in the following table. Analyze the data and draw conclusions. Replicate I Replicate Worker Bottle Type Permanent EndAisle Cooler Permanent EndAisle Cooler Plastic 3.45 4.4 5.8 3.36 4.9 5.3 8-mm glass 4.7 4.38 5.48 3.5 4.6 4.85 38-mm glass 4. 4.6 5.67 3.68 4.37 5.58 Plastic 4.8 5. 6. 4.4 4.7 5.88 8-mm glass 4.5 5.5 6.5 4.44 4.65 6. 38-mm glass 4.96 5.7 6.3 4.39 4.75 6.38 3 Plastic 4.8 3.94 5.4 3.65 4.8 4.49 8-mm glass 4.3 4.53 4.99 4.4 4.8 4.59 38-mm glass 4.7 4.86 4.85 3.88 4.48 4.9 Design Expert Output Response: Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 8.38 6.9 3.6 <. significant A.33.6.95.68 B 7.9 8.95 7. <. C 7.9 3.96 47.33 <. AB. 4.7.33.8583 AC. 4.7.3.8638 BC.59 4.4 4.76.49 ABC.43 8.53.64.738 Residual.6 7.84 Lack of Fit. Pure Error.6 7.84 Cor Total 3.64 53 The Model F-value of 3.6 implies the model is significant. There is only 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, C, BC are significant model terms. Factors B and C, shelf type and worker, and the BC interaction are significant. For the shortest time regardless of worker chose the permanent shelves. This can easily be seen in the interaction plot below. DESIGN-EXPERT Plot Time 6.56 Interaction Graph Shelf Type X = C: Worker Y = B: Shelf Type Design Points 5.73 B Permanent B End Aisle B3 Cooler 4.948 Actual Factor A: Bottle Type = 8mm glass Tim e 4.54 3.36 3 Worker 9-4 A medical researcher is studying the effect of lidocaine on the enzyme level in the heart muscle of beagle dogs. Three different commercial brands of lidocaine (A), three dosage levels (B), and three dogs (C) are used in the experiment, and two replicates of a 3 3 factorial design are run. The observed enzyme levels follow. Analyze the data from this experiment. Replicate I Replicate Lidocaine Dosage Dog Dog Brand Strength 3 3 86 84 85 84 85 86 94 99 98 95 97 9 3 6 98 5 4 3 85 84 86 8 8 84 95 98 97 93 99 95 3 8 4 9 3 84 83 8 83 8 79 95 97 93 9 96 93 3 5 6 8 Design Expert Output Response: Enzyme Level ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 449.33 6 7.7 6.99 <. significant A 3. 5.5.5.359 B 46.78 3.39 9.55 <. C 8. 4..38.695 AB 69.56 4 7.39.7.768 AC 3.33 4.83.8.987 BC 36.89 4 9..9.4738 ABC 6.78 8 7.6.75.65 9-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residual Lack of Fit 74.5. 7.7 Pure Error 74.5 7.7 Cor Total 4764.83 53 The Model F-value of 6.99 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. The dosage is significant. 9-5 Compute the I and J components of the two-factor interactions for Example 9-. A 34 88 44 B -55-348 -89 76 7 88 I totals = 74,75,6 J totals = -8,3,-8 I(AB) = 6.78 J(AB) = 674. SS AB = 63.9 A -9-58 - C 399 3 394 6-5 -4 I totals = -,34,-77 J totals = 5,4,- I(AC) = 6878.78 J(AC) = 635. SS AC = 753.9 B -93-35 -6 C -55-33 533-4 -39 74 I totals = -5,79,38 J totals =-53,87,3 I(BC) = 473. J(BC) = 858.34 SS BC = 854.34 9-6 An experiment is run in a chemical process using a 3 factorial design. The design factors are temperature and pressure, and the response variable is yield. The data that result from this experiment are shown below. Pressure, psig Temperature, C 4 8 47.58, 48.77 64.97, 69. 8.9, 7.6 9 5.86, 8.43 88.47, 84.3 93.95, 88.54 7.8, 9.77 96.57, 88.7 76.58, 83.4 (a) Analyze the data from this experiment by conducting an analysis of variance. What conclusions can you draw? 9-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 387.3 8 398.39 4.37.5 significant A 96.93 548.47 6..9 B 53.56 75.78 8.5.9 AB 586.64 4 46.66.6.536 Pure Error 89.98 9 9. Cor Total 47. 7 The Model F-value of 4.37 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. Temperature and pressure are significant. Their interaction is not. An alternate analysis is performed below with the A and B treated as numeric factors: Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 373.7 5 64.65 7.9.7 significant A 85.76 85.76.93.63 B 97.9 97.9 6.68.5 A 46.8 46.8 3.6.6 B 5.64 5.64.64.3 AB 47.78 47.78 6.8.98 Residual 933.83 77.8 Lack of Fit 3.86 3 37.95.4.7454 not significant Pure Error 89.98 9 9. Cor Total 47. 7 The Model F-value of 7.9 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, AB are significant model terms. (b) Graphically analyze the residuals. Are there any concerns about underlying assumptions or model adequacy? 9-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 5.85 99 Normal % probability 95 9 8 7 5 3 5 Residuals 7.645-7.64 5-5.85-5. 85-7.645 7.645 5.85 48.8 59.9 7.4 8.53 9.65 Residual Predicted Residuals vs. Pressure Residuals vs. Temperature 5.85 5.85 7.645 7.645 Residuals Residuals -7.64 5-7.64 5-5.85-5.85 3 3 Pressure Temperature The plot of residuals versus pressure shows a decreasing funnel shape indicating a non-constant variance. (c) Verify that if we let the low, medium and high levels of both factors in this experiment take on the levels -,, and +, then a least squares fit to a second order model for yield is y 86. 8. 4x 8. 4x 7. 7x 7. 86x 7. 69x x The coefficients can be found in the following table of computer output. Design Expert Output Final Equation in Terms of Coded Factors: Yield = +86.8 +8.4 * A +.4 * B -7.84 * A -7.7-7.69 * B * A * B 9-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (d) Confirm that the model in part (c) can be written in terms of the natural variables temperature (T) and pressure (P) as y 335. 638. 56T8. 59P. 7T. 96P. 384TP The coefficients can be found in the following table of computer output. Design Expert Output Final Equation in Terms of Actual Factors: Yield = -335.65 +8.58737 * Pressure +8.5585 * Temperature -.96 * Pressure -.77 * Temperature -.38437 * Pressure * Temperature (e) Construct a contour plot for yield as a function of pressure and temperature. Based on the examination of this plot, where would you recommend running the process.. Yield 85 95. B: Temperature 9. 85 8 75 85. 65 7 55 6 5 8.... 3. 4. 9 A: Pres s ure Run the process in the oval region indicated by the yield of 9. 9-7 (a) Confound a 3 3 design in three blocks using the ABC component of the three-factor interaction. Compare your results with the design in Figure 9-7. L = X + X + X 3 Block Block Block 3 9-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The new design is a 8 rotation around the Factor B axis. (b) Confound a 3 3 design in three blocks using the AB C component of the three-factor interaction. Compare your results with the design in Figure 9-7. L = X + X + X 3 Block Block Block 3 The new design is a 8 rotation around the Factor C axis. (c) Confound a 3 3 design three blocks using the ABC component of the three-factor interaction. Compare your results with the design in Figure 9-7. L = X + X + X 3 Block Block Block 3 The new design is a 9 rotation around the Factor C axis along with switching layer and layer in the C axis. (d) After looking at the designs in parts (a), (b), and (c) and Figure 9-7, what conclusions can you draw? All four designs are relatively the same. The only differences are rotations and swapping of layers. 9-8 Confound a 3 4 design in three blocks using the AB CD component of the four-factor interaction. 9-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY L = X + X + X 3 + X 4 Block Block Block 3 9-9 Consider the data from the first replicate of Problem 9-3. Assuming that all 7 observations could not be run on the same day, set up a design for conducting the experiment over three days with AB C confounded with blocks. Analyze the data. Block Block Block 3 = 3.45 = 4.7 = 4. = 4.38 = 4.6 = 4.4 = 5. = 4.4 = 5.7 = 4.3 = 4.7 = 4.8 = 4.96 = 4.8 = 4.5 = 4.86 = 3.94 = 4.53 = 6.5 = 4.99 = 6. = 5.4 = 6.3 = 4.85 = 5.67 = 5.8 = 5.48 Totals-> = 44.3 43. 43.8 Design Expert Output Response: Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block.3. Model 3.7 8.73 4.7.44 significant A.48.4.4.873 B 8.9 4.46 6.. C.57.78 4.57.6 AB.3 4.33.9.84 AC.87 4..7.3774 BC.45 4..66.64 Residual.3 6.7 Cor Total 4.43 6 The Model F-value of 4.7 implies the model is significant. There is only a 4.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. 9-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 9- Outline the analysis of variance table for the 3 4 design in nine blocks. Is this a practical design? Source DF A B C D AB 4 AC 4 AD 4 BC 4 BD 4 CD 4 ABC (AB C,ABC,AB C ) 6 ABD (ABD,AB D,ABD ) 6 ACD (ACD,ACD,AC D ) 6 BCD (BCD,BC D,BCD ) 6 ABCD 6 Blocks (ABC,AB C,AC D,BC D ) 8 Total 8 Any experiment with 8 runs is large. Instead of having three full levels of each factor, if two levels of each factor could be used, then the overall design would have 6 runs plus some center points. This twolevel design could now probably be run in or 4 blocks, with center points in each block. Additional curvature effects could be determined by augmenting the experiment with the axial points of a central composite design and additional enter points. The overall design would be less than 8 runs. 9- Consider the data in Problem 9-3. If ABC is confounded in replicate I and ABC is confounded in replicate II, perform the analysis of variance. L = X + X + X 3 L = X + X + X Block Block Block 3 Block Block Block 3 = 3.45 = 4.8 = 4.8 = 3.36 = 3.5 = 3.68 = 5.5 = 4.53 = 4.38 = 4.44 = 4.39 = 4.4 = 4.85 = 5.67 = 6.3 = 4.7 = 4.65 = 4.75 = 5.48 = 6.5 = 4.99 = 6.38 = 5.88 = 6. = 4.3 = 4.7 = 4.5 = 3.88 = 3.65 = 4.4 = 4.6 = 5.7 = 4.86 = 4.49 = 4.59 = 4.9 = 4.96 = 4.7 = 4. = 4.85 = 5.58 = 5.3 = 3.94 = 4.4 = 5. = 4.37 = 4.9 = 4.6 = 6. = 5.4 = 5.8 = 4.8 = 4.48 = 4.8 The sums of squares for A, B, C, AB, AC, and BC are calculated as usual. The only sums of squares presenting difficulties with calculations are the four components of the ABC interaction (ABC, ABC, AB C, and AB C ). ABC is computed using replicate I and ABC is computed using replicate II. AB C and AB C are computed using data from both replicates. We will show how to calculate AB C and AB C from both replicates. Form a two-way table of A x B at each level of C. Find the I(AB) and J(AB) totals for each third of the A x B table. A C B I J 6.8 7.59 7.88 6.7 7.55 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 8.33 8.64 8.63 7.5 7.7.3.33.5 6.54 5.77 9. 8.96 9.35 3.4 3.5 9.9 9.8 9.9 3.97 3.9.9.45.4 3.7 3.57 7.73 8.34 8.5 6.9 6.9 8. 8.6 9.34 7.3 6. 9.63 9.58 9.75 5.65 6.65 The I and J components for each third of the above table are used to form a new table of diagonal totals. C I(AB) J(AB).67 7.5 6.54 7.55 7.7 5.77 3.4 3.97 3.7 3.5 3.9 3.57 6.9 7.3 5.65 6.9 6. 6.65 I Totals: I Totals: 85.6,85.6,83.3 85.99,85.3,83. J Totals: J Totals: 85.73,83.6,84.3 83.35,85.6,85.3 Now, AB C = I[C x I(AB)] = (85. 6 ) (85. 6 ) (83. 3 ) ( 5364. ) 65. 8 54 and, AB C = J[C x I(AB)]= (85. ) (83. ) (84. ) 73 6 3 ( 5364. ) 37. 8 54 If it were necessary, we could find ABC as ABC = I[C x J(AB)] and ABC as J[C x J(AB)]. However, these components must be computed using the data from the appropriate replicate. The analysis of variance table: Source SS DF MS F Replicates.6696 Blocks within Replicates.38 4 A.44.5 5. B 7.754 8.8757 7. C 7.663 3.836 93.68 AB.6 4.9 < AC.93 4.73 < BC.679 4.498.6 ABC (rep I).45.6 < ABC (rep II)..5.5 AB C.37.754.6 AB C.65.633.55 Error.8998.49 Total 3.369 53 9- Consider the data from replicate I in Problem 9-3. Suppose that only a one-third fraction of this design with I=ABC is run. Construct the design, determine the alias structure, and analyze the design. 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The design is,,,,,,,,. The alias structure is: A = BC = AB C B = AC = AB C C = AB = ABC AB = AC = BC C A B 3.45 5.48 4.6 6. 5.5 4.96 3.94 4.3 4.85 Source SS DF A.5 B.3 C.8 AB.3 Total 5.66 8 9-3 From examining Figure 9-9, what type of design would remain if after completing the first 9 runs, one of the three factors could be dropped? A full 3 factorial design results. 9-4 Construct a 3 4 IV The 7 runs for this design are as follows: design with I=ABCD. Write out the alias structure for this design. A = AB C D = BCD B = AB CD = ACD C = ABC D = ABD D = ABCD = ABC AB = ABC D = CD AB = AC D = BC D AC = AB CD = BD AC = AB D = BC D BC = AB C D = AD BC = AB D = AC D BD = AB C = ACD CD = ABC = ABCD AD = AB C = BCD 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 9-5 Verify that the design in Problem 9-4 is a resolution IV design. The design in Problem 9-4 is a Resolution IV design because no main effect is aliased with a component of a two-factor interaction, but some two-factor interaction components are aliased with each other. 9-6 Construct a 3 5- design with I=ABC and I=CDE. Write out the alias structure for this design. What is the resolution of this design? The complete defining relation for this design is : I = ABC = CDE = ABC DE = ABD E This is a resolution III design. The defining contrasts are L = X + X + X 3 and L = X 3 + X 4 + X 5. To find the alias of any effect, multiply the effect by I and I. For example, the alias of A is: A = AB C = ACDE = AB CDE = AB DE = BC = AC D E = BC DE = BD E 9-7 Construct a 3 9-6 design, and verify that is a resolution III design. Use the generators I = AC D, I = AB C E, I = BC F, I = AB CG, I = ABCH, and I = ABJ To find the alias of any effect, multiply the effect by I and I. For example, the alias of C is: C = C(BC F ) = BF, At least one main effect is aliased with a component of a two-factor interaction. 9-8 Construct a 4 x 3 design confounded in two blocks of 6 observations each. Outline the analysis of variance for this design. Design is a 4 x 3, with ABC at two levels, and Z at 4 levels. Represent Z with two pseudo-factors D and E as follows: 9-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Pseudo- Factors Z D E Z = () Z = d Z 3 = e Z 4 = de The 4 x 3 is now a 5 in the factors A, B, C, D and E. Confound ABCDE with blocks. We have given both the letter notation and the digital notation for the treatment combinations. Block Block () = a = ab = b = ac = c = bc = abc = abcd = bcd = abce = bce = cd = acd = ce = ace = de = 3 ade = 3 abde = 3 bde = 3 bcde = 3 abcde = 3 be = abd = ad = abe = ae = d = acde = 3 e = bd = cde = 3 Source DF A B C Z (D+E+DE) 3 AB AC AZ (AD+AE+ADE) 3 BC BZ (BD+BE+BDE) 3 CZ (CD+CE+CDE) 3 ABC ABZ (ABD+ABE+ABDE) 3 ACZ (ACD+ACE+ACDE) 3 BCZ (BCD+BCE+BCDE) 3 ABCZ (ABCD+ABCE) Blocks (or ABCDE) Total 3 9-9 Outline the analysis of variance table for a 3 factorial design. Discuss how this design may be confounded in blocks. 9-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Suppose we have n replicates of a 3 factorial design. A and B are at levels, and C and D are at 3 levels. Source DF Components for Confounding A A B B C C D D AB AB AC AC AD AD BC BD BD CD,CD CD 4 ABC ABC ABD ABD ACD,ACD ACD 4 BCD,BCD BCD 4 ABCD,ABCD ABCD 4 Error 36(n-) Total 36n- Confounding in this series of designs is discussed extensively by Margolin (967). The possibilities for a single replicate of the 3 design are: blocks of 8 observations 6 blocks of 6 observations 3 blocks of observations 9 blocks of 4 observations 4 blocks of 9 observations For example, one component of the four-factor interaction, say ABCD, could be selected to confound the design in 3 blocks for observations each, while to confound the design in blocks of 8 observations 3 each we would select the AB interaction. Cochran and Cox (957) and Anderson and McLean (974) discuss confounding in these designs. 9- Starting with a 6-run 4 design, show how two three-level factors can be incorporated in this experiment. How many two-level factors can be included if we want some information on two-factor interactions? Use column A and B for one three-level factor and columns C and D for the other. Use the AC and BD columns for the two, two-level factors. The design will be of resolution V. 9- Starting with a 6-run 4 design, show how one three-level factor and three two-level factors can be accommodated and still allow the estimation of two-factor interactions. Use columns A and B for the three-level factor, and columns C and D and ABCD for the three two-level factors. This design will be of resolution V. 9- In Problem 9-6, you met Harry and Judy Peterson-Nedry, two friends of the author who have a winery and vineyard in Newberg, Oregon. That problem described the application of two-level fractional factorial designs to their 985 Pinor Noir product. In 987, they wanted to conduct another Pinot Noir experiment. The variables for this experiment were Variable Clone of Pinot Noir Berry Size Levels Wadenswil, Pommard Small, Large 9-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Fermentation temperature 8F, 85F, 9/8F, 9F Whole Berry None, % Maceration Time days, days Yeast Type Assmanhau, Champagne Oak Type Troncais, Allier Harry and Judy decided to use a 6-run two-level fractional factorial design, treating the four levels of fermentation temperature as two two-level variables. As in Problem 8-6, they used the rankings from a taste-test panel as the response variable. The design and the resulting average ranks are shown below: Berry Ferm. Whole Macer. Yeast Oak Average Run Clone Size Temp. Berry Time Type Type Rank - - - - - - - - 4 + - - - - + + + 3 - + - - + - + + 6 4 + + - - + + - - 9 5 - - + - + + + - 6 + - + - + - - + 7 - + + - - + - + 5 8 + + + - - - + - 5 9 - - - + + + - + + - - + + - + - - + - + - + + - 6 + + - + - - - + 3 3 - - + + - - + + 8 4 + - + + - + - - 4 5 - + + + + - - - 7 6 + + + + + + + + 3 (a) Describe the aliasing in this design. The design is a resolution IV design such that the main effects are aliased with three factor interactions. Design Expert Output Term Aliases Intercept ABCG ABDH ABEF ACDF ACEH ADEG AFGH BCDE BCFH BDFG BEGH CDGH CEFG DEFH A BCG BDH BEF CDF CEH DEG FGH ABCDE B ACG ADH AEF CDE CFH DFG EGH C ABG ADF AEH BDE BFH DGH EFG D ABH ACF AEG BCE BFG CGH EFH E ABF ACH ADG BCD BGH CFG DFH F ABE ACD AGH BCH BDG CEG DEH G ABC ADE AFH BDF BEH CDH CEF H ABD ACE AFG BCF BEG CDG DEF AB CG DH EF ACDE ACFH ADFG AEGH BCDF BCEH BDEG BFGH AC BG DF EH ABDE ABFH ADGH AEFG BCDH BCEF CDEG CFGH AD BH CF EG ABCE ABFG ACGH AEFH BCDG BDEF CDEH DFGH AE BF CH DG ABCD ABGH ACFG ADFH BCEG BDEH CDEF EFGH AF BE CD GH ABCH ABDG ACEG ADEH BCFG BDFH CEFH DEFG AG BC DE FH ABDF ABEH ACDH ACEF BDGH BEFG CDFG CEGH AH BD CE FG ABCF ABEG ACDG ADEF BCGH BEFH CDFH DEGH (b) Analyze the data and draw conclusions. All of the main effects except Yeast and Oak are significant. The Macer Time is the most significant. None of the interactions were significant. 9-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Rank Normal plot A: Clone B: Berry Size C: Ferm T emp D: Ferm T emp E: Wh o le B erry F: Macer Time G: Yeast H: Oak Normal % probability 99 95 9 8 7 5 3 5 A E D C B F -.7 5 -. 6.63 5.3 8. Effect Design Expert Output Response: Rank ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 38.75 6 54.79 43.83 <. significant A 3.5 3.5 4..8 B 9. 9. 7..5 C 9. 9. 7..5 D.5.5 9.8. E.5.5 9.8. F 56. 56. 4.8 <. Residual.5 9.5 Cor Total 34. 5 The Model F-value of 43.83 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev.. R-Squared.9669 Mean 8.5 Adj R-Squared.9449 C.V. 3.5 Pred R-Squared.8954 PRESS 35.56 Adeq Precision 9.7 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 8.5.8 7.87 9.3 A-Clone -.38.8 -. -.74. B-Berry Size.75.8..38. C-Ferm Temp.75.8..38. D-Ferm Temp.88.8.4.5. E-Whole Berry -.87.8 -.5 -.4. F-Macer Time 4..8 3.37 4.63. Final Equation in Terms of Coded Factors: Rank = +8.5 -.38 * A +.75 * B +.75 * C +.88 * D -.87 * E +4. * F 9-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) What comparisons can you make between this experiment and the 985 Pinot Noir experiment from Problem 8-6? The experiment from Problem 8-6 indicates that yeast, barrel, whole cluster and the clone x yeast interactions were significant. This experiment indicates that maceration time, whole berry, clone and fermentation temperature are significant. 9-3 An article by W.D. Baten in the 956 volume of Industrial Quality Control described an experiment to study the effect of three factors on the lengths of steel bars. Each bar was subjected to one of two heat treatment processes, and was cut on one of four machines at one of three times during the day (8 am, am, or 3 pm). The coded length data are shown below (a) Analyze the data from this experiment assuming that the four observations in each cell are replicates. The Machine effect (C) is significant, the Heat Treat Process (B) is also significant, while the Time of Day (A) is not significant. None of the interactions are significant. Time of Day 8am am 3 pm Heat Treat Process Machine 3 4 6 9 7 9 6 6 3 5 5 4 7 3 4 6 6 5-4 5 3 4 5 4 6 3 8 7 3 7 9-4 8 6 3 6 4 9 4-3 - 6 3 5 4-5 9 6 6 4 6 4 8 6 8 7-4 3 3 7 4-4 7 Design Expert Output Response: Length ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59.33 3 5.67 4.3 <. significant A 6.7 3.4..83 B 4.67 4.67 6.86.7 C 393.4 3 3.4. <. AB 3.77.89.9.55 AC 4.5 6 7..3.3537 BC 3.8 3 4.36.7.554 ABC 48.98 6 8.6.3.63 Pure Error 447.5 7 6. Cor Total 37.83 95 The Model F-value of 4.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..49 R-Squared.5688 Mean 3.96 Adj R-Squared.43 9-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY C.V. 6.98 Pred R-Squared.334 PRESS 795.56 Adeq Precision 7. Coefficient Standard 95% CI 95% CI Term Estimate DF Error Low High VIF Intercept 3.96.5 3.45 4.47 A[]..36 -.7.73 A[] -.65.36 -.36.7 B-Process -.67.5 -.7 -.6. C[] -.54.44 -.4.34 C[].9.44.4.8 C[3] -3.8.44-3.96 -. A[]B..36 -.7.73 A[]B.6.36 -..3 A[]C[].3.6 -.9.57 A[]C[] -.7.6 -.5 -.8 A[]C[] -.39.6 -.63.86 A[]C[] -..6 -.35.4 A[]C[3].4.6 -..48 A[]C[3].77.6 -.47. BC[] -.5.44 -.3.63 BC[] -.46.44 -.34.4 BC[3].46.44 -.4.34 A[]BC[] -.94.6 -.34.5 A[]BC[] -.44.6 -.68.8 A[]BC[]..6 -.3.36 A[]BC[] -..6 -.35.4 A[]BC[3] -.43.6 -.67.8 A[]BC[3].6.6 -.64.85 Final Equation in Terms of Coded Factors: Length = +3.96 +. * A[] -.65 * A[] -.67 * B -.54 * C[] +.9 * C[] -3.8 * C[3] +. * A[]B +.6 * A[]B +.3 * A[]C[] -.7 * A[]C[] -.39 * A[]C[] -. * A[]C[] +.4 * A[]C[3] +.77 * A[]C[3] -.5 * BC[] -.46 * BC[] +.46 * BC[3] -.94 * A[]BC[] -.44 * A[]BC[] +. * A[]BC[] -. * A[]BC[] -.43 * A[]BC[3] +.6 * A[]BC[3] (b) Analyze the residuals from this experiment. Is there any indication that there is an outlier in one cell? If you find an outlier, remove it and repeat the analysis from part (a). What are your conclusions? Standard Order 84, Time of Day at 3:pm, Heat Treat #, Machine #, and length of, appears to be an outlier. 9-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 4.5 99 Normal % probability 95 9 8 7 5 3 5 Residuals.85 -.8 75-3.5 6 5 3 3 3 3-6. 5-6. 5-3.5 6 5 -.8 75.85 4.5 -.5.69 3.88 6.6 8.5 Residual Predicted The following analysis was performed with the outlier described above removed. As with the original analysis, Machine is significant and Heat Treat Process is also significant, but now Time of Day, factor A, is also significant with an F-value of 3.5 (the P-value is just above.5). Design Expert Output Response: Length ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 66.58 3 7.4 4.89 <. significant A 34.3 7. 3.6.533 B 33.6 33.6 5.94.73 C 4.89 3 37.3 4.65 <. AB 6.4 8..47.36 AC 5.9 6 8.37.5.9 BC 8.38 3.79.5.684 ABC 67. 6.7..76 Pure Error 395.4 7 5.57 Cor Total. 94 The Model F-value of 4.89 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..36 R-Squared.63 Mean 4. Adj R-Squared.4878 C.V. 59. Pred R-Squared.3 PRESS 75.7 Adeq Precision 7.447 Coefficient Standard 95% CI 95% CI Term Estimate DF Error Low High VIF Intercept 4.5.4 3.z56 4.53 A[] -.76.34 -.76.6 A[] -.73.34 -.4 -.5 B-Process -.58.4 -.6 -.96. C[] -.63.4 -.46. C[].8.43.33 3.3 C[3] -3.7.4-4. -.34 A[]B -.76.34 -.76.6 A[]B.5.34 -.6. A[]C[].4.59 -.77.59 A[]C[] -.8.59 -.36-6.78E-3 A[]C[] -.65.6 -.83.54 A[]C[] -.36.6 -.55.8 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A[]C[3] A[]C[3].33.86.59.59 -.85 -.3.5.4 BC[] -.34.4 -.7.5 BC[] -..43 -.5.65 BC[3].37.4 -.46. A[]BC[] -6.944E-3.59 -.8.7 A[]BC[] -.35.59 -.53.83 A[]BC[] -.5.6 -.33.4 A[]BC[] -.36.6 -.55 -.8 A[]BC[3] -.34.59 -.5.84 A[]BC[3].69.59 -.49.87 Final Equation in Terms of Coded Factors: Length = +4.5 -.76 * A[] -.73 * A[] -.58 * B -.63 * C[] +.8 * C[] -3.7 * C[3] -.76 * A[]B +.5 * A[]B +.4 * A[]C[] -.8 * A[]C[] -.65 * A[]C[] -.36 * A[]C[] +.33 * A[]C[3] +.86 * A[]C[3] -.34 * BC[] -. * BC[] +.37 * BC[3] -6.944E-3 * A[]BC[] -.35 * A[]BC[] -.5 * A[]BC[] -.36 * A[]BC[] -.34 * A[]BC[3] +.69 * A[]BC[3] The following residual plots are acceptable. Both the normality and constant variance assumptions are satisfied Normal plot of residuals Residuals vs. Predicted 4.5 99 Norm al % probability 95 9 8 7 5 3 5 Residuals.375.5 -.8 75 3 3 3 3-4 -4 -.8 75.5.375 4.5 -.5.7 3.9 6. 8.33 Residual Predicted 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Suppose that the observations in the cells are the lengths (coded) of bars processed together in heat treating and then cut sequentially (that is, in order) on the three machines. Analyze the data to determine the effects of the three factors on mean length. The analysis with all effects and interactions included: Design Expert Output Response: Length ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 47.58 3 6.4 A 6.57 3.8 B.67.67 C 98.35 3 3.78 AB 5.94.97 AC.54 6.76 BC 3.7 3.9 ABC.4 6.4 Pure Error. Cor Total 47.58 3 The by removing the three factor interaction from the model and applying it to the error, the analysis identifies factor C as being significant and B as being mildly significant. Design Expert Output Response: Length ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 35.34 7 7.96 3.9.5 not significant A 6.57 3.8.6.757 B.67.67 5.3.63 C 98.35 3 3.78 6.6.8 AB 5.94.97.46.35 AC.54 6.76.86.57 BC 3.7 3.9.53.6756 Residual.4 6.4 Cor Total 47.58 3 When removing the remaining insignificant factors from the model, C, Machine, is the most significant factor while B, Heat Treat Process, is moderately significant. Factor A, Time of Day, is not significant. Design Expert Output Response: Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 9. 4 7.6 3.43 <. significant B.67.67 5.6.335 C 98.35 3 3.78 6.5 <. Residual 38.56 9.3 Cor Total 47.58 3 The Model F-value of 3.43 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..4 R-Squared.7387 Mean 3.96 Adj R-Squared.6837 C.V. 35.99 Pred R-Squared.583 PRESS 6.53 Adeq Precision 9.74 9-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Term Coefficient Estimate DF Standard Error 95% CI Low 95% CI High VIF Intercept 3.96.9 3.35 4.57 B-Process -.67.9 -.8 -.58. C[] -.54.5 -.6.5 C[].9.5.86.97 C[3] -3.8.5-4.4 -.3 Final Equation in Terms of Coded Factors: Avg = +3.96 -.67 * B -.54 * C[] +.9 * C[] -3.8 * C[3] The following residual plots are acceptable. Both the normality and uniformity of variance assumptions are verified. Normal plot of residuals Residuals vs. Predicted.9667 99 Normal % probability 95 9 8 7 5 3 5 Residuals.9375 -.46667 -.83 - - -.83 -.46667.9375.9667..79 3.38 4.96 6.54 Residual Predicted (d) Calculate the log variance of the observations in each cell. Analyze the average length and the log variance of the length for each of the bars cut at each machine/heat treatment process combination. What conclusions can you draw? Factor B, Heat Treat Process, has an affect on the log variance of the observations while Factor A, Time of Day, and Factor C, Machine, are not significant at the 5 percent level. However, A is significant at the percent level, so Tome of Day has some effect on the variance. Design Expert Output Response: Log(Var) ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.79.5.5.648 not significant A.58.9.86.966 B.5.5 4.89.47 C.59 3..95.757 AB.49.4.4.34 BC.64 3...538 Residual.. 9-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Cor Total 4. 3 The Model F-value of.5 implies there is a 6.48% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..3 R-Squared.6967 Mean.65 Adj R-Squared.486 C.V. 49. Pred R-Squared -.33 PRESS 4.86 Adeq Precision 5.676 Coefficient Standard 95% CI 95% CI Term Estimate DF Error Low High VIF Intercept.65.65.5.79 A[] -.54.9 -.5.5 A[] -.6.9 -.36.43 B-Process.4.65.8E-3.9. C[].. -.5.47 C[].66. -.8.3 C[3] -.9. -.44.5 A[]B -..9 -.4 3.37E-3 A[]B.4.9 -.65.34 BC[] -.8. -.4.68 BC[] -.5. -.39.98 BC[3].4. -..39 Final Equation in Terms of Coded Factors: Log(Var) = +.65 -.54 * A[] -.6 * A[] +.4 * B +. * C[] +.66 * C[] -.9 * C[3] -. * A[]B +.4 * A[]B -.8 * BC[] -.5 * BC[] +.4 * BC[3] The following residual plots are acceptable. Both the normality and uniformity of variance assumptions are verified. Normal plot of residuals Residuals vs. Predicted.3446 99 Norm al % probability 95 9 8 7 5 3 5.56 Residuals -.37556 -. 6 4 -.4 4 9 7 -.4497 -.64 -.37556.56.3446 -...5.84.6 Residual Predicted 9-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (e) Suppose the time at which a bar is cut really cannot be controlled during routine production. Analyze the average length and the log variance of the length for each of the bars cut at each machine/heat treatment process combination. What conclusions can you draw? The analysis of the average length is as follows: Design Expert Output Response: Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 37.43 7 5.35 A 3.56 3.56 B 3.78 3.93 AB.9 3.36 Pure Error. Cor Total 37.43 7 Because the Means Square of the AB interaction is much less than the main effects, it is removed from the model and placed in the error. The average length is strongly affected by Factor B, Machine, and moderately affected by Factor A, Heat Treat Process. The interaction effect was small and removed from the model. Design Expert Output Response: Avg ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 36.34 4 9.9 5.. significant A 3.56 3.56 9.78.5 B 3.78 3.93 3.7.97 Residual.9 3.36 Cor Total 37.43 7 The Model F-value of 5. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..6 R-Squared.979 Mean 3.96 Adj R-Squared.93 C.V. 5.3 Pred R-Squared.799 PRESS 7.75 Adeq Precision 3.89 Coefficient Standard 95% CI 95% CI Term Estimate DF Error Low High VIF Intercept 3.96. 3.8 4.64 A-Process -.67. -.34.. B[] -.54.37 -.7.63 B[].9.37.74 3.9 B[3] -3.8.37-4.6 -.9 Final Equation in Terms of Coded Factors: Avg = +3.96 -.67 * A -.54 * B[] +.9 * B[] -3.8 * B[3] The following residual plots are acceptable. Both the normality and uniformity of variance assumptions are verified. 9-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted.458333 99 Normal % probability 95 9 8 7 5 3 5.967 Residuals -. 9 6 7 -.4 58 3 3 3 -.458333 -.967.967.458333..79 3.38 4.96 6.54 Residual Predicted The Log(Var) is analyzed below: Design Expert Output Response: Log(Var) ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3 7.46 A.9.9 B.3 3.44 AB.98 3.33 Pure Error. Cor Total.3 7 Because the AB interaction has the smallest Mean Square, it was removed from the model and placed in the error. From the following analysis of variance, neither Heat Treat Process, Machine, nor the interaction affect the log variance of the length. Design Expert Output Response: Log(Var) ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model. 4.56.7.344 not significant A.9.9.8.96 B.3 3.44.34.47 Residual.98 3.33 Cor Total.3 7 The "Model F-value" of.7 implies the model is not significant relative to the noise. There is a 34.4 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..8 R-Squared.6949 Mean.79 Adj R-Squared.88 C.V..9 Pred R-Squared -.693 PRESS.69 Adeq Precision 3.99 Coefficient Standard 95% CI 95% CI Term Estimate DF Error Low High VIF Intercept.79.64.59.99 A-Process..64 -.96.3. 9-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B[] B[].5.3.. -. -.3.5.38 B[3] -.. -.55.5 Final Equation in Terms of Coded Factors: Log(Var) = +.79 +. * A +.5 * B[] +.3 * B[] -. * B[3] The following residual plots are acceptable. Both the normality and uniformity of variance assumptions are verified. Normal plot of residuals Residuals vs. Predicted.6958 99 Normal % probability 95 9 8 7 5 3 5.8479 Residuals -.8479 -. 6 9 5 8 -.6958 -.8479.8479.6958.48.6.76.9.5 Residual Predicted 9-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter Fitting Regression Models Solutions - The tensile strength of a paper product is related to the amount of hardwood in the pulp. Ten samples are produced in the pilot plant, and the data obtained are shown in the following table. Strength Percent Hardwood Strength Percent Hardwood 6 8 7 5 88 5 75 5 93 5 8 95 8 84 3 (a) Fit a linear regression model relating strength to percent hardwood. Minitab Output Regression Analysis: Strength versus Hardwood The regression equation is Strength = 44 +.88 Hardwood Predictor Coef SE Coef T P Constant 43.84.5 57.4. Hardwood.8786.65 6.. S =.3 R-Sq = 97.% R-Sq(adj) = 96.6% PRESS = 66.665 R-Sq(pred) = 94.9% Regression Plot Strength = 43.84 +.87864 Hardwood S =.3 R-Sq = 97. % R-Sq(adj) = 96.6 % 9 Strength 8 7 6 Hardwood 3 (b) Test the model in part (a) for significance of regression. Minitab Output -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Analysis of Variance Source DF SS MS F P Regression 6. 6. 6.. Residual Error 8 38.8 4.9 Lack of Fit 4 3.7 3.4.54.76 Pure Error 4 5. 6.3 Total 9 3.9 3 rows with no replicates No evidence of lack of fit (P >.) (c) Find a 95 percent confidence interval on the parameter. The 95 percent confidence interval is: ˆ t, n p se ˆ ˆ t seˆ, n p.65.8786.36.65.8786.36.69.473 - A plant distills liquid air to produce oxygen, nitrogen, and argon. The percentage of impurity in the oxygen is thought to be linearly related to the amount of impurities in the air as measured by the pollution count in part per million (ppm). A sample of plant operating data is shown below. Purity(%) 93.3 9. 9.4 9.7 94. 94.6 93.6 93. 93. 9.9 9. 9.3 9. 9.6 9.9 Pollution count (ppm)..45.36.59.8.75..99.83..47.8.3.75.68 (a) Fit a linear regression model to the data. Minitab Output Regression Analysis: Purity versus Pollution The regression equation is Purity = 96.5 -.9 Pollution Predictor Coef SE Coef T P Constant 96.4546.48 5.4. Pollutio -.9.356-9.49. S =.477 R-Sq = 87.4% R-Sq(adj) = 86.4% PRESS = 3.43946 R-Sq(pred) = 8.77% -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Regression Plot Purity = 96.4546 -.996 Pollution S =.47745 R-Sq = 87.4 % R-Sq(adj) = 86.4 % 95 94 Purity 93 9 9 9..5 Pollution. (b) Test for significance of regression. Minitab Output Analysis of Variance Source DF SS MS F P Regression 6.49 6.49 9.3. Residual Error 3.379.83 Total 4 8.869 No replicates. Cannot do pure error test. No evidence of lack of fit (P >.) (c) Find a 95 percent confidence interval on. The 95 percent confidence interval is: ˆ t, n p se ˆ ˆ t seˆ, n p.356 -.9.64.356 -.9.64 3.56.48-3 Plot the residuals from Problem - and comment on model adequacy. -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot of the Residuals (response is Strength) Normal Score - -3 - - Residual 3 Residuals Versus the Fitted Values (response is Strength) 3 Residual - - -3 6 7 8 Fitted Value 9-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Order of the Data (response is Strength) 3 Residual - - -3 3 4 5 6 Observation Order 7 8 9 There is nothing unusual about the residual plots. The underlying assumptions have been met. -4 Plot the residuals from Problem - and comment on model adequacy. Normal Probability Plot of the Residuals (response is Purity) Normal Score - - -. -.5 Residual..5-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Purity).5 Residual. -.5 -. 9.5 9.5 9.5 Fitted Value 93.5 94.5 Residuals Versus the Order of the Data (response is Purity).5 Residual. -.5 -. 4 6 8 Observation Order 4 There is nothing unusual about the residual plots. The underlying assumptions have been met. -5 Using the results of Problem -, test the regression model for lack of fit. Minitab Output Analysis of Variance Source DF SS MS F P Regression 6. 6. 6.. Residual Error 8 38.8 4.9-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Lack of Fit 4 3.7 3.4.54.76 Pure Error 4 5. 6.3 Total 9 3.9 3 rows with no replicates No evidence of lack of fit (P >.) -6 A study was performed on wear of a bearing y and its relationship to x = oil viscosity and x = load. The following data were obtained. y x x 93.6 85 3 5.5 86 7. 58 9 43. 3 33. 357 5 4. 5 (a) Fit a multiple linear regression model to the data. Minitab Output Regression Analysis: Wear versus Viscosity, Load The regression equation is Wear = 35 -.7 Viscosity -.54 Load Predictor Coef SE Coef T P VIF Constant 35.99 74.75 4.7.8 Viscosit -.7.69 -.9.356.6 Load -.539.8953 -.7.84.6 S = 5.5 R-Sq = 86.% R-Sq(adj) = 77.% PRESS = 696.7 R-Sq(pred) =.3% (b) Test for significance of regression. Minitab Output Analysis of Variance Source DF SS MS F P Regression 6.6 68.8 9.35.5 Residual Error 3 95.4 65. Total 5 4. No replicates. Cannot do pure error test. Source DF Seq SS Viscosit 4.4 Load 9. * Not enough data for lack of fit test (c) Compute t statistics for each model parameter. What conclusions can you draw? Minitab Output Regression Analysis: Wear versus Viscosity, Load The regression equation is Wear = 35 -.7 Viscosity -.54 Load Predictor Coef SE Coef T P VIF Constant 35.99 74.75 4.7.8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Viscosit -.7.69 -.9.356.6 Load -.539.8953 -.7.84.6 S = 5.5 R-Sq = 86.% R-Sq(adj) = 77.% PRESS = 696.7 R-Sq(pred) =.3% The t-tests are shown in part (a). Notice that overall regression is significant (part(b)), but neither variable has a large t-statistic. This could be an indicator that the regressors are nearly linearly dependent. -7 The brake horsepower developed by an automobile engine on a dynomometer is thought to be a function of the engine speed in revolutions per minute (rpm), the road octane number of the fuel, and the engine compression. An experiment is run in the laboratory and the data that follow are collected. Brake Horsepower rpm Road Octane Number Compression 5 9 8 94 95 9 4 88 9 9 96 9 6 86 78 5 96 46 3 94 98 37 3 9 33 8 88 5 4 34 86 97 3 8 9 3 5 89 4 (a) Fit a multiple linear regression model to the data. Minitab Output Regression Analysis: Horsepower versus rpm, Octane, Compression The regression equation is Horsepower = - 66 +.7 rpm + 3.3 Octane +.87 Compression Predictor Coef SE Coef T P VIF Constant -66.3 9.67 -.87. rpm.73.4483.39.44. Octane 3.348.8444 3.7.6. Compress.8674.5345 3.49.8. S = 8.8 R-Sq = 8.7% R-Sq(adj) = 73.4% PRESS = 494.5 R-Sq(pred) =.33% (b) Test for significance of regression. What conclusions can you draw? Minitab Output Analysis of Variance Source DF SS MS F P Regression 3 589.73 863.4..3 Residual Error 8 6.7 77.66 Total 3. r No replicates. Cannot do pure error test. Source DF Seq SS rpm 59.35 Octane 3.56 Compress 947.83 Lack of fit test Possible interactions with variable Octane (P-Value =.8) Possible lack of fit at outer X-values (P-Value =.) Overall lack of fit test is significant at P =. -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Based on t tests, do you need all three regressor variables in the model? Yes, all of the regressor variables are important. -8 Analyze the residuals from the regression model in Problem -7. Comment on model adequacy. Normal Probability Plot of the Residuals (response is Horsepow) Normal Score - - - Residual Residuals Versus the Fitted Values (response is Horsepow) Residual - 3 4 Fitted Value 5 6 7-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Order of the Data (response is Horsepow) Residual - 4 6 Observation Order 8 The normal probability plot is satisfactory, as is the plot of residuals versus run order (assuming that observation order is run order). The plot of residuals versus predicted response exhibits a slight bow shape. This could be an indication of lack of fit. It might be useful to consider adding some ineraction terms to the model. -9 The yield of a chemical process is related to the concentration of the reactant and the operating temperature. An experiment has been conducted with the following results. Yield Concentration Temperature 8. 5 89. 8 83. 5 9. 8 79. 5 87. 8 84. 5 9. 8 (a) Suppose we wish to fit a main effects model to this data. Set up the X X matrix using the data exactly as it appears in the table.. 5. 8. 5. 8. 5. 8. 5. 8........ 5 8 5 8 8 5 3 8 5 8 98 3 98 96 (b) Is the matrix you obtained in part (a) diagonal? Discuss your response. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The X X is not diagonal, even though an orthogonal design has been used. The reason is that we have worked with the natural factor levels, not the orthogonally coded variables. (c) Suppose we write our model in terms of the usual coded variables.5 x Conc,.5 x 65 Temp 5 Set up the X X matrix for the model in terms of these coded variables. Is this matrix diagonal? Discuss your response. 8 8 8 The X X matrix is diagonal because we have used the orthogonally coded variables. (d) Define a new set of coded variables. x Conc,. x 5 Temp 3 Set up the X X matrix for the model in terms of this set of coded variables. Is this matrix diagonal? Discuss your response. 8 4 4 4 4 4 4 The X X is not diagonal, even though an orthogonal design has been used. The reason is that we have not used orthogonally coded variables. (e) Summarize what you have learned from this problem about coding the variables. If the design is orthogonal, use the orthogonal coding. This not only makes the analysis somewhat easier, but it also results in model coefficients that are easier to interpret because they are both dimensionless and uncorrelated. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - Consider the 4 factorial experiment in Example 6-. Suppose that the last observation in missing. Reanalyze the data and draw conclusions. How do these conclusions compare with those from the original example? The regression analysis with the one data point missing indicates that the same effects are important. Minitab Output Regression Analysis: Rate versus A, B, C, D, AB, AC, AD, BC, BD, CD The regression equation is Rate = 69.8 +.5 A +.5 B + 4.63 C + 7. D -.5 AB - 9.38 AC + 8. AD +.87 BC -.5 BD -.87 CD Predictor Coef SE Coef T P VIF Constant 69.75.5 46.5. A.5.5 7... B.5.5.83.45. C 4.65.5 3.8.37. D 7..5 4.67.. AB -.5.5 -.7.876. AC -9.375.5-6.5.3. AD 8..5 5.33.6. BC.875.5.58.59. BD -.5.5 -.33.756. CD -.875.5 -.58.59. S = 5.477 R-Sq = 97.6% R-Sq(adj) = 9.6% PRESS = 75. R-Sq(pred) = 65.9% Analysis of Variance Source DF SS MS F P Regression 4893.33 489.33 6.3.8 Residual Error 4. 3. Total 4 53.33 No replicates. Cannot do pure error test. Source DF Seq SS A 44.4 B 4. C 6.86 D 758.88 AB.6 AC 5.63 AD 94.5 BC 6.7 BD.7 CD. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot of the Residuals (response is Rate) Normal Score - - -5 Residual 5 Residuals Versus the Fitted Values (response is Rate) 5 Residual -5 4 5 6 7 Fitted Value 8 9 Residuals Versus the Order of the Data (response is Rate) 5 Residual -5 4 6 8 Observation Order 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The residual plots are acceptable; therefore, the underlying assumptions are valid. - Consider the 4 factorial experiment in Example 6-. Suppose that the last two observations are missing. Reanalyze the data and draw conclusions. How do these conclusions compare with those from the original example? The regression analysis with the one data point missing indicates that the same effects are important. Minitab Output Regression Analysis: Rate versus A, B, C, D, AB, AC, AD, BC, BD, CD The regression equation is Rate = 7.4 +. A +.87 B + 6.5 C + 8.6 D -.66 AB - 9.78 AC + 7.59 AD +.5 BC +. BD +.75 CD Predictor Coef SE Coef T P VIF Constant 7.375.673 4.66. A.94.33 7.63.5. B.875.673.7.84.7 C 6.5.673 3.74.33.7 D 8.65.673 5.5.4.7 AB -.656.33 -.5.654. AC -9.78.33-7.39.5. AD 7.594.33 5.74.. BC.5.673.49.3.7 BD.5.673.67.549.7 CD.75.673.45.684.7 S = 4.73 R-Sq = 98.7% R-Sq(adj) = 94.% PRESS = 493.6 R-Sq(pred) = 7.% Analysis of Variance Source DF SS MS F P Regression 4943.7 494.3.7.4 Residual Error 3 67.9.4 Total 3 5.36 No replicates. Cannot do pure error test. Source DF Seq SS A 543.5 B.5 C 77.63 D 76. AB.7 AC 7.53 AD 738. BC 4.9 BD 6. CD 4.5-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot of the Residuals (response is Rate) Normal Score - - -3 - - Residual 3 Residuals Versus the Fitted Values (response is Rate) 3 Residual - - -3 4 5 6 7 Fitted Value 8 9 Residuals Versus the Order of the Data (response is Rate) 3 Residual - - -3 4 6 8 Observation Order 4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The residual plots are acceptable; therefore, the underlying assumptions are valid. - Given the following data, fit the second-order polynomial regression model y x x x x xx y x x 6.. 4.. 75.5 4. 6.5 4. 63.5 4. 55.5. 6.5..5 3. 6..5 3.5.5 7..5 7.5.5 After you have fit the model, test for significance of regression. Minitab Output Regression Analysis: y versus x, x, x^, x^, xx The regression equation is y = 4.4-38. x +.7 x + 35. x^ +. x^ - 9.99 xx Predictor Coef SE Coef T P VIF Constant 4.4 6.59.9.394 x -38.3 4.45 -.94.383 89.6 x.7.69.6.953 5. x^ 34.98.56.6.56 3.9 x^.66 3.58 3.5.3 4.7 xx -9.986 8.74 -.4.97 5. S = 6.4 R-Sq = 99.4% R-Sq(adj) = 98.9% PRESS = 37.7 R-Sq(pred) = 96.4% r Analysis of Variance Source DF SS MS F P Regression 5 359.6 78.5 9.3. Residual Error 6 9. 36.5 Lack of Fit 3 9. 3.4.7.67 Pure Error 3 8. 4.7 Total 353.7 7 rows with no replicates Source DF Seq SS x 55. x 95.3 x^.9 x^ 5.8 xx 47.6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot of the Residuals (response is y) Normal Score - - -5 Residual 5 Residuals Versus the Fitted Values (response is y) 5 Residual -5 7 Fitted Value 7 Residuals Versus the Order of the Data (response is y) 5 Residual -5 4 6 Observation Order 8-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3 (a) Consider the quadratic regression model from Problem -. Compute t statistics for each model parameter and comment on the conclusions that follow from the quantities. Minitab Output Predictor Coef SE Coef T P VIF Constant 4.4 6.59.9.394 x -38.3 4.45 -.94.383 89.6 x.7.69.6.953 5. x^ 34.98.56.6.56 3.9 x^.66 3.58 3.5.3 4.7 xx -9.986 8.74 -.4.97 5. x is the only model parameter that is statistically significant with a t-value of 3.5. A logical model might also include x to preserve model hierarchy. (b) Use the extra sum of squares method to evaluate the value of the quadratic terms, the model. x, x and x x to The extra sum of squares due to is SS R SS,, SS, SS SS,, R R R, SS sum of squares of regression for the model in Problem - = 359.6 R SS R F SS R =345.3, 359.6 345.3 59. 3 SS 3 59.3 3 36.5 R, MSE 5.389 Since F.5,3,6 4. 76, then the addition of the quadratic terms to the model is significant. The P-values indicate that it s probably the term x that is responsible for this. R -4 Relationship between analysis of variance and regression. Any analysis of variance model can be expressed in terms of the general linear model y = X +, where the X matrix consists of zeros and ones. Show that the single-factor model y, i=,,3, j=,,3,4 can be written in general linear model form. Then ij i ij (a) Write the normal equations the model in Chapter 3. ( XX ) ˆ Xy and compare them with the normal equations found for The normal equations are ( XX ) ˆ Xy -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 4 4 4 4 4 4 4 ˆ y ˆ y ˆ y 4 ˆ 3 y.... 3. which are in agreement with the results of Chapter 3. (b) Find the rank of XX. Can ( XX ) be obtained? XXis a 4 x 4 matrix of rank 3, because the last three columns add to the first column. Thus (X X) - does not exist. (c) Suppose the first normal equation is deleted and the restriction n ˆ i i is added. Can the resulting system of equations be solved? If so, find the solution. Find the regression sum of squares ˆXy, and compare it to the treatment sum of squares in the single-factor model. 3 Imposing n ˆi yields the normal equations i 3 4 4 4 4 4 4 4 4ˆ y ˆ y ˆ y 4ˆ 3 y.... 3. The solution to this set of equations is y.. ˆ y ˆ y This solution was found be solving the last three equations for ˆ i, yielding substituting in the first equation to find ˆ y.. The regression sum of squares is a i SS R ˆ X y =.. i... y y y.... i. y.. i y i..... y an a i y n y an ˆ i y i ˆ, and then with a degrees of freedom. This is the same result found in Chapter 3. For more discussion of the relationship between analysis of variance and regression, see Montgomery and Peck (99). a i y i. n. -5 Suppose that we are fitting a straight line and we desire to make the variance of as small as possible. Restricting ourselves to an even number of experimental points, where should we place these points so as to minimize V ˆ? (Note: Use the design called for in this exercise with great caution because, even though it minimized V ˆ, it has some undesirable properties; for example, see Myers and -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - Montgomery (995). Only if you are very sure the true functional relationship is linear should you consider using this design. Since S xx V ˆ, we may minimize ˆ V by making S xx as large as possible. S xx is maximized by spreading out the x j s as much as possible. The experimenter usually has a region of interest for x. If n is even, n/ of the observations should be run at each end of the region of interest. If n is odd, then run one of the observations in the center of the region and the remaining (n-)/ at either end. -6 Weighted least squares. Suppose that we are fitting the straight line x y, but the variance of the y s now depends on the level of x; that is, n i w x y V i i,...,,, where the w i are known constants, often called weights. Show that if we choose estimates of the regression coefficients to minimize the weighted sum of squared errors given by n i i i i x y w, the resulting least squares normal equations are n i n i i i i n i i i y w x w w ˆ ˆ n i n i i i i i n i i i i y x w x w x w ˆ ˆ The least squares normal equations are found: ˆ ˆ ˆ ˆ n i i i n i i i n i i i w x x y L w x y L w x y L which simplify to i n i i n i n i i i i n i i n i n i i i y x w w x w x y w w x w ˆ ˆ ˆ ˆ -7 Consider the 4 IV design discussed in Example -5.

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - (a) Suppose you elect to augment the design with the single run selected in that example. Find the variances and covariances of the regression coefficients in the model (ignoring blocks): 4 3 34 4 4 3 3 x x x x x x x x y 9 7 7 9 9 9 9 9 9 X X'.4375.375.65.65.65.65.65.375.4375.65.65.65.65.65.65.65.5.65.65.5.65.65.5.65.65.5.65.65.5 ) ( X'X (b) Are there any other runs in the alternate fraction that Any other run from the alternate fraction will dealias AB from CD. (c) Suppose you augment the design with four runs suggested in Example -5. Find the variance and the covariances of the regression coefficients (ignoring blocks) for the model in part (a). Choose 4 runs that are one of the quarter fractions not used in the principal half fraction. X X'

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY X' X 4 4 4 4 4 4 4 4 X' X.833.7.79.536.357.938.33.33.938.79.7.536.357.536.536.4.49.357.357.49.785 (d) Considering parts (a) and (c), which augmentation strategy would you prefer and why? If you only have the resources to run one more run, then choose the one-run augmentation. But if resources are not scarce, then augment the design in multiples of two runs, to keep the design orthogonal. Using four runs results in smaller variances of the regression coefficients and a simpler covariance structure. 7 4-8 Consider the III. Suppose after running the experiment, the largest observed effects are A + BD, B + AD, and D + AB. You wish to augment the original design with a group of four runs to dealias these effects. (a) Which four runs would you make? Take the first four runs of the original experiment and change the sign on A. Design Expert Output Factor Factor Factor 3 Factor 4 Factor 5 Factor 6 Factor 7 Std Run Block A:x B:x C:x3 D:x4 E:x5 F:x6 G:x7 Block -. -. -.... -. Block. -. -. -. -... 3 3 Block -.. -. -.. -.. 4 4 Block.. -.. -. -. -. 5 5 Block -. -... -. -.. 6 6 Block. -.. -.. -. -. 7 7 Block -... -. -.. -. 8 8 Block....... 9 9 Block... -. -. -. -. Block. -. -.. -. -. -. Block -. -... -. -. -. Block -. -. -. -. -. -. -. Main effects and interactions of interest are: x x x4 xx xx4 xx4 - - - - -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (b) Find the variances and covariances of the regression coefficients in the model y x x 4 x4 xx 4 x x4 4 x x4 X' X 4 4 4 4 4 4 X' X.833.7.79.536.74.78.7.74.536.938.33.33.938.536.74.43.67.74.536.67.43 (c) Is it possible to dealias these effects with fewer than four additional runs? It is possible to dealias these effects in only two runs. By utilizing Design Expert s design augmentation function, the runs 9 and (Block ) were generated as follows: Design Expert Output Factor Factor Factor 3 Factor 4 Factor 5 Factor 6 Factor 7 Std Run Block A:x B:x C:x3 D:x4 E:x5 F:x6 G:x7 Block -. -. -.... -. Block. -. -. -. -... 3 3 Block -.. -. -.. -.. 4 4 Block.. -.. -. -. -. 5 5 Block -. -... -. -.. 6 6 Block. -.. -.. -. -. 7 7 Block -... -. -.. -. 8 8 Block....... 9 9 Block -.. -.. -. -. -. Block. -. -. -. -. -. -. -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter Response Surface Methods and Other Approaches to Process Optimization Solutions - A chemical plant produces oxygen by liquefying air and separating it into its component gases by fractional distillation. The purity of the oxygen is a function of the main condenser temperature and the pressure ratio between the upper and lower columns. Current operating conditions are temperature ) - C and pressure ratio ).. Using the following data find the path of steepest ascent. ( ( Temperature (x ) Pressure Ratio (x ) Purity -5. 8.8-5.3 83.5-5. 84.7-5.3 85. -. 84. -. 84.5 -. 83.9 -. 84.3 Design Expert Output Response: Purity ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.4.57 6.7.5 significant A.89.89 48.7.3 B.5.5 4.7.8 Curvature.8.8.33.35 not significant Residual.4 4.6 Lack of Fit.4.4.6.495 not significant Pure Error. 3.67 Cor Total 3.46 7 The Model F-value of 6.7 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..4 R-Squared.99 Mean 84. Adj R-Squared.8935 C.V..9 Pred R-Squared.73 PRESS. Adeq Precision.7 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 84.. 83.66 84.34 A-Temperature.85..5.9. B-Pressure Ratio.5. -.9.59. Center Point..7 -.8.68. Final Equation in Terms of Coded Factors: Purity = +84. +.85 * A +.5 * B Final Equation in Terms of Actual Factors: Purity = +8.4 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.7 * Temperature +.5 * Pressure Ratio From the computer output use the model ŷ 84. 85x. 5x as the equation for steepest ascent. Suppose we use a one degree change in temperature as the basic step size. Thus, the path of steepest ascent passes through the point (x =, x =) and has a slope.5/.85. In the coded variables, one degree of temperature is equivalent to a step of x /5=.. Thus, x (.5/.85).=.59. The path of steepest ascent is: Coded Variables Natural Variables x x Origin -...59.59 Origin +..59-9.59 Origin +5..95-5.95 Origin +7.4.43-3.43 - An industrial engineer has developed a computer simulation model of a two-item inventory system. The decision variables are the order quantity and the reorder point for each item. The response to be minimized is the total inventory cost. The simulation model is used to produce the data shown in the following table. Identify the experimental design. Find the path of steepest descent. Item Item Order Reorder Order Reorder Total Quantity (x) Point (x) Quantity (x3) Point (x4) Cost 5 5 4 65 4 45 5 4 67 4 5 3 4 663 4 5 5 8 654 45 3 4 648 45 5 8 634 5 3 8 69 4 45 3 8 686 35 75 6 68 35 75 6 674 35 75 6 68 The design is a 4- fractional factorial with generator I=ABCD, and three center points. Design Expert Output Response: Total Cost ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 388. 6 646.67 63.6.3 significant A 684.5 684.5 66.96.38 C 44.5 44.5 37.4.3 D 45. 45. 44..7 AC 39. 39. 38.35.85 AD 64.5 64.5 5.88.47 CD 684.5 684.5 66.96.38 Curvature 85.5 85.5 79.78.3 significant Residual 3.67 3. Lack of Fit...4.7446 not significant Pure Error 8.67 4.33 Cor Total 476.8 The Model F-value of 63.6 implies the model is significant. There is only -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3. R-Squared.99 Mean 664.7 Adj R-Squared.9765 C.V..48 Pred R-Squared.9593 PRESS 9.5 Adeq Precision 4.573 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 659..3 655.4 66.6 A-Item QTY 9.5.3 5.65.85. C-Item QTY 3.5.3 9.65 6.85. D-Item Reorder 7.5.3 3.9.. AC -7..3 -.6-3.4. AD -5.75.3-9.35 -.5. CD 9.5.3 5.65.85. Center Point 9.33.6.44 6.. Final Equation in Terms of Coded Factors: Total Cost = +659. +9.5 * A +3.5 * C +7.5 * D -7. * A * C -5.75 * A * D +9.5 * C * D Final Equation in Terms of Actual Factors: Total Cost = +75. +5.75 * Item QTY +. * Item QTY -.9875 * Item Reorder -.4 * Item QTY * Item QTY -.4375 * Item QTY * Item Reorder +.85 * Item QTY * Item Reorder +.9 * Item QTY * Item Reorder The equation used to compute the path of steepest ascent is ŷ 659 9. 5x 3. 5x3 7. 5x4. Notice that even though the model contains interaction, it is relatively common practice to ignore the interactions in computing the path of steepest ascent. This means that the path constructed is only an approximation to the path that would have been obtained if the interactions were considered, but it s usually close enough to give satisfactory results. It is helpful to give a general method for finding the path of steepest ascent. Suppose we have a first-order model in k variables, say ŷ ˆ k ˆ i x i i The path of steepest ascent passes through the origin, x=, and through the point on a hypersphere of radius, R where ŷ is a maximum. Thus, the x s must satisfy the constraint k i x i R To find the set of x s that maximize ŷ subject to this constraint, we maximize -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY k L ˆ ˆ i xi i i k x i R where is a LaGrange multiplier. From L / xi L /, we find ˆ i xi It is customary to specify a basic step size in one of the variables, say x j, and then calculate as = ˆ j / x j. Then this value of can be used to generate the remaining coordinates of a point on the path of steepest ascent. We demonstrate using the data from this problem. Suppose that we use - units in as the basic step size. Note that a decrease in is called for, because we are looking for a path of steepest decent. Now - units in is equal to -/ = -.5 units change in x. Thus, = ˆ / x = 9.5/(-.5) = -8.5 Consequently, x x ˆ 3 3. 5 76 8. 5 ˆ 4 7. 5 75 8. 5 3. 4. are the remaining coordinates of points along the path of steepest decent, in terms of the coded variables. The path of steepest decent is shown below: Coded Variables Natural Variables x x x 3 x 4 3 4 Origin 35 75 6 -.5 -.76 -.45 - -7.9-8. Origin + -.5 -.76 -.45 35 57.9 5.89 Origin + -. -.43 -.8 35 39.8 43.78-3 Verify that the following design is a simplex. Fit the first-order model and find the path of steepest ascent. Position x x x 3 y - 8.5-9.8 3 - - 7.4 4.5-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 x x 3 3 x The graphical representation of the design identifies a tetrahedron; therefore, the design is a simplex. Design Expert Output Response: y ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4.49 3 4.83 A 3.64 3.64 B.6.6 C.4.4 Pure Error. Cor Total 4.49 3 Std. Dev. R-Squared. Mean 9.55 Adj R-Squared C.V. Pred R-Squared N/A PRESS N/A Adeq Precision. Case(s) with leverage of.: Pred R-Squared and PRESS statistic not defined Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 9.55 A-x.35. B-x.55. C-x3.6. Final Equation in Terms of Coded Factors: y = +9.55 +.35 * A +.55 * B +.6 * C Final Equation in Terms of Actual Factors: y = +9.55 +.95459 * x +.3889 * x +.6 * x3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The first order model is ŷ 9. 55. 35x. 55x. 6x3. To find the path of steepest ascent, let the basic step size be x 3. Then using the results obtained in the previous problem, we obtain ˆ 3 x3 or. =.6 which yields. 6. Then the coordinates of points on the path of steepest ascent are defined by Therefore, in the coded variables we have: x x ˆ. 96 6. 6 ˆ. 4 4. 6.. Coded Variables x x x 3 Origin.6.4. Origin +.6.4. Origin +..48. -4 For the first-order model ŷ 6. 5x. 8x. x3 find the path of steepest ascent. The variables are coded as x. Let the basic step size be 3 ˆ 3 x3 or. = x ˆ. x ˆ. Therefore, in the coded variables we have i. 5 75... 8 4.. Then. Coded Variables x x x 3 Origin.75 -.4. Origin +.75 -.4. Origin +.5 -.8. -5 The region of experimentation for three factors are time ( 4 T 8 min), temperature ( T 3 C), and pressure ( P 5 psig). A first-order model in coded variables has been fit to yield data from a 3 design. The model is -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ŷ 3 5x x. 5x 3. 5 Is the point T = 85, T = 35, P=6 on the path of steepest ascent? The coded variables are found with the following: 6 T 5 x T P 35 x x 3 5 5 T 5 5 x. 5 ˆ x or.5 = ˆ. 5 x. 5 ˆ 3 3. 5 x3. 75 Coded Variables Natural Variables x x x 3 T T P Origin 6 5 35.5.5.75 5 6.5.65 Origin +.5.5.75 65 56.5 37.65 Origin +5.5.65.875 85 8.5 48.5 The point T =85, T =35, and P=6 is not on the path of steepest ascent. 3-6 The region of experimentation for two factors are temperature ( T 3 F) and catalyst feed rate ( C 3 lb/h). A first order model in the usual coded variables has been fit to a molecular weight response, yielding the following model. (a) Find the path of steepest ascent. ŷ 5x 4x T x x C T x ˆ 5 x or 5 ˆ 4 x. 3 5 Coded Variables Natural Variables x x T C Origin.3 3. -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Origin +.3 3 3. Origin +5 5.6 7 36. (a) It is desired to move to a region where molecular weights are above 5. Based on the information you have from the experiment, in this region, about how may steps along the path of steepest ascent might be required to move to the region of interest? ŷ x ˆ ˆ 5. 3437 8 x. # Steps 5 3. 63 4 37. 8-7 The path of steepest ascent is usually computed assuming that the model is truly first-order.; that is, there is no interaction. However, even if there is interaction, steepest ascent ignoring the interaction still usually produces good results. To illustrate, suppose that we have fit the model ŷ 5x x 8x 3x using coded variables (- x +) (a) Draw the path of steepest ascent that you would obtain if the interaction were ignored. Path of Steepest Ascent for Main Effects Model - - X -3-4 -5 3 4 5 X (b) Draw the path of steepest ascent that you would obtain with the interaction included in the model. Compare this with the path found in part (a). -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Path of Steepest Ascent for Full Model - - X -3-4 -5 - - 3 X -8 The data shown in the following table were collected in an experiment to optimize crystal growth as a function of three variables x, x, and x 3. Large values of y (yield in grams) are desirable. Fit a second order model and analyze the fitted surface. Under what set of conditions is maximum growth achieved? x x x 3 y - - - 66 - - 7 - - 78-6 - - 8-7 - 75 -.68.68 8 -.68 68.68 63 -.68 65.68 8 3 8 88 85 Design Expert Output Response: Yield ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 366. 9 46.89.9.94 not significant A.8.8..7377 B 5.3 5.3.4.7 C 3.5 3.5.6.694-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A 4.55 4.55..39 B 6.45 6.45.96.6 C 38.46 38.46 7.4.34 AB 66. 66..36.5644 AC 55.3 55.3.3.598 BC 7.3 7.3.9.36 Residual 86.95 86.9 Lack of Fit.6 5.3.7.4353 not significant Pure Error 859.33 5 7.87 Cor Total 55.95 9 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a.94 % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 3.64 R-Squared.663 Mean 83.5 Adj R-Squared.3598 C.V. 6.43 Pred R-Squared -.634 PRESS 8855.3 Adeq Precision 3.88 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.67 5.56 88.7 3.6 A-x.7 3.69-6.95 9.5. B-x.36 3.69-6.86 9.59. C-x3 -.49 3.69-9.7 6.73. A -3.77 3.59 -.77 4.4. B -.43 3.59 -.44-4.4. C -9.6 3.59-7.6 -.59. AB.87 4.8-7.87 3.6. AC -.63 4.8-3.37 8.. BC -4.63 4.8-5.37 6.. Final Equation in Terms of Coded Factors: Yield = +.67 +.7 * A +.36 * B -.49 * C -3.77 * A -.43 * B -9.6 * C +.87 * A * B -.63 * A * C -4.63 * B * C Final Equation in Terms of Actual Factors: Yield = +.6669 +.746 * x +.363 * x -.49445 * x3-3.76749 * x -.4955 * x -9.63 * x3 +.875 * x * x -.65 * x * x3-4.65 * x * x3 There are so many nonsignificant terms in this model that we should consider eliminating some of them. A reasonable reduced model is shown below. Design Expert Output -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Yield ANOVA for Response Surface Reduced Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 343. 4 785.75 4.95.95 significant B 5.3 5.3.6.695 C 3.5 3.5.9.6673 B 5.3 5.3 3.33.4 C 39.7 39.7 7.8.36 Residual 379.95 5 58.66 Lack of Fit 5.6 5.6.88.5953 not significant Pure Error 859.33 5 7.87 Cor Total 55.95 9 The Model F-value of 4.95 implies the model is significant. There is only a.95% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..6 R-Squared.569 Mean 83.5 Adj R-Squared.454 C.V. 5.7 Pred R-Squared.46 PRESS 4735.5 Adeq Precision 5.778 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 97.58 4.36 88.9 6.88 B-x.36 3.4-5.9 8.63. C-x3 -.49 3.4-8.76 5.77. B -.6 3.3-9.9-5.. C -9.3 3.3-6.6 -.9. Final Equation in Terms of Coded Factors: Yield = +97.58 +.36 * B -.49 * C -.6 * B -9.3 * C Final Equation in Terms of Actual Factors: Yield = +97.586 +.363 * x -.49445 * x3 -.5546 * x -9.73 * x3 The contour plot identifies a maximum near the center of the design space. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Yield X = B: x Y = C: x3. 8 Yield 85 8 Design Points Actual Factor A: x =. C: x3.5. 85 Prediction97.68 95% Low 69.73 95% High6.9 SE mean4.35584 SE pred 3.38 X.6 Y -.8 6 -.5 95 85 8 9 -. -. -.5..5. B: x -9 The following data were collected by a chemical engineer. The response y is filtration time, x is temperature, and x is pressure. Fit a second-order model. x x y - - 54-45 - 3 47 -.44 5.44 53 -.44 47.44 5 4 39 44 4 4 Design Expert Output Response: y ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 64. 4 66.6.57.94 not significant A 3. 3..5.4955 B 5.7 5.7..3467 A 8.39 8.39 3.6.3 AB 44. 44. 5.6.455 Residual 5.78 8 5.7 Lack of Fit 9.98 4 47.74.9.48 significant Pure Error 4.8 4 3.7 Cor Total 47. The "Model F-value" of.57 implies the model is not significant relative to the noise. There is a.94 % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 5.7 R-Squared.56 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Mean 45. Adj R-Squared.3433 C.V..7 Pred R-Squared -.549 PRESS 76.73 Adeq Precision 4.955 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 4.9.83 38.69 47.4 A-Temperature.8.79 -.85 5.4. B-Pressure -.79.79-5.93.34. A 3.39.9 -. 7.79. AB 6..54.5.85. Final Equation in Terms of Coded Factors: Time = +4.9 +.8 * A -.79 * B +3.39 * A +6. * A * B Final Equation in Terms of Actual Factors: Time = +4.934 +.833 * Temperature -.7989 * Pressure +3.393 * Temperature +6. * Temperature * Pressure The lack of fit test in the above analysis is significant. Also, the residual plot below identifies an outlier which happens to be standard order number 8. Normal plot of residuals 99 95 Normal % probability 9 8 7 5 3 5-5.3 -.677.69568 6.6599.65 Residual We chose to remove this run and re-analyze the data. Design Expert Output Response: y ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 47.34 4.84 3.3. significant A 3. 3. 3.88.895-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B 3.63 3.63 39.5.4 A 55.7 55.7 45.95.3 AB 44. 44. 4.6.3 Residual 3.66 7 3.38 Lack of Fit 8.86 3.95.8.556 not significant Pure Error 4.8 4 3.7 Cor Total 43. The Model F-value of 3.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..84 R-Squared.945 Mean 44.5 Adj R-Squared.938 C.V. 4.3 Pred R-Squared.89 PRESS 8.66 Adeq Precision 8.43 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 4.68.73 38.95 4.4 A-Temperature.8.65 -.6.8. B-Pressure -4.8.77-6.64-3.. A 4.88.7 3.8 6.59. AB 6..9 3.83 8.7. Final Equation in Terms of Coded Factors: Time = +4.68 +.8 * A -4.8 * B +4.88 * A +6. * A * B Final Equation in Terms of Actual Factors: Time = +4.67673 +.833 * Temperature -4.8374 * Pressure +4.888 * Temperature +6. * Temperature * Pressure The lack of fit test is satisfactory as well as the following normal plot of residuals: Normal plot of residuals 99 95 Norm al % probability 9 8 7 5 3 5 -.67673 -.4673.837.737 3.337 Residual -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) What operating conditions would you recommend if the objective is to minimize the filtration time?. Prediction 33.95 95% Low 7.885 95% High 38.56 SE mean.97 SE pred.458 X -.68 Y.5. 34 36 38 Time B: Press ure. 4 5 4 46 -.5 48 46 5 5 44 -. -. -.5..5. A: Temperature (b) What operating conditions would you recommend if the objective is to operate the process at a mean filtration time very close to 46?. 34 Time 36.5 38 B: Press ure. 4 5 4 46 -.5 48 46 5 5 44 -. -. -.5..5. A: Temperature There are two regions that enable a filtration time of 46. Either will suffice; however, higher temperatures and pressures typically have higher operating costs. We chose the operating conditions at the lower pressure and temperature as shown. - The hexagon design that follows is used in an experiment that has the objective of fitting a secondorder model. x x y 68-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Fit the second-order model..5 75. 74 -.5 75. 65-6 -.5-75. 63.5-75. 7 58 6 57 55 69 Design Expert Output Response: y ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 45.6 5 49.5.89.5 not significant A 85.33 85.33 3.3.9 B 9. 9..35.58 A 5. 5..97.369 B 9.83 9.83 5..753 AB...39.859 Residual 9.47 5 5.89 Lack of Fit.67.67.36.583 not significant Pure Error 8.8 4 9.7 Cor Total 374.73 The "Model F-value" of.89 implies the model is not significant relative to the noise. There is a 5. % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 5.9 R-Squared.6545 Mean 63.55 Adj R-Squared.39 C.V. 8. Pred R-Squared -.5 PRESS 569.63 Adeq Precision 3.75 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59.8.8 53.95 65.65 A-x 5.33.94 -..89. B-x.73.94-5.8 9.8. A 4. 4.6-6.74 5.4. B 9.53 4.6 -.4.48. AB.5 5.88-3.95 6.6. Final Equation in Terms of Coded Factors: y = +59.8 +5.33 * A +.73 * B +4. * A +9.53 * B +.5 * A * B (a) Perform the canonical analysis. What type of surface has been found? The full quadratic model is used in the following analysis because the reduced model is singular. -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Solution Variable Critical Value X -.67658 X -.589 Predicted Value at Solution 58.849 Eigenvalues and Eigenvectors Variable 9.5957 4.38 X.64.9943 X.9943 -.64 Since both eigenvalues are positive, the response is a minimum at the stationary point. (c) What operating conditions on x and x lead to the stationary point? The stationary point is (x,x ) = (-.6766, -.583) (d) Where would you run this process if the objective is to obtain a response that is as close to 65 as possible?.87 y 75 7.43 65 B: x. 5 6 -.43 7 -.87 -. -.5..5. A: x Any value of x and x that give a point on the contour with value of 65 would be satisfactory. - An experimenter has run a Box-Behnken design and has obtained the results below, where the response variable is the viscosity of a polymer. Level Temp. Agitation Rate Pressure x x x 3 High. 5 + + + Middle 75 7.5 Low 5 5. 5 - - - -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Fit the second-order model. Run x x x 3 y - - 535-58 3-596 4 563 5 - - 645 6-458 7-35 8 6 9 - - 595-648 - 53 656 3 653 4 599 5 6 Design Expert Output Response: Viscosity ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 8965.58 9 996.4 9.54.5 significant A 73. 73..67.449 B 65. 65. 5.85.6 C 548. 548. 5.8.79 A 769.3 769.3 9.9.66 B 44. 44..35.985 C 479. 479. 4.5.868 AB 5. 5..46.84 AC 4774.5 4774.5 45.74. BC 6.5 6.5..39 Residual 58.75 5 43.75 Lack of Fit 3736.75 3 45.58.68.394 not significant Pure Error 48. 74. Cor Total 9487.33 4 The Model F-value of 9.54 implies the model is significant. There is only a.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3.3 R-Squared.945 Mean 575.33 Adj R-Squared.846 C.V. 5.6 Pred R-Squared.3347 PRESS 63.5 Adeq Precision.45 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 64. 8.65 576.5 67.95 A-Temperatue 9.37.4-9.99 38.74. B-Agitation Rate 7.6.4 -.74 56.99. C-Pressure -6..4-55.36 3.36. A -75. 6.8-8. -3.78. B 9.5 6.8-3.7 6.7. C -35.75 6.8-78.97 7.47. AB -9.5 6.5-6... AC 9.5 6.5 67.73 5.77. BC 7.75 6.5-3.77 59.7. Final Equation in Terms of Coded Factors: -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Viscosity = +64. +9.37 * A +7.6 * B -6. * C -75. * A +9.5 * B -35.75 * C -9.5 * A * B +9.5 * A * C +7.75 * B * C Final Equation in Terms of Actual Factors: Viscosity = -69.5 +7.35 * Temperatue -9.55 * Agitation Rate -.6 * Pressure -. * Temperatue +3. * Agitation Rate -.43 * Pressure -.3 * Temperatue * Agitation Rate +.874 * Temperatue * Pressure +.4 * Agitation Rate * Pressure (b) Perform the canonical analysis. What type of surface has been found? The system is a saddle point. Solution Variable Critical Value X.849596 X -.8737 X3.75865 Predicted Value at Solution 586.34437 Eigevalues and Eigevectors Variable.99.58-4.694 X -.739.588.833 X.999 -.897.973 X3.883.8888 -.57368 (c) What operating conditions on x, x, and x 3 lead to the stationary point? The stationary point is (x, x, x 3 ) = (.8496, -.8767,.7586). This is outside the design region. It would be necessary to either examine contour plots or use numerical optimization methods to find desired operating conditions. (d) What operating conditions would you recommend if it is important to obtain a viscosity that is as close to 6 as possible? -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot V i sco si ty X = A: Temperatue Y = C: Pressure Design Points.5 Actual Factor B: Agitation Rate = 7.5 C: Pressure 5. 4. 45 5 55 Viscosity 6 3 6 7.5 55 5 5. 5. 6.5 75. 87.5. A: Tem peratue Any point on either of the contours showing a viscosity of 6 is satisfactory. - Consider the three-variable central composite design shown below. Analyze the data and draw conclusions, assuming that we wish to maximize conversion (y ) with activity (y ) between 55 and 6. Time Temperature Catalyst Conversion (%) Activity Run (min) (C) (%) y y -. -. -. 74. 53.. -. -. 5. 6.9 3 -.. -. 88. 53.4 4.. -. 7. 6.6 5 -. -.. 7. 57.3 6. -.. 9. 67.9 7 -... 66. 59.8 8... 97. 67.8 9... 8. 59.... 75. 6.4... 76. 59.... 83. 6.6 3 -.68.. 76. 59. 4.68.. 79. 65.9 5. -.68. 85. 6. 6..68. 97. 6.7 7.. -.68 55. 57.4 8...68 8. 63. 9... 8. 6.8... 9. 58.9 Quadratic models are developed for the Conversion and Activity response variables as follows: -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Conversion ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 555.73 9 83.97.76. significant A 4.44 4.44.65.439 B.96.96.. C 55.64 55.64 3.63.7 A 48.47 48.47.8.77 B 4.48 4.48 5.6.396 C 388.59 388.59 7.47.9 AB 36.3 36.3.6.34 AC 35.3 35.3 46.53 <. BC.. 5.4.45 Residual.47.5 Lack of Fit 56.47 5.9.34.869 not significant Pure Error 66. 5 33. Cor Total 87.8 9 The Model F-value of.76 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 4.7 R-Squared.999 Mean 78.3 Adj R-Squared.8479 C.V. 6. Pred R-Squared.7566 PRESS 676. Adeq Precision 4.39 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 8.9.9 76.8 85.38 A-Time.3.8 -.8 3.87. B-Temperature 4.4.8. 6.88. C-Catalyst 6..8 3.36 9.5. A -.83.4-4.6.93. B.94.4.7 5.7. C -5.9.4-7.96 -.4. AB.3.67 -.59 5.84. AC.38.67 7.66 5.9. BC -3.87.67-7.59 -.6. Final Equation in Terms of Coded Factors: Conversion = +8.9 +.3 * A +4.4 * B +6. * C -.83 * A +.94 * B -5.9 * C +.3 * A * B +.38 * A * C -3.87 * B * C Final Equation in Terms of Actual Factors: Conversion = +8.98 +.845 * Time +4.457 * Temperature +6.396 * Catalyst -.83398 * Time +.93899 * Temperature -5.974 * Catalyst +.5 * Time * Temperature +.375 * Time * Catalyst -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3.875 * Temperature * Catalyst Design Expert Output Response: Activity ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 56. 9 8.47 9.6.9 significant A 75.35 75.35 56.4 <. B.89.89.8.65 C 67.9 67.9.85.9 A.5.5 3.3.4 B.8.8.6.8753 C.47.47.5.946 AB...39.548 AC.. 3.6E-3.953 BC.78.78.5.67 Residual 3.8 3. Lack of Fit 7.43 5 5.49 7.5.6 significant Pure Error 3.65 5.73 Cor Total 87.8 9 The Model F-value of 9.6 implies the model is significant. There is only a.9% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..76 R-Squared.898 Mean 6.5 Adj R-Squared.7945 C.V..9 Pred R-Squared.536 PRESS 4.43 Adeq Precision.9 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59.85.7 58.5 6.45 A-Time 3.58.48.5 4.65. B-Temperature.5.48 -.8.3. C-Catalyst.3.48.7 3.9. A.83.46 -..87. B.75.46 -.96.. C.57.46 -.98.9. AB -.39.6 -.78.. AC -.38.6 -.43.35. BC.3.6 -.8.7. Final Equation in Terms of Coded Factors: Conversion = +59.85 +3.58 * A +.5 * B +.3 * C +.83 * A +.75 * B +.57 * C -.39 * A * B -.38 * A * C +.3 * B * C Final Equation in Terms of Actual Factors: Conversion = +59.84984 +3.5837 * Time +.546 * Temperature +.997 * Catalyst +.8349 * Time +.7477 * Temperature -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.5794 * Catalyst -.3875 * Time * Temperature -.375 * Time * Catalyst +.35 * Temperature * Catalyst Because many of the terms are insignificant, the reduced quadratic model is fit as follows: Design Expert Output Response: Activity ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 53. 3 84.4 39.63 <. significant A 75.35 75.35 8.34 <. C 67.9 67.9 3.89 <. A 9.94 9.94 4.67.463 Residual 34.7 6.3 Lack of Fit 3.4.77 3.78.766 not significant Pure Error 3.65 5.73 Cor Total 87.8 9 The Model F-value of 39.63 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..46 R-Squared.884 Mean 6.5 Adj R-Squared.859 C.V..4 Pred R-Squared.63 PRESS 6.4 Adeq Precision.447 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59.95.4 59.6 6.83 A-Time 3.58.39.75 4.4. C-Catalyst.3.39.39 3.7. A.8.38.5.63. Final Equation in Terms of Coded Factors: Activity = +59.95 +3.58 * A +.3 * C +.8 * A Final Equation in Terms of Actual Factors: Activity = +59.948 +3.5837 * Time +.997 * Catalyst +.83 * Time -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Conversion X = A: Time Y = C: Catalyst Design Points Actual Factor B: Temperature = -...5 74 76 78 Conversion 86 84 8 88 9 9 DESIGN-EXPERT Plot Activity X = A: Time Y = C: Catalyst Design Points Actual Factor B: Temperature = -...5 Activity 64 66 C: Catalyst. 8 78 76 C: Catalyst. 58 6 6 -.5 74 7 7 68 66 64 6 6 58 56 54 -.5 56 -. -. -. -.5..5. -. -.5..5. A: Tim e A: Tim e DESIGN-EXPERT Plot Overlay Plot X = A: Time Y = C: Catalyst. Overlay Plot Design Points Actual Factor B: Temperature = -..5 Conv ersion: 8 C: Catalyst. Activ ity : 6 -.5 -. -. -.5..5. A: Tim e The contour plots visually describe the models while the overlay plots identifies the acceptable region for the process. -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3 A manufacturer of cutting tools has developed two empirical equations for tool life in hours (y ) and for tool cost in dollars (y ). Both models are linear functions of steel hardness (x ) and manufacturing time (x ). The two equations are ŷ ŷ 5x 3 3x x 4x and both equations are valid over the range -.5x.5. Unit tool cost must be below $7.5 and life must exceed hours for the product to be competitive. Is there a feasible set of operating conditions for this process? Where would you recommend that the process be run? The contour plots below graphically describe the two models. The overlay plot identifies the feasible operating region for the process..5 Life.5 Cost 3 3.75 8.75 6 8 7.5 6 4 B: Tim e. 4 6 8 4 B: Tim e. -.75 -.75 8 6 4 -.5 -.5 -.75..75.5 -.5 -.5 -.75..75.5 A: H ardn es s A: H ardn es s.5 Overlay Plot.75 Cost: 7.5 B: Tim e. Life: -.75 -.5 -.5 -.75..75.5 5x 3 3x A: H ardn es s x 4x 7. 5-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -4 A central composite design is run in a chemical vapor deposition process, resulting in the experimental data shown below. Four experimental units were processed simultaneously on each run of the design, and the responses are the mean and variance of thickness, computed across the four units. x x y s - - 36.6 6.689-445. 4.3-4. 7.88 6.7 8.586.44 58. 3.3 -.44 4.4 6.644.44 497.6 7.649 -.44 397.6.74 53.6 7.836 495.4 9.36 5. 7.956 487.3 9.7 (a) Fit a model to the mean response. Analyze the residuals. Design Expert Output Response: Mean Thick ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 47644.6 5 958.85 6.. significant A 573.36 573.36 38.9.8 B 56.9 56.9 5.8.3 A 795.58 795.58 4.73.76 B 555.74 555.74 9.39. AB 756.5 756.5 4.66.74 Residual 3546.83 6 59.4 Lack of Fit 46.4 3 8.68.7.59 not significant Pure Error 84.79 3 36.6 Cor Total 59.9 The Model F-value of 6. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 4.3 R-Squared.937 Mean 47.3 Adj R-Squared.873 C.V. 5.5 Pred R-Squared.63 PRESS 9436.37 Adeq Precision.6 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 55.88.6 476.3 535.6 A-x 53. 8.6 3.9 74.5. B-x 43.68 8.6.64 64.7. A -.9 9.6-44.4.6.4 B -9.45 9.6-5.97-5.93.4 AB 6.5.6-3.5 56.. Final Equation in Terms of Coded Factors: Mean Thick = +55.88 +53. * A +43.68 * B -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -.9 * A -9.45 * B +6.5 * A * B Final Equation in Terms of Actual Factors: Mean Thick = +55.875 +53.94 * x +43.67767 * x -.9 * x -9.45 * x +6.5 * x * x Normal plot of residuals Residuals vs. Predicted 4.75 99 Normal % probability 95 9 8 7 5 3 5 Residuals.4493.73533 -. -4.3779-4.3779 -..73533.4493 4.75 384.98 433.38 48.78 53.7 578.57 Residual Predicted A modest deviation from normality can be observed in the Normal Plot of Residuals; however, not enough to be concerned. (b) Fit a model to the variance response. Analyze the residuals. Design Expert Output Response: Var Thick ANOVA for Response Surface FI Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 65.8 3.93 35.86 <. significant A 4.46 4.46 67.79 <. B 5. 5. 4.87. AB 9.3 9.3 4.93.48 Residual 4.89 8.6 Lack of Fit 3.3 5.63.6.537 not significant Pure Error.77 3.59 Cor Total 7.69 The Model F-value of 35.86 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..78 R-Squared.938 Mean 9.7 Adj R-Squared.948 C.V. 8.53 Pred R-Squared.89 PRESS 7.64 Adeq Precision 8.57-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Coefficient Estimate DF Standard Error 95% CI Low 95% CI High VIF Intercept 9.7.3 8.64 9.69 A-x.8.8.64.9. B-x -.38.8 -. -.74. AB -.5.39 -.4 -.6. Final Equation in Terms of Coded Factors: Var Thick = +9.7 +.8 * A -.38 * B -.5 * A * B Final Equation in Terms of Actual Factors: Var Thick = +9.658 +.7645 * x -.3788 * x -.575 * x * x Normal plot of residuals Residuals vs. Predicted.74553 99 Normal % probability 95 9 8 7 5 3 5 Residuals.6878 -.9776 -.849 -.398 -.398 -.849 -.9776.6878.74553 5.95 8.4.4.3 4.33 Residual Predicted The residual plots are not acceptable. A transformation should be considered. If not successful at correcting the residual plots, further investigation into the two apparently unusual points should be made. (c) Fit a model to the ln(s ). Is this model superior to the one you found in part (b)? Design Expert Output Response: Var Thick Transform: Natural log Constant: ANOVA for Response Surface FI Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.67 3. 36.94 <. significant A.46.46 74.99 <. B.4.4.8.4 AB.79.79 3.4.69 Residual.49 8 6.8E-3 Lack of Fit.4 5 4.887E-3.6.793 not significant Pure Error.4 3 8.7E-3 Cor Total.7-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The Model F-value of 36.94 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..78 R-Squared.937 Mean.8 Adj R-Squared.974 C.V. 3.57 Pred R-Squared.8797 PRESS.87 Adeq Precision 8.854 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.8.3.3.4 A-x.4.8.8.3. B-x -.3.8 -. -.68. AB -.4.39 -.3 -.5. Final Equation in Terms of Coded Factors: Ln(Var Thick) = +.8 +.4 * A -.3 * B -.4 * A * B Final Equation in Terms of Actual Factors: Ln(Var Thick) = +.8376 +.3874 * x -.365 * x -.479 * x * x Normal plot of residuals Residuals vs. Predicted.93684 99 Normal % probability 95 9 8 7 5 3 5 Residuals.385439 -.5985 -.755 -.59 -.59 -.755 -.5985.385439.93684.85.6.7.48.69 Residual Predicted The residual plots are much improved following the natural log transformation; however, the two runs still appear to be somewhat unusual and should be investigated further. They will be retained in the analysis. (d) Suppose you want the mean thickness to be in the interval 45±5. Find a set of operating conditions that achieve the objective and simultaneously minimize the variance. -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY. Mean Thick 575. Ln(Var Thick).5 55.5 B: x. 4 5 55 B: x.. 4..3 -.5 45 475 -.5.4.5 45 4.6 -. -. -.5..5. -. -. -.5..5. A: x A: x. Overlay Plot.5 B: x. Ln(Var Thick):. 4 Mean Thick: 475 -.5 Mean Thick: 45 -. -. -.5..5. A: x The contour plots describe the two models while the overlay plot identifies the acceptable region for the process. (e) Discuss the variance minimization aspects of part (d). Have you minimized total process variance? The within run variance has been minimized; however, the run-to-run variation has not been minimized in the analysis. This may not be the most robust operating conditions for the process. -5 Verify that an orthogonal first-order design is also first-order rotatable. To show that a first order orthogonal design is also first order rotatable, consider V( ŷ ) k ˆ ˆ ˆ V( x ) V( ) x V( ˆ i i i i ) i k i -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3 since all covariances between i and j are zero, due to design orthogonality. Furthermore, we have: n ) ˆ V(... ) ˆ V( ) ˆ V( ˆ V k, so k i x i n n ŷ ) V( k i x i n n ŷ ) V( which is a function of distance from the design center (i.e. x=), and not direction. Thus the design must be rotatable. Note that n is, in general, the number of points in the exterior portion of the design. If there are n c centerpoints, then ) n ( n ) ˆ V( c. -6 Show that augmenting a k design with n c center points does not affect the estimates of the i (i=,,..., k), but that the estimate of the intercept is the average of all k + n c observations. In general, the X matrix for the k design with n c center points and the y vector would be: k X The upper half of the matrix is the usual notation of the k design The lower half of the matrix represents the center points (n c rows) c k n n n y y y y k +n c observations. k k c k n X X' g k g g g y X' Grand total of all k +n c observations usual contrasts from k

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Therefore, ˆ g k n k c observations, while ˆ g i i, which does k nc not depend on the number of center points, since in computing the contrasts g i, all observations at the center are multiplied by zero., which is the average of all -7 The rotatable central composite design. It can be shown that a second-order design is rotatable if n a b x u iu x ju if a or b (or both) are odd and if / 4 composite design these conditions lead to the factorial portion. n f n u 4 iu x n 3 x u iux ju. Show that for the central for rotatability, where n f is the number of points in The balance between + and - in the factorial columns and the orthogonality among certain column in the X matrix for the central composite design will result in all odd moments being zero. To solve for use the following relations: n n 4 4 x iu n f, u u x iu x ju n f then n f 4 n u 4 4 x 4 iu 4 n n n f f 3 3( n f n u f ) x iu x ju -8 Verify that the central composite design shown below blocks orthogonally. Block Block Block 3 x x x 3 x x x 3 x x x 3 -.633.633 - -.633 - - -.633 - - - -.633 - - - - -.633 Note that each block is an orthogonal first order design, since the cross products of elements in different columns add to zero for each block. To verify the second condition, choose a column, say column x. Now k u xu 3. 334, and n= -33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY For blocks and, So m m n u x m 4, n m =6 x x m u n 4 3. 334 m.3 =.3 6 6 and condition is satisfied by blocks and. For block 3, we have m x m 5. 334, n m = 8, so m n u x x m u 5. 334 3. 334 nm n 8.4 =.4 And condition is satisfied by block 3. Similar results hold for the other columns. -9 Blocking in the central composite design. Consider a central composite design for k = 4 variables in two blocks. Can a rotatable design always be found that blocks orthogonally? To run a central composite design in two blocks, assign the n f factorial points and the n center points to block and the k axial points plus n center points to block. Both blocks will be orthogonal first order designs, so the first condition for orthogonal blocking is satisfied. The second condition implies that m m x x im im block n f n k n block c c -34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY However, xim n f in block and x im m m in block, so Which gives: n n n f f c k nc n f k n n f n Since 4 if the design is to be rotatable, then the design must satisfy n f c c n f n f k n n f n c c It is not possible to find rotatable central composite designs which block orthogonally for all k. For example, if k=3, the above condition cannot be satisfied. For k=, there must be an equal number of center points in each block, i.e. n c = n c. For k=4, we must have n c = 4 and n c =. - How could a hexagon design be run in two orthogonal blocks? The hexagonal design can be blocked as shown below. There are n c = n c = n c center points with n c even. 6 n 3 5 4 Put the points,3,and 5 in block and,4,and 6 in block. Note that each block is a simplex. - Yield during the first four cycles of a chemical process is shown in the following table. The variables are percent concentration (x ) at levels 3, 3, and 3 and temperature (x ) at 4, 4, and 44F. Analyze by EVOP methods. Conditions Cycle () () (3) (4) (5) -35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.7 59.8 6. 64. 57.5 59. 6.8 6.5 64.6 58.3 3 56.6 59. 59. 6.3 6. 4 6.5 59.8 64.5 6. 6. Cycle: n= Phase Calculation of Averages Calculation of Standard Deviation Operating Conditions () () (3) (4) (5) (i) Previous Cycle Sum Previous Sum S= (ii) Previous Cycle Average Previous Average = (iii) New Observation 6.7 59.8 6. 64. 57.5 New S=Range x f k,n (iv) Differences Range= (v) New Sums 6.7 59.8 6. 64. 57.5 New Sum S= (vi) New Averages 6.7 59.8 6. 64. 57.5 New average S = New Sum S/(n-)= Calculation of Effects A y3 y4 y y5 B y3 y4 y y5 AB y3 y4 y y5 CIM y3 y4 y y5 4y 3.55-3.55 -.85 -. Calculation of Error Limits For New Average: S n For New Effects: S n.78 For CIM: S n Cycle: n= Phase Calculation of Averages Calculation of Standard Deviation Operating Conditions () () (3) (4) (5) (i) Previous Cycle Sum 6.7 59.8 6. 64. 57.5 Previous Sum S= (ii) Previous Cycle Average 6.7 59.8 6. 64. 57.5 Previous Average = (iii) New Observation 59. 6.8 6.5 64.6 58.3 New S=Range x f k,n=.38 (iv) Differences.6-3. -.3 -.4 -.8 Range=4.6 (v) New Sums 9.8.6.7 8.8 5.8 New Sum S=.38 (vi) New Averages 59.9 6.3 6.35 64.4 57.9 New average S = New Sum S/(n-)=.38 Calculation of Effects A y3 y4 y y5 B y3 y4 y y5 AB y3 y4 y y5 CIM y3 y4 y y5 4y 3.8-3.3.8.7 Calculation of Error Limits.95 For New Average: S n.95 For New Effects: S n.78.74 For CIM: S n Cycle: n=3 Phase Calculation of Averages Calculation of Standard Deviation Operating Conditions () () (3) (4) (5) (i) Previous Cycle Sum 9.8.6.7 8.8 5.8 Previous Sum S=.38-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (ii) Previous Cycle Average 59.9 6.3 6.35 64.4 57.9 Previous Average =.38 (iii) New Observation 56.6 59. 59. 6.3 6. New S=Range x f k,n=.8 (iv) Differences 3.3..35. -3. Range=6.5 (v) New Sums 76.4 8.7 8.7 9. 76.9 New Sum S=3.66 (vi) New Averages 58.8 6.57 6.57 63.7 58.97 New average S = New Sum S/(n-)=.38 Calculation of Effects A y3 y4 y y5 B y3 y4 y y5 AB y3 y4 y y5 CIM y3 y4 y y5 4y.37 -.37 -.77.7 Calculation of Error Limits. For New Average: S n. For New Effects: S n.78.74 For CIM: S n Cycle: n=4 Phase Calculation of Averages Calculation of Standard Deviation Operating Conditions () () (3) (4) (5) (i) Previous Cycle Sum 76.4 8.7 8.7 9. 76.9 Previous Sum S=3.66 (ii) Previous Cycle Average 58.8 6.57 6.57 63.7 58.97 Previous Average =.83 (iii) New Observation 6.5 59.8 64.5 6. 6. New S=Range x f k,n=.45 (iv) Differences -.7.77-3.93.7 -.3 Range=6.63 (v) New Sums 36.9 4.5 45. 5. 37. New Sum S=6. (vi) New Averages 59.3 6.38 6.55 63.3 59.5 New average S = New Sum S/(n-)=.4 Calculation of Effects A y3 y4 y y5 B y3 y4 y y5 AB y3 y4 y y5 CIM y3 y4 y y5 4y.48 -.3 -.8.46 Calculation of Error Limits.4 For New Average: S n.4 For New Effects: S n.78.8 For CIM: S n From studying cycles 3 and 4, it is apparent that A (and possibly B) has a significant effect. A new phase should be started following cycle 3 or 4. - Suppose that we approximate a response surface with a model of order d, such as y=x +, when the true surface is described by a model of order d >d ; that is E(y)= X + X. (a) Show that the regression coefficients are biased, that is, that E( )= +A, where A=(X X ) - X X. A is usually called the alias matrix. -37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -38 Aβ β β X X X X X β X X X β X X β X X X y X X X X y X X β ' ' ' ' ' ' ' ' ' ' E E E ˆ where ' ' X X X X A (a) If d = and d =, and a full k is used to fit the model, use the result in part (a) to determine the alias structure. In this situation, we have assumed the true surface to be first order, when it is really second order. If a full factorial is used for k=, then X = X = and A = Then, E = E ˆ ˆ ˆ The pure quadratic terms bias the intercept. (b) If d =, d = and k=3, find the alias structure assuming that a 3- design is used to fit the model. X = 3 X = 33 3 3 and A = Then, E = E 3 3 3 3 3 33 3 3 ˆ ˆ ˆ ˆ (d) If d =, d =, k=3, and the simplex design in Problem -3 is used to fit the model, determine the alias structure and compare the results with part (c).

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY X = 3 X = 33 3 3 and A = Then, E = E ˆ ˆ ˆ ˆ 3 3 33 3 3 3 3 3 Notice that the alias structure is different from that found in the previous part for the 3- design. In general, the A matrix will depend on which simplex design is used. -3 In an article ( Let s All Beware the Latin Square, Quality Engineering, Vol., 989, pp. 453-465) J.S. Hunter illustrates some of the problems associated with 3 k-p fractional factorial designs. Factor A is the amount of ethanol added to a standard fuel and factor B represents the air/fuel ratio. The response variable is carbon monoxide (CO) emission in g/m. The design is shown below. Design Observations A B x x y y - - 66 6-78 8-9 94-7 67 8 8 75 78-68 66 66 69 6 58 Notice that we have used the notation system of,, and to represent the low, medium, and high levels for the factors. We have also used a geometric notation of -,, and. Each run in the design is replicated twice. (a) Verify that the second-order model 7. x 4. 5x 4. x 9. x ŷ 78. 5 4. 5x x is a reasonable model for this experiment. Sketch the CO concentration contours in the x, x space. In the computer output that follows, the coded factors model is in the -,, + scale. Design Expert Output Response: CO Emis ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 64. 5 34.8 5.95 <. significant A 43. 43. 38. <. -39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B 588. 588. 9.4 <. A 8. 8..7.39 B 64. 64..4.8 AB 648. 648..65 <. Residual 76.5 6.37 Lack of Fit 3. 3..94.944 not significant Pure Error 46.5 9 5.7 Cor Total 7.5 7 The Model F-value of 5.95 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..5 R-Squared.955 Mean 7.83 Adj R-Squared.9363 C.V. 3.47 Pred R-Squared.9 PRESS 69.7 Adeq Precision.95 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 78.5.33 75.6 8.4 A-Ethanol 4.5.73.9 6.9. B-Air/Fuel Ratio -7..73-8.59-5.4. A -4.5.6-7.5 -.75. B -4..6-6.75 -.5. AB -9..89 -.94-7.6. Final Equation in Terms of Coded Factors: CO Emis = +78.5 +4.5 * A -7. * B -4.5 * A -4. * B -9. * A * B CO Emis. 65 7.5 B: Air/Fuel Ratio 75. 8 -.5 85 65 -. - -.5.5 A: Ethanol (b) Now suppose that instead of only two factors, we had used four factors in a 3 4- fractional factorial design and obtained exactly the same data in part (a). The design would be as follows: Design Observations A B C D x x x 3 x 4 y y -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY - - - - 66 6-78 8 + - + + 9 94 - + 7 67 - + 8 8 + - 75 78 - + + 68 66 + + - 66 69 + + - 6 58 Confirm that this design is an L 9 orthogonal array. This is the same as the design in Table -. (c) Calculate the marginal averages of the CO response at each level of the four factors A, B, C, and D. Construct plots of these marginal averages and interpret the results. Do factors C and D appear to have strong effects? Do these factors really have any effect on CO emission? Why is their apparent effect strong? One Factor Plot One Factor Plot 94 94 85 85 CO Emis 76 CO Emis 76 67 67 58 58 A: A B: B -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY One Factor Plot One Factor Plot 94 94 85 85 CO Emis 76 CO Emis 76 67 67 58 58 C: C D: D Both Factors C and D appear to have an effect on CO emission. This is probably because both C and D are aliased with components of interaction involving A and B, and there is a strong AB interaction. (a) The design in part (b) allows the model y to be fitted. Suppose that the true model is y 4 4 i x 4 i i 4 i iixi i xi ii xi i i ij x x Show that if j represents the least squares estimates of the coefficients in the fitted model, then E E E E E E E E E ˆ 3 4 34 ˆ 3 4 / ˆ 3 4 34 / ˆ 3 3 4 / ˆ 4 4 3/ ˆ 3 4 / ˆ 3 4 34 / ˆ 33 33 4 / 4 ˆ 44 44 3 / 3 ij i j -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -43 Let X = 3 4 33 44 and X = 3 4 3 4 34 Then, A X X X X ' ' = A = 3 3 44 4 4 33 34 4 3 4 3 3 4 4 3 34 4 3 4 3 34 4 3 34 4 3 4 3 44 33 4 3 44 33 4 3 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ E -4 Suppose that you need to design an experiment to fit a quadratic model over the region i x, i=, subject to the constraint x x. If the constraint is violated, the process will not work properly. You can afford to make no more than n= runs. Set up the following designs: (a) An inscribed CCD with center points at x x x x -.5 -.5.5 -.5 -.5.5.5.5 -.77.77

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -.77.77 (a)* An inscribed CCD with center points at x x 5 so that a larger design could be fit within the constrained region. x x - -.5 - -.5.5.5 -.664 -.5.64 -.5 -.5 -.664 -.5.64 -.5 -.5 -.5 -.5 -.5 -.5 -.5 -.5 (a) An inscribed 3 factorial with center points at x x. 5 x x - - -.5 -.5 - - -.5 -.5 -.5.5 -.5 -.5 -.5.5.5.5 -.5 -.5 -.5 -.5 -.5 -.5 (a) A D-optimal design. x x - - - - - - -44

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.5.5 - - - - (a) A modified D-optimal design that is identical to the one in part (c), but with all replicate runs at the design center. x x - - - - - -.5.5 (a) Evaluate the ( X X) criteria for each design. X X (a) (a)* (b) (c) (d).5.548.77.6.94 (a) Evaluate the D-efficiency for each design relative to the D-optimal design in part (c). (a) (a)* (b) (c) (d) D-efficiency 4.5%.64% 49.4%.% 87.3% (a) Which design would you prefer? Why? The offset CCD, (a)*, is the preferred design based on the D-efficiency. Not only is it better than the D- optimal design, (c), but it maintains the desirable design features of the CCD. -45

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -5 Consider a 3 design for fitting a first-order model. (a) Evaluate the D-criterion ( X X) for this design. ( X X) =.44E-4 (b) Evaluate the A-criterion tr ( XX) for this design. tr ( XX) =.5 (c) Find the maximum scaled prediction variance for this design. Is this design G-optimal? v x NVar ŷx Nx XX x 4. Yes, this is a G-optimal design. -6 Repeat Problem -5 using a first order model with the two-factor interaction. ( X X) = 4.768E-7 tr ( XX) =.875 v x NVar ŷx Nx XX x 7. Yes, this is a G-optimal design. -7 A chemical engineer wishes to fit a calibration curve for a new procedure used to measure the concentration of a particular ingredient in a product manufactured in his facility. Twelve samples can be prepared, having known concentration. The engineer s interest is in building a model for the measured concentrations. He suspects that a linear calibration curve will be adequate to model the measured concentration as a function of the known concentrations; that is, where x is the actual concentration. Four experimental designs are under consideration. Design consists of 6 runs at known concentration and 6 runs at known concentration. Design consists of 4 runs at concentrations, 5.5, and. Design 3 consists of 3 runs at concentrations, 4, 7, and. Finally, design 4 consists of 3 runs at concentrations and and 6 runs at concentration 5.5. (a) Plot the scaled variance of prediction for all four designs on the same graph over the concentration range. Which design would be preferable, in your opinion? -46

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3.5 Scaled Variance of Prediction 3.5.5 Design 4 Design 3 Design Design.5 3 5 7 9 Because it has the lowest scaled variance of prediction at all points in the design space with the exception of 5.5, Design is preferred. (b) For each design calculate the determinant of to the D criterion? ( X X). Which design would be preferred according Design ( X X).343.54 3.67 4.686 Design would be preferred. (c) Calculate the D-efficiency of each design relative to the best design that you found in part b. Design D-efficiency.% 8.65% 3 74.55% 4 7.7% (a) For each design, calculate the average variance of prediction over the set of points given by x =,.5,,.5,...,. Which design would you prefer according to the V-criterion? Average Variance of Prediction Design Actual Coded.374.4.5556.96 3.6664.389 4.747.45-47

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design is still preferred based on the V-criterion. (e) Calculate the V-efficiency of each design relative to the best design you found in part (d). (f) What is the G-efficiency of each design? Design V-efficiency.% 88.% 3 8.4% 4 78.7% Design G-efficiency.% 8.% 3 7.4% 4 66.7% -8 Rework Problem -7 assuming that the model the engineer wishes to fit is a quadratic. Obviously, only designs, 3, and 4 can now be considered. 4.5 Scaled Variance of Prediction 4 3.5 3.5.5.5 Design 4 Design 3 Design 3 5 7 9 Based on the plot, the preferred design would depend on the region of interest. Design 4 would be preferred if the center of the region was of interest; otherwise, Design would be preferred. Design ( X X) 4.74E-7 3 6.35E-7 4 5.575E-7 Design is preferred based on ( X X). -48

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design 4 is preferred. Design D-efficiency.% 3 9.46% 4 94.49% Average Variance of Prediction Design Actual Coded.44.34 3.393.994 4.4.869 Design V-efficiency 9.89% 3 93.74% 4.% Design G-efficiency.% 3 79.% 4 75.% -9 An experimenter wishes to run a three-component mixture experiment. The constraints are the components proportions are as follows:. x. x.4 x 3.4.3.7 (a) Set up an experiment to fit a quadratic mixture model. Use n=4 runs, with 4 replicates. Use the D- criteria. (a) Draw the experimental design region. Std x x x3..3.5.3.3.4 3.3.5.55 4...7 5.4..4 6.4..5 7...6 8.75.5.475 9.35.75.475.3..6..3.5.3.3.4 3...7 4.4..5-49

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A: x.5.4..4 B: x..7 C: x3 (c) Set up an experiment to fit a quadratic mixture model with n= runs, assuming that three of these runs are replicated. Use the D-criterion. Std x x x 3.3.5.55..3.5 3.3.3.4 4...7 5.4..4 6.4..5 7...6 8.75.5.475 9.35.75.475...7.4..5.4..4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A: x.5.4..4 B: x..7 C: x3 (d) Comment on the two designs you have found. The design points are the same for both designs except that the edge center on the x-x3 edge is not included in the second design. None of the replicates for either design are in the center of the experimental region. The experimental runs are fairly uniformly spaced in the design region. -3 Myers and Montgomery (995) describe a gasoline blending experiment involving three mixture components. There are no constraints on the mixture proportions, and the following run design is used. Design Point x x x 3 y(mpg) 4.5, 5. 4.8, 3.9 3.7, 3.6 4 ½ ½ 5. 5 ½ ½ 4.3 6 ½ ½ 3.5 7 /3 /3 /3 4.8, 4. 8 /3 /6 /6 4. 9 /6 /3 /6 3.9 /6 /6 /3 3.7 (a) What type of design did the experimenters use? A simplex centroid design was used. (b) Fit a quadratic mixture model to the data. Is this model adequate? Design Expert Output Response: y ANOVA for Mixture Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F -5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Model Linear Mixture 4. 3.9 5.84.96 3.9 9.6.435.88 significant AB.5.5.69.489 AC.8.8.38.5569 BC.77.77.36.5664 Residual.73 8. Lack of Fit.5 4..4.83 not significant Pure Error.4 4.3 Cor Total 5.95 3 The Model F-value of 3.9 implies the model is significant. There is only a 4.35% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..47 R-Squared.79 Mean 4.6 Adj R-Squared.574 C.V..93 Pred R-Squared.44 PRESS 5.7 Adeq Precision 5.674 Coefficient Standard 95% CI 95% CI Component Estimate DF Error Low High A-x 4.74.3 4. 5.49 B-x 4.3.3 3.57 5.5 C-x3 3.8.3.43 3.9 AB.5.8 -.68 5.7 AC..8-3.8 5.3 BC -.9.8-5.8 3. Final Equation in Terms of Pseudo Components: y = +4.74 * A +4.3 * B +3.8 * C +.5 * A * B +. * A * C -.9 * B * C Final Equation in Terms of Real Components: y = +4.7443 * x +4.398 * x +3.7765 * x3 +.5364 * x * x +.364 * x * x3 -.8636 * x * x3 The quadratic terms appear to be insignificant. The analysis below is for the linear mixture model: Design Expert Output Response: y ANOVA for Mixture Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.9.96.64.7 significant Linear Mixture 3.9.96.64.7 Residual.3.8 Lack of Fit.79 7..37.885 not significant Pure Error.4 4.3 Cor Total 5.95 3 The Model F-value of.64 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..43 R-Squared.659 Mean 4.6 Adj R-Squared.597 C.V..78 Pred R-Squared.396 PRESS 3.6 Adeq Precision 8.75-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Coefficient Standard 95% CI 95% CI Component Estimate DF Error Low High A-x 4.93.5 4.38 5.48 B-x 4.35.5 3.8 4.9 C-x3 3.9.5.64 3.74 Adjusted Adjusted Approx t for H Component Effect DF Std Error Effect= Prob > t A-x.6.33 3.49.5 B-x.9.33.87.4 C-x3 -.45.33-4.36. Final Equation in Terms of Pseudo Components: y = +4.93 * A +4.35 * B +3.9 * C Final Equation in Terms of Real Components: y = +4.9348 * x +4.3548 * x +3.948 * x3 (c) Plot the response surface contours. What blend would you recommend to maximize the MPG? A: x. 4.8 4.6.. 4.4 4. 4 3.8 3.6 3.4. B: x. y. C: x3 To maximize the miles per gallon, the recommended blend is x =, x =, and x 3 =. -3 Consider the bottle filling experiment in Example 6-. Suppose that the percent carbonation (A) is a noise variable (in coded units ). z (a) Fit the response model to these data. Is there a robust design problem? From the analysis below, the AB interaction appears to have some importance. Because of this, there is opportunity for improvement in the robustness of the process. Design Expert Output -53

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Fill Height ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 73. 7.43 6.69.3 significant A 36. 36. 57.6 <. B.5.5 3.4.5 C.5.5 9.6. AB.5.5 3.6.943 AC.5.5.4.5447 BC...6.45 ABC...6.45 Pure Error 5. 8.63 Cor Total 78. 5 The Model F-value of 6.69 implies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..79 R-Squared.9359 Mean. Adj R-Squared.8798 C.V. 79.6 Pred R-Squared.7436 PRESS. Adeq Precision 3.46 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept...54.46 A-Carbination.5..4.96. B-Pressure.3..67.58. C-Speed.88..4.33. AB.38. -.8.83. AC.3. -.33.58. BC.5. -..7. ABC.5. -..7. Final Equation in Terms of Coded Factors: Fill Height = +. +.5 * A +.3 * B +.88 * C +.38 * A * B +.3 * A * C +.5 * B * C +.5 * A * B * C Final Equation in Terms of Actual Factors: Fill Height = -5.5 +. * Carbination +7.8 * Pressure +.8 * Speed -.75 * Carbination * Pressure -.5 * Carbination * Speed -.4 * Pressure * Speed +4.E-3 * Carbination * Pressure * Speed (b) Find the mean model and either the variance model or the POE. The mean model in coded terms is: yx,z.. 3B. 88C. BC E z 5 Contour plots of the mean model and POE are shown below: -54

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Fill Height X = B: Pressure Y = C: Speed. Fill Height.5 3 DESIGN-EXPERT Plot POE(Fill Height) X = B: Pressure Y = C: Speed. POE(Fill Height). Coded Factor A: Carbination =..5 Coded Factor A: Carbination =..5.4 C: Speed..5.5 C: Speed..6.8 -.5 -.5 -.5 -. -. -.5..5. -. -. -.5..5. B: Press ure B: Press ure (c) Find a set of conditions that result in mean fill deviation as close to zero as possible with minimum transmitted variability from carbonation. The overlay plot below identifies a region that meets these requirements. The Pressure should be set at its low level and the Speed should be set between approximately. and.5 in coded terms. DESIGN-EXPERT Plot Overlay Plot X = B: Pressure Y = C: Speed Coded Factor A: Carbination =...5 Overlay Plot C: Speed. Fill Height:.5 -.5 Fill Height: -.5 -. -. -.5..5. B: Press ure -3 Consider the experiment in Problem -. Suppose that temperature is a noise variable ( z in coded units). Fit response models for both responses. Is there a robust design problem with respect to both responses? Find a set of conditions that maximize conversion with activity between 55 and 6, and that minimize the variability transmitted from temperature. The following is the analysis of variance for the Conversion response. Because of a significant BC interaction, there is some opportunity for improvement in the robustness of the process with regards to Conversion. Design Expert Output -55

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Conversion ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 555.73 9 83.97.76. significant A 4.44 4.44.65.439 B.96.96.. C 55.64 55.64 3.63.7 A 48.47 48.47.8.77 B 4.48 4.48 5.6.396 C 388.59 388.59 7.47.9 AB 36.3 36.3.6.34 AC 35.3 35.3 46.53 <. BC.. 5.4.45 Residual.47.5 Lack of Fit 56.47 5.9.34.869 not significant Pure Error 66. 5 33. Cor Total 87.8 9 The Model F-value of.76 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 4.7 R-Squared.999 Mean 78.3 Adj R-Squared.8479 C.V. 6. Pred R-Squared.7566 PRESS 676. Adeq Precision 4.39 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 8.9.9 76.8 85.38 A-Time.3.8 -.8 3.87. B-Temperature 4.4.8. 6.88. C-Catalyst 6..8 3.36 9.5. A -.83.4-4.6.93. B.94.4.7 5.7. C -5.9.4-7.96 -.4. AB.3.67 -.59 5.84. AC.38.67 7.66 5.9. BC -3.87.67-7.59 -.6. Final Equation in Terms of Coded Factors: Conversion = +8.9 +.3 * A +4.4 * B +6. * C -.83 * A +.94 * B -5.9 * C +.3 * A * B +.38 * A * C -3.87 * B * C Final Equation in Terms of Actual Factors: Conversion = +8.98 +.845 * Time +4.457 * Temperature +6.396 * Catalyst -.83398 * Time +.93899 * Temperature -5.974 * Catalyst +.5 * Time * Temperature +.375 * Time * Catalyst -3.875 * Temperature * Catalyst -56

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The following is the analysis of variance for the Activity response. Because there is not a significant interaction term involving temperature, there is no opportunity for improvement in the robustness of the process with regards to Activity. Design Expert Output Response: Activity ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 56. 9 8.47 9.6.9 significant A 75.35 75.35 56.4 <. B.89.89.8.65 C 67.9 67.9.85.9 A.5.5 3.3.4 B.8.8.6.8753 C.47.47.5.946 AB...39.548 AC.. 3.6E-3.953 BC.78.78.5.67 Residual 3.8 3. Lack of Fit 7.43 5 5.49 7.5.6 significant Pure Error 3.65 5.73 Cor Total 87.8 9 The Model F-value of 9.6 implies the model is significant. There is only a.9% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..76 R-Squared.898 Mean 6.5 Adj R-Squared.7945 C.V..9 Pred R-Squared.536 PRESS 4.43 Adeq Precision.9 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59.85.7 58.5 6.45 A-Time 3.58.48.5 4.65. B-Temperature.5.48 -.8.3. C-Catalyst.3.48.7 3.9. A.83.46 -..87. B.75.46 -.96.. C.57.46 -.98.9. AB -.39.6 -.78.. AC -.38.6 -.43.35. BC.3.6 -.8.7. Final Equation in Terms of Coded Factors: Conversion = +59.85 +3.58 * A +.5 * B +.3 * C +.83 * A +.75 * B +.57 * C -.39 * A * B -.38 * A * C +.3 * B * C Final Equation in Terms of Actual Factors: Conversion = +59.84984 +3.5837 * Time +.546 * Temperature +.997 * Catalyst +.8349 * Time -57

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY +.7477 +.5794 * Temperature * Catalyst -.3875 * Time * Temperature -.375 * Time * Catalyst +.35 * Temperature * Catalyst Because many of the terms are insignificant, the reduced quadratic model is fit as follows: Design Expert Output Response: Activity ANOVA for Response Surface Quadratic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 53. 3 84.4 39.63 <. significant A 75.35 75.35 8.34 <. C 67.9 67.9 3.89 <. A 9.94 9.94 4.67.463 Residual 34.7 6.3 Lack of Fit 3.4.77 3.78.766 not significant Pure Error 3.65 5.73 Cor Total 87.8 9 The Model F-value of 39.63 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..46 R-Squared.884 Mean 6.5 Adj R-Squared.859 C.V..4 Pred R-Squared.63 PRESS 6.4 Adeq Precision.447 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 59.95.4 59.6 6.83 A-Time 3.58.39.75 4.4. C-Catalyst.3.39.39 3.7. A.8.38.5.63. Final Equation in Terms of Coded Factors: Activity = +59.95 +3.58 * A +.3 * C +.8 * A Final Equation in Terms of Actual Factors: Activity = +59.948 +3.5837 * Time +.997 * Catalyst +.83 * Time Contour plots of the mean models for the responses along with POE for Conversion are shown below: -58

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Conversion X = A: Time Y = C: Catalyst Design Points Actual Factor B: Temperature = -...5 74 76 78 Conversion 86 84 8 88 9 9 DESIGN-EXPERT Plot Activity X = A: Time Y = C: Catalyst Design Points Actual Factor B: Temperature = -...5 Activity 64 66 C: Catalyst. 8 78 76 C: Catalyst. 58 6 6 -.5 74 7 7 68 66 64 6 6 58 56 54 -.5 56 -. -. -. -.5..5. -. -.5..5. A: Tim e A: Tim e DESIGN-EXPERT Plot POE(Conversion) X = A: Time Y = C: Catalyst Design Points Actual Factor B: Temperature = -...5 POE(Conversion) 8.5 8 7.5 7 6.5 6 5.5 C: Catalyst. 5 -.5 5 5.5 6 -. -. -.5..5. A: Tim e The overlay plot shown below identifies a region near the center of the design space that meets the constraints for the process. -59

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Overlay Plot X = A: Time Y = B: Temperature. Overlay Plot Design Points Actual Factor C: Catalyst =. B: Temperature.5. POE(Conv ersion): 8 Conv ersion: 8 6 Activ ity : 6 -.5 -. -. -.5..5. A: Tim e -33 An experiment has been run in a process that applies a coating material to a wafer. Each run in the experiment produced a wafer, and the coating thickness was measured several times at different locations on the wafer. Then the mean y, and standard deviation y of the thickness measurement was obtained. The data [adapted from Box and Draper (987)] are shown in the table below. Run Speed Pressure Distance Mean (y ) Std Dev (y ) -. -. -. 4..5. -. -..3 8.4 3. -. -. 3.7 4.8 4 -.. -. 86. 3.5 5.. -. 36.6 8.4 6.. -. 34.7 6. 7 -.. -..3 7.6 8.. -. 56.3 4.6 9.. -. 7.7 3.6 -. -.. 8... -...7 7.7. -.. 357. 3.9 3 -... 7.3 5. 4... 37.. 5... 5.7 9.5 6 -... 64. 63.5 7... 47. 88.6 8... 73.7. 9 -. -...7 33.8. -.. 39.7 3.5. -.. 4. 8.5 -... 99. 9.4 3... 485.3 44.7 4... 673.7 58. 5 -... 76.7 55.5 6... 5. 38.9 7.... 4.4 (a) What type of design did the experimenters use? Is this a good choice of design for fitting a quadratic model? The design is a 3 3. A better choice would be a 3 central composite design. The CCD gives more information over the design region with fewer points. -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (b) Build models of both responses. The model for the mean is developed as follows: Design Expert Output Response: Mean ANOVA for Response Surface Reduced Cubic Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.89E+6 7.84E+5 6.45 <. significant A 5.64E+5 5.64E+5 85.6 <. B.55E+5.55E+5 7.75 <. C 3.E+5 3.E+5.4 <. AB 534.8 534.8 7.8.6 AC 6837.5 6837.5.43. BC 794.8 794.8 7.48.3 ABC 5483.6 5483.6 8..4 Residual 57874.57 9 346.3 Cor Total.347E+6 6 The Model F-value of 6.45 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 55.9 R-Squared.957 Mean 34.67 Adj R-Squared.94 C.V. 7.54 Pred R-Squared.956 PRESS.7E+5 Adeq Precision 33.333 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 34.67.6 9.44 336.9 A-Speed 77. 3. 49.78 4.4. B-Pressure 9.4 3. 8.9 36.65. C-Distance 3.47 3. 4.4 58.7. AB 66.3 5.93 3.69 99.38. AC 75.46 5.93 4. 8.8. BC 43.58 5.93.4 76.93. ABC 8.79 9.5 4.95 3.63. Final Equation in Terms of Coded Factors: Mean = +34.67 +77. * A +9.4 * B +3.47 * C +66.3 * A * B +75.46 * A * C +43.58 * B * C +8.79 * A * B * C Final Equation in Terms of Actual Factors: Mean = +34.6737 +77. * Speed +9.4 * Pressure +3.47 * Distance +66.3333 * Speed * Pressure +75.45833 * Speed * Distance +43.58333 * Pressure * Distance +8.7875 * Speed * Pressure * Distance The model for the Std. Dev. response is as follows. A square root transformation was applied to correct problems with the normality assumption. -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Std. Dev. Transform: Square root Constant: ANOVA for Response Surface Linear Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.75 3 38.9 4.34.45 significant A 6.5 6.5.84.878 B 6.3 6.3.94. C 73.9 73.9 8.5.86 Residual 6.7 3 8.96 Cor Total 3.9 6 The Model F-value of 4.34 implies the model is significant. There is only a.45% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..99 R-Squared.366 Mean 6. Adj R-Squared.783 C.V. 49.88 Pred R-Squared.359 PRESS 79.5 Adeq Precision 7.78 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 6..58 4.8 7.9 A-Speed.96.7 -.5.4. B-Pressure..7 -.5.67. C-Distance.3.7.57 3.49. Final Equation in Terms of Coded Factors: Sqrt(Std. Dev.) = +6. +.96 * A +. * B +.3 * C Final Equation in Terms of Actual Factors: Sqrt(Std. Dev.) = +6.73 +.95796 * Speed +.96 * Pressure +.643 * Distance Because Factor A is insignificant, it is removed from the model. The reduced linear model analysis is shown below: Design Expert Output Response: Std. Dev. Transform: Square root Constant: ANOVA for Response Surface Reduced Linear Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.3 5. 5.4.6 significant B 6.3 6.3.84.5 C 73.9 73.9 7.97.94 Residual.68 4 9.8 Cor Total 3.9 6 The Model F-value of 5.4 implies the model is significant. There is only a.6% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 3.5 R-Squared.34 Mean 6. Adj R-Squared.59 C.V. 5.74 Pred R-Squared.476 PRESS 75.4 Adeq Precision 6.373 Coefficient Standard 95% CI 95% CI -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Intercept Estimate 6. DF Error.59 Low 4.79 High 7. VIF B-Pressure..7 -.7.69. C-Distance.3.7.54 3.5. Final Equation in Terms of Coded Factors: Sqrt(Std. Dev.) = +6. +. * B +.3 * C Final Equation in Terms of Actual Factors: Sqrt(Std. Dev.) = +6.73 +.96 * Pressure +.643 * Distance The following contour plots graphically represent the two models: DESIGN-EXPERT Plot Mean X = B: Pressure Y = C: Distance Design Points Actual Factor A: Speed =. C: Distance..5. 4 Mean 45 95 9 85 8 75 7 65 6 55 5 DESIGN-EXPERT Plot Sq rt(std. De v.) X = B: Pressure Y = C: Distance Design Points Actual Factor A: Speed =. C: Distance..5. Std. Dev. 6 55 5 45 4 35 3 8 75 7 65 5 -.5 35 -.5 3 5 5 -. -. -.5..5. -. -. -.5..5. B: Press ure B: Press ure (c) Find a set of optimum conditions that result in the mean as large as possible with the standard deviation less than 6. The overlay plot identifies a region that meets the criteria of the mean as large as possible with the standard deviation less than 6. The optimum conditions in coded terms are approximately Speed =., Pressure =.75 and Distance =.5. -63

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Overlay Plot X = B: Pressure Y = C: Distance Design Points Actual Factor A: Speed =...5 Overlay Plot Std. Dev.: 6 Mean: 7 C: Distance. -.5 -. -. -.5..5. B: Press ure -34 A variation of Example 6-. In example 6- we found that one of the process variables (B=pressure) was not important. Dropping this variable produced two replicates of a 3 design. The data are shown below. C D A(+) A(-) y s - - 45, 48 7, 65 57.75.9 + - 68, 8 6, 65 68.5 7.5 - + 43, 45, 4 73. 4.67 + + 75, 7 86, 96 8.75 34.9 Assume that C and D are controllable factors and that A is a noise factor. (a) Fit a model to the mean response. The following is the analysis of variance with all terms in the model: Design Expert Output Response: Mean ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.5 3. A 9.64 9.64 B 6.64 6.64 AB.77.77 Pure Error. Cor Total 3.5 3 Based on the above analysis, the AB interaction is removed from the model and used as error. Design Expert Output Response: Mean ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F -64

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Model 99.8 49.64 95.45.55 not significant A 9.64 9.64..577 B 6.64 6.64 69.9.387 Residual.77.77 Cor Total 3.5 3 The Model F-value of 95.45 implies there is a 5.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..87 R-Squared.9974 Mean 7.9 Adj R-Squared.993 C.V..5 Pred R-Squared.959 PRESS.5 Adeq Precision 3.67 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 7.9.44 64.63 75.75 A-Concentration 4.8.44 -.75.37. B-Stir Rate 7.9.44.63.75. Final Equation in Terms of Coded Factors: Mean = +7.9 +4.8 * A +7.9 * B Final Equation in Terms of Actual Factors: Mean = +7.875 +4.85 * Concentration +7.875 * Stir Rate The following is a contour plot of the mean model:. Mean 8.5 75 B: Stir Rate. 7 -.5 65 6 -. -. -.5..5. A: Concentration (b) Fit a model to the ln(s ) response. The following is the analysis of variance with all terms in the model: Design Expert Output Response: Variance Transform: Natural log Constant: ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] -65

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Source Sum of Squares DF Mean Square F Value Prob > F Model 4.4 3.47 A.74.74 B.3.3 AB.64.64 Pure Error. Cor Total 4.4 3 Based on the above analysis, the AB interaction is removed from the model and applied to the residual error. Design Expert Output Response: Variance Transform: Natural log Constant: ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 3.77.89.94.385 not significant A.74.74.7.3477 B.3.3 3.7.36 Residual.64.64 Cor Total 4.4 3 The "Model F-value" of.94 implies the model is not significant relative to the noise. There is a 38.5 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..8 R-Squared.8545 Mean 5.5 Adj R-Squared.5634 C.V. 5.6 Pred R-Squared -.384 PRESS.8 Adeq Precision 3.954 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 5.5.4.6.34 A-Concentration -.66.4-5.75 4.43. B-Stir Rate.7.4-4.38 5.8. Final Equation in Terms of Coded Factors: Ln(Variance) = +5.5 -.66 * A +.7 * B Final Equation in Terms of Actual Factors: Ln(Variance) = +5.585 -.65945 * Concentration +.73 * Stir Rate The following is a contour plot of the variance model in the untransformed form: -66

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY..5 656 55 5 45 4 35 3 Variance B: Stir Rate. 5 5 -.5 -. -. -.5..5. A: Concentration (c) Find operating conditions that result in the mean filtration rate response exceeding 75 with minimum variance. The overlay plot shown below identifies the region required by the process:. Overlay Plot.5 Mean: 75 B: Stir Rate. Variance: 3 -.5 -. -. -.5..5. A: Concentration (d) Compare your results with those from Example -6 which used the transmission of error approach. How similar are the two answers. The results are very similar. Both require the Concentration to be held at the high level while the stirring rate is held near the middle. -67

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter Experiments with Random Factors Solutions - A textile mill has a large number of looms. Each loom is supposed to provide the same output of cloth per minute. To investigate this assumption, five looms are chosen at random and their output is noted at different times. The following data are obtained: Loom Output (lb/min) 4. 4. 4. 4. 4. 3.9 3.8 3.9 4. 4. 3 4. 4. 4. 4. 3.9 4 3.6 3.8 4. 3.9 3.7 5 3.8 3.6 3.9 3.8 4. (a) Explain why this is a random effects experiment. Are the looms equal in output? Use =.5. The looms used in the experiment are a random sample of all the looms in the manufacturing area. The following is the analysis of variance for the data: Minitab Output ANOVA: Output versus Loom Factor Type Levels Values Loom random 5 3 4 5 Analysis of Variance for Output Source DF SS MS F P Loom 4.346.854 5.77.3 Error.96.48 Total 4.6376 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Loom.4 () + 5() Error.48 () (b) Estimate the variability between looms. ˆ MS Model n MS E. 854. 48. 4 5 (c) Estimate the experimental error variance. ˆ MS E. 48 (d) Find a 95 percent confidence interval for. MS L n MS Model E F,a,n a. 88 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY MS U n MS Model E F L L. 44,a,n a U U. 794 3. 85 (e) Analyze the residuals from this experiment. Do you think that the analysis of variance assumptions are satisfied? There is nothing unusual about the residual plots; therefore, the analysis of variance assumptions are satisfied. Normal Probability Plot of the Residuals (response is Output) Normal Score - - -. -.. Residual.. Residuals Versus the Fitted Values (response is Output).. Residual. -. -. 3.8 3.9 Fitted Value 4. 4. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus Loom (response is Output).. Residual. -. -. 3 Loom 4 5 - A manufacturer suspects that the batches of raw material furnished by her supplier differ significantly in calcium content. There are a large number of batches currently in the warehouse. Five of these are randomly selected for study. A chemist makes five determinations on each batch and obtains the following data: Batch Batch Batch 3 Batch 4 Batch 5 3.46 3.59 3.5 3.8 3.9 3.48 3.46 3.64 3.4 3.46 3.56 3.4 3.46 3.37 3.37 3.39 3.49 3.5 3.46 3.3 3.4 3.5 3.49 3.39 3.38 (a) Is there significant variation in calcium content from batch to batch? Use =.5. Yes, as shown in the Minitab Output below, there is a difference. Minitab Output ANOVA: Calcium versus Batch Factor Type Levels Values Batch random 5 3 4 5 Analysis of Variance for Calcium Source DF SS MS F P Batch 4.96976.444 5.54.4 Error.876.438 Total 4.84576 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Batch.397 () + 5() Error.438 () (b) Estimate the components of variance. MS.. Model MS E 444 438. 397 n 5-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY MS E. 438 (c) Find a 95 percent confidence interval for MS L n MS MS U n MS Model E Model E. F,a,n a F L L,a,n a. 35. 54 U U 9. 76. 97 (d) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? There are five residuals that stand out in the normal probability plot. From the Residual vs. Batch plot, we see that one point per batch appears to stand out. A natural log transformation was applied to the data but did not change the results of the residual analysis. Further investigation should probably be performed to determine if these points are outliers. Normal Probability Plot of the Residuals (response is Calcium) Normal Score - - -.. Residual. -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Calcium). Residual. -. 3.35 3.37 3.39 3.4 3.43 3.45 Fitted Value 3.47 3.49 3.5 3.53 Residuals Versus Batch (response is Calcium). Residual. -. 3 Batch 4 5-3 Several ovens in a metal working shop are used to heat metal specimens. All the ovens are supposed to operate at the same temperature, although it is suspected that this may not be true. Three ovens are selected at random and their temperatures on successive heats are noted. The data collected are as follows: Oven Temperature 49.5 498.3 498. 493.5 493.6 488.5 484.65 479.9 477.35 3 49. 484.8 488.5 473. 47.85 478.65 (a) Is there significant variation in temperature between ovens? Use =.5. The analysis of variance shown below identifies significant variation in temperature between the ovens. Minitab Output General Linear Model: Temperature versus Oven Factor Type Levels Values -5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Oven random 3 3 Analysis of Variance for Temperat, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Oven 594.53 594.53 97.7 8.6.5 Error 43.8 43.8 34.48 Total 4 8.34 Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term Oven () + 4.9333() Error () Error Terms for Tests, using Adjusted SS Source Error DF Error MS Synthesis of Error MS Oven. 34.48 () Variance Components, using Adjusted SS Source Estimated Value Oven 53.7 Error 34.48 (b) Estimate the components of variance. n n i 5 6 36 ni 5 493 a ni 5. MS.. Model MS E 97 7 34 48 533. n 493. MS E 34. 48 (c) Analyze the residuals from this experiment. Draw conclusions about model adequacy. There is a funnel shaped appearance in the plot of residuals versus predicted value indicating a possible non-constant variance. There is also some indication of non-constant variance in the plot of residuals versus oven. The inequality of variance problem is not severe. Normal Probability Plot of the Residuals (response is Temperat) Normal Score - - - Residual -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Temperat) Residual - 48 485 Fitted Value 49 495 Residuals Versus Oven (response is Temperat) Residual - Oven 3-4 An article in the Journal of the Electrochemical Society (Vol. 39, No., 99, pp. 54-53) describes an experiment to investigate the low-pressure vapor deposition of polysilicon. The experiment was carried out in a large-capacity reactor at Sematech in Austin, Texas. The reactor has several wafer positions, and four of these positions are selected at random. The response variable is film thickness uniformity. Three replicates of the experiments were run, and the data are as follows: Wafer Position Uniformity.76 5.67 4.49.43.7.9 3.34.97.47 4.94.36.65 (a) Is there a difference in the wafer positions? Use =.5. Yes, there is a difference. -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output ANOVA: Uniformity versus Wafer Position Factor Type Levels Values Wafer Po fixed 4 3 4 Analysis of Variance for Uniformi Source DF SS MS F P Wafer Po 3 6.98 5.466 8.9.8 Error 8 5.75.65 Total.4373 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Wafer Po () + 3Q[] Error.65 () (b) Estimate the variability due to wafer positions. (c) Estimate the random error component. MSTreatment MSE n 5466.. 65 5844. 3. 65 (d) Analyze the residuals from this experiment and comment on model adequacy. Variability in film thickness seems to depend on wafer position. These observations also show up as outliers on the normal probability plot. Wafer position number appears to have greater variation in uniformity than the other positions. Normal Probability Plot of the Residuals (response is Uniformi) Normal Score - - - Residual -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Uniformi) Residual - 3 Fitted Value 4 Residuals Versus Wafer Po (response is Uniformi) Residual - Wafer Po 3 4-5 Consider the vapor deposition experiment described in Problem -4. (a) Estimate the total variability in the uniformity response. ˆ ˆ. 5848. 65. 37 (b) How much of the total variability in the uniformity response is due to the difference between positions in the reactor? ˆ ˆ ˆ. 5848. 7845. 37 (c) To what level could the variability in the uniformity response be reduced, if the position-to-position variability in the reactor could be eliminated? Do you believe this is a significant reduction? -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The variability would be reduced from.37 to ˆ. 65 which is a reduction of approximately:. 37. 65 7%. 37-6 An article in the Journal of Quality Technology (Vol. 3, No., 98, pp. -4) describes and experiment that investigates the effects of four bleaching chemicals on pulp brightness. These four chemicals were selected at random from a large population of potential bleaching agents. The data are as follows: Chemical Pulp Brightness 77.99 74.466 9.746 76.8 8.876 8.5 79.36 8.94 8.346 73.385 3 79.47 78.7 9.596 8.8 8.66 4 78. 78.358 77.544 77.364 77.386 (a) Is there a difference in the chemical types? Use =.5. The computer output shows that the null hypothesis cannot be rejected. Therefore, there is no evidence that there is a difference in chemical types. Minitab Output ANOVA: Brightness versus Chemical Factor Type Levels Values Chemical random 4 3 4 Analysis of Variance for Brightne Source DF SS MS F P Chemical 3 53.98 7.99.75.538 Error 6 383.99 4. Total 9 437.97 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Chemical -. () + 5() Error 3.999 () (b) Estimate the variability due to chemical types. MS Treatment MSE n 7. 994 3. 999. 5 which agrees with the Minitab output. Because the variance component cannot be negative, this likely means that the variability due to chemical types is zero. (c) Estimate the variability due to random error. 3999. (d) Analyze the residuals from this experiment and comment on model adequacy. -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Two data points appear to be outliers in the normal probability plot of effects. These outliers belong to chemical types and 3 and should be investigated. There seems to be much less variability in brightness with chemical type 4. Normal Probability Plot of the Residuals (response is Brightne) Normal Score - - -5 Residual 5 Residuals Versus the Fitted Values (response is Brightne) Residual 5-5 78 79 8 Fitted Value 8 8 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Residuals Versus Chemical (response is Brightne) Residual 5-5 Chemical 3 4-7 Consider the one-way balanced, random effects method. Develop a procedure for finding a (- ) percent confidence interval for /( ). We know that PL U PL U PL U L U P L U -8 Refer to Problem -. (a) What is the probability of accepting H if is four times the error variance? 4. 6 n 5 4 a N a 55. 35, from the OC curve. 4 (b) If the difference between looms is large enough to increase the standard deviation of an observation by percent, we wish to detect this with a probability of at least.8. What sample size should be used? a 4 N a 55 5. P( accept).. P n.. n n 44 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Trial and Error yields: n P(accept) 5.79.6 45.3.3 4 65.67. Choose n 4, therefore N 7-9 An experiment was performed to investigate the capability of a measurement system. Ten parts were randomly selected, and two randomly selected operators measured each part three times. The tests were made in random order, and the data below resulted. (a) Analyze the data from this experiment. Minitab Output ANOVA: Measurement versus Part, Operator Operator Operator Measurements Measurements Part -------------------------- ------------------------- Number 3 3 5 49 5 5 48 5 5 5 5 5 5 5 3 53 5 5 54 5 5 4 49 5 5 48 5 5 5 48 49 48 48 49 48 6 5 5 5 5 5 5 7 5 5 5 5 5 5 8 5 5 49 53 48 5 9 5 5 5 5 48 49 47 46 49 46 47 48 Factor Type Levels Values Part random 3 4 5 6 7 8 9 Operator random Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.7. 8.8. Operator.47.47.69.47 Part*Operator 9 5.47.6.4.97 Error 4 6..5 Total 59 64.85 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Part.73333 3 (4) + 3(3) + 6() Operator -.67 3 (4) + 3(3) + 3() 3 Part*Operator -.9938 4 (4) + 3(3) 4 Error.5 (4) (b) Find point estimates of the variance components using the analysis of variance method. MS E 5. -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY MS AB MS A MS E n MS B MS an MS bn All estimates agree with the Minitab output. AB.. 6859 5, assume = 3. 85. 6859 ˆ. 7333 3. 46667. 6859 ˆ, assume = AB 3 - Reconsider the data in Problem 5-6. Suppose that both factors, machines and operators, are chosen at random. (a) Analyze the data from this experiment. Machine Operator 3 4 9 8 5 9 8 4 9 3 6 4 4 5 9 7 The following Minitab output contains the analysis of variance and the variance component estimates: Minitab Output ANOVA: Strength versus Operator, Machine Factor Type Levels Values Operator random 3 3 Machine random 4 3 4 Analysis of Variance for Strength Source DF SS MS F P Operator 6.333 8.67.77. Machine 3.458 4.53.56.66 Operator*Machine 6 44.667 7.444.96.5 Error 45.5 3.79 Total 3 6.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Operator 9.93 3 (4) + (3) + 8() Machine -.5486 3 (4) + (3) + 6() 3 Operator*Machine.864 4 (4) + (3) 4 Error 3.797 (4) (b) Find point estimates of the variance components using the analysis of variance method. MS AB MS E 37967. MS n E.. 7 44444 3 7967 8639. -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY MS MS an B AB MS MS bn A AB 4578. 7. 44444 3 ( ), assume 86667. 7. 44444 9. 98 4 ( ) These results agree with the Minitab variance component analysis. - Reconsider the data in Problem 5-3. Suppose that both factors are random. (a) Analyze the data from this experiment. Minitab Output General Linear Model: Response versus Row, Column Factor Type Levels Values Row random 3 3 Column random 4 3 4 Column Factor Row Factor 3 4 36 39 36 3 8 3 3 37 33 34 Analysis of Variance for Response, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Row 58.5 58.5 9.5 6.4 ** Column 3 8.97 8.97 9.639. ** Row*Column 6 8.833 8.833 4.86 ** Error... Total 638.5 ** Denominator of F-test is zero. Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term Row (4) + (3) + 4.() Column (4) + (3) + 3.() 3 Row*Column (4) + (3) 4 Error (4) Error Terms for Tests, using Adjusted SS Source Error DF Error MS Synthesis of Error MS Row * 4.86 (3) Column * 4.86 (3) 3 Row*Column * * (4) Variance Components, using Adjusted SS Source Estimated Value Row 7.36 Column.6 Row*Column 4.856 Error. (b) Estimate the variance components. -5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Because the experiment is unreplicated and the interaction term was included in the model, there is no estimate of MS E, and therefore, no estimate of. MS AB MS n MS B MS AB an MS A MS bn These estimates agree with the Minitab output. AB E ˆ 4. 856 ˆ 4. 856 9. 6389 4. 856 ˆ. 6 3 9. 5 4. 856 7. 36 4 - Suppose that in Problem 5- the furnace positions were randomly selected, resulting in a mixed model experiment. Reanalyze the data from this experiment under this new assumption. Estimate the appropriate model components. The following analysis assumes a restricted model: Minitab Output ANOVA: Density versus Position, Temperature Factor Type Levels Values Position random Temperat fixed 3 8 85 85 Analysis of Variance for Density Temperature ( C) Position 8 85 85 57 63 565 565 8 5 583 43 59 58 988 56 547 6 538 5 4 53 Source DF SS MS F P Position 76 76 6.. Temperat 94534 4767 55.5. Position*Temperat 88 49.9.47 Error 537 448 Total 7 95869 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Position 745.83 4 (4) + 9() Temperat 3 (4) + 3(3) + 6Q[] 3 Position*Temperat -.83 4 (4) + 3(3) 4 Error 447.56 (4) MS AB MS n MS E ˆ 447. 56 E 49 448 ˆ assume 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ MS A bn MS E ˆ 76 448 745. 83 3 3 These results agree with the Minitab output. -3 Reanalyze the measurement systems experiment in Problem -9, assuming that operators are a fixed factor. Estimate the appropriate model components. The following analysis assumes a restricted model: Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 3 4 5 6 7 8 9 Operator fixed Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.7. 7.33. Operator.47.47.69.47 Part*Operator 9 5.47.6.4.97 Error 4 6..5 Total 59 64.85 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Part.5836 4 (4) + 6() Operator 3 (4) + 3(3) + 3Q[] 3 Part*Operator -.994 4 (4) + 3(3) 4 Error.5 (4) MS E ˆ. 5 MS MS E. 685. 5 ˆ ˆ assume n 3 MS A MS E. 85. 5 ˆ ˆ. 58364 bn 3 AB These results agree with the Minitab output. -4 In problem 5-6, suppose that there are only four machines of interest, but the operators were selected at random. (a) What type of model is appropriate? A mixed model is appropriate. (b) Perform the analysis and estimate the model components. The following analysis assumes a restricted model: Minitab Output ANOVA: Strength versus Operator, Machine Factor Type Levels Values -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Operator random 3 3 Machine fixed 4 3 4 Analysis of Variance for Strength Source DF SS MS F P Operator 6.333 8.67.4. Machine 3.458 4.53.56.66 Operator*Machine 6 44.667 7.444.96.5 Error 45.5 3.79 Total 3 6.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Operator 9.547 4 (4) + 8() Machine 3 (4) + (3) + 6Q[] 3 Operator*Machine.86 4 (4) + (3) 4 Error 3.79 (4) ˆ MS MS AB A MS E ˆ 3. 79 MS n MS bn These results agree with the Minitab output. E E 7. 444 3. 79 ˆ. 86 8. 67 3. 79 ˆ 9. 547 4-5 By application of the expectation operator, develop the expected mean squares for the two-factor factorial, mixed model. Use the restricted model assumptions. Check your results with the expected mean squares given in Table - to see that they agree. The sums of squares may be written as SS AB n SS A bn a yi.. y... i b, SSB an y. j. y... j a b yij. yi.. y. j. y... a b n, SSE yijk y... i j i j k y Using the model ijk i j ijk y y y y i... j. ij.... ij, we may find that i i. j j.... j. i. Using the assumptions for the restricted form of the mixed model,.,. j. Substituting these expressions into the sums of squares yields.. i.. ij ij., which imply that -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY SS SS SS SS A B AB E bn an n a i b j a b i j ) a b n ijk ij. i j k j i.. j. ij i..... i.... ij. Using the assumption that E ijk, V ( ijk ) E ijk i' j' k' squares by its degrees of freedom and take the expectation to produce E E MS E MS E MS A B AB MS E i... j., and bn E a an b n a b i j j a b i E i.... a b ij i. i j, we may divide each sum of Note that EMS B and EMS E are the results given in Table 8-3. We need to simplify MS A E. Consider E MS AB E E MS A MS E E crossproducts E MS A A a since ij is, a bn a bn a i MSA n a i i i i a i bn a a a a a b i NID. Consider E E E MS E MS AB AB MS n AB MS AB a b a b n n a a i j a b b i j i. E ij i. b a b a E and -9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Thus EMS A and MS AB E agree with table -8. -6 Consider the three-factor factorial design in Example -6. Propose appropriate test statistics for all main effects and interactions. Repeat for the case where A and B are fixed and C is random. If all three factors are random there are no exact tests on main effects. We could use the following: A : F B : F MS MS MS MS MS C : F MS A AB B AB C AC MS MS MS MS ABC AC ABC MS MS If A and B are fixed and C is random, the expected mean squares are (assuming the restricted for m of the model): F F R R a b c n Factor i j k l E(MS) i b c n j a c n BC ABC bn an k a b n abn c n ij n ik b n bn jk a n an ijk n n ijkl BC i bcn a j acn b cn These are exact tests for all effects. ji a b -7 Consider the experiment in Example -7. Analyze the data for the case where A, B, and C are random. Minitab Output ANOVA: Drop versus Temp, Operator, Gauge Factor Type Levels Values Temp random 3 6 75 9 Operator random 4 3 4 Gauge random 3 3 Analysis of Variance for Drop Source DF SS MS F P Temp 3.36 5.68.3.7 x -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Operator 3 43.8 4.7.63.66 x Gauge 7.9 3.6.6.938 x Temp*Operator 6.97. 4.59. Temp*Gauge 4 37.89 34.47.49.99 Operator*Gauge 6 9.47 34.9.5.8 Temp*Operator*Gauge 66. 3.84.65.788 Error 36 77.5.4 Total 7 395.3 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Temp.44 * (8) + (7) + 8(5) + 6(4) + 4() Operator -4.544 * (8) + (7) + 6(6) + 6(4) + 8() 3 Gauge -.64 * (8) + (7) + 6(6) + 8(5) + 4(3) 4 Temp*Operator 3.359 7 (8) + (7) + 6(4) 5 Temp*Gauge.579 7 (8) + (7) + 8(5) 6 Operator*Gauge 3.5 7 (8) + (7) + 6(6) 7 Temp*Operator*Gauge -3.78 8 (8) + (7) 8 Error.43 (8) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS Temp 6.97.63 (4) + (5) - (7) Operator 7.9 3.6 (4) + (6) - (7) 3 Gauge 5.98 55.54 (5) + (6) - (7) Since all three factors are random there are no exact tests on main effects. Minitab uses an approximate F test for the these factors. -8 Derive the expected mean squares shown in Table -4. F R R R a b c n Factor i j k l E(MS) i b c n n bn j a c n an acn k a b n an abn ij c n n cn ik b n n bn jk a n an ijk n n ijkl cn bcn a i -9 Consider a four-factor factorial experiment where factor A is at a levels, factor B is at b levels, factor C is at c levels, factor D is at d levels, and there are n replicates. Write down the sums of squares, the degrees of freedom, and the expected mean squares for the following cases. Do exact tests exist for all effects? If not, propose test statistics for those effects that cannot be directly tested. Assume the restricted model on all cases. You may use a computer package such as Minitab. The four factor model is: -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY yijklh i j k l ij ik il jk jl kl ijklh ijk ijl jkl ikl ijkl To simplify the expected mean square derivations, let capital Latin letters represent the factor effects or variance components. For example, A bcdn i, or B acdn. a (a) A, B, C, and D are fixed factors. F F F F R a b c d n Factor i j k l h E(MS) i b c d n A j a c d n B k a b d n C l a b c n D ( ) ij c d n AB ( ) ik b d n AC ( ) il b c n AD ( ) jk a d n BC ( ) jl a c n BD ( ) kl a b n CD ( ) ijk d n ABC ( ) ijl c n ABD ( ) jkl a n BCD ( ) ikl b n ACD ( ) ijkl n ABCD ( ijkl) h There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B fixed H L C fixed H L D fixed H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3.49.49 B.3.3..9 C.3.3.9.767 D.3.3..9 A*B 3.3 3.3.5.6 A*C 3.3 3.3.5.6 A*D 3.3 3.3.5.6 B*C 3.3 3.3.5.6 B*D 3.3 3.3.5.6 C*D 3.3 3.3.5.6 A*B*C 3.3 3.3.5.6 A*B*D 8.3 8.3.7.5 -

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A*C*D 3.3 3.3.5.6 B*C*D 3.3 3.3.5.6 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A 6 (6) + 6Q[] B 6 (6) + 6Q[] 3 C 6 (6) + 6Q[3] 4 D 6 (6) + 6Q[4] 5 A*B 6 (6) + 8Q[5] 6 A*C 6 (6) + 8Q[6] 7 A*D 6 (6) + 8Q[7] 8 B*C 6 (6) + 8Q[8] 9 B*D 6 (6) + 8Q[9] C*D 6 (6) + 8Q[] A*B*C 6 (6) + 4Q[] A*B*D 6 (6) + 4Q[] 3 A*C*D 6 (6) + 4Q[3] 4 B*C*D 6 (6) + 4Q[4] 5 A*B*C*D 6 (6) + Q[5] 6 Error.38 (6) (b) A, B, C, and D are random factors. R R R R R a b c d n Factor i j k l h E(MS) i b c d n ABCD ACD ABD ABC AD AC AB A j a c d n ABCD BCD ABD ABC BD BC AB B k a b d n ABCD ACD BCD ABC AB BC CD C l a b c n ABCD ACD BCD ABD BD AD CD D ( ) ij c d n ABCD ABC ABD AB ( ) ik b d n ABCD ABC ACD AC ( ) il b c n ABCD ABD ACD AD ( ) jk a d n ABCD ABC BCD BC ( ) jl a c n ABCD ABD BCD BD ( ) kl a b n ABCD ACD BCD CD ( ) ijk d n ABCD ABC ( ) ijl c n ABCD ABD ( ) jkl a n ABCD BCD ( ) ikl b n ABCD ACD ( ) ijkl n ABCD ( ijkl) h No exact tests exist on main effects or two-factor interactions. For main effects use statistics such as: AF : MS MS MS MS MS MS MS MS A ABC ABD ACD AB AC AD ABCD For testing two-factor interactions use statistics such as: AB: F MS MS AB ABC MS MS ABCD ABD The results can also be generated in Minitab as follows: -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A random H L B random H L C random H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3 ** B.3.3 ** C.3.3.36.843 x D.3.3 ** A*B 3.3 3.3..796 x A*C 3.3 3.3..667 x A*D 3.3 3.3..796 x B*C 3.3 3.3..667 x B*D 3.3 3.3..796 x C*D 3.3 3.3..667 x A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5 A*C*D 3.3 3.3..5 B*C*D 3.3 3.3..5 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A.75 * (6) + (5) + 4(3) + 4() + 4() + 8(7) + 8(6) + 8(5) + 6() B.375 * (6) + (5) + 4(4) + 4() + 4() + 8(9) + 8(8) + 8(5) + 6() 3 C -.5 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(8) + 8(6) + 6(3) 4 D.375 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(9) + 8(7) + 6(4) 5 A*B -3.5 * (6) + (5) + 4() + 4() + 8(5) 6 A*C. * (6) + (5) + 4(3) + 4() + 8(6) 7 A*D -3.5 * (6) + (5) + 4(3) + 4() + 8(7) 8 B*C. * (6) + (5) + 4(4) + 4() + 8(8) 9 B*D -3.5 * (6) + (5) + 4(4) + 4() + 8(9) C*D. * (6) + (5) + 4(4) + 4(3) + 8() A*B*C. 5 (6) + (5) + 4() A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D. 5 (6) + (5) + 4(3) 4 B*C*D. 5 (6) + (5) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS A.56 * (5) + (6) + (7) - () - () - (3) + (5) B.56 * (5) + (8) + (9) - () - () - (4) + (5) 3 C.4 3.3 (6) + (8) + () - () - (3) - (4) + (5) 4 D.56 * (7) + (9) + () - () - (3) - (4) + (5) 5 A*B.98 8.3 () + () - (5) 6 A*C.33 3.3 () + (3) - (5) 7 A*D.98 8.3 () + (3) - (5) -4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 8 B*C.33 3.3 () + (4) - (5) 9 B*D.98 8.3 () + (4) - (5) C*D.33 3.3 (3) + (4) - (5) (c) A is fixed and B, C, and D are random. F R R R R a b c d n Factor i j k l h E(MS) i b c d n ABCD ACD ABD ABC AD AC AB A j a c d n BCD ABD BC B k a b d n BCD BC CD C l a b c n BCD BDCD D ( ) ij c d n ABCD ABC ABD AB ( ) ik b d n ABCD ABC ACD AC ( ) il b c n ABCD ABD ACD AD ( ) jk a d n BCD BC ( ) jl a c n BCD BD ( ) kl a b n BCD CD ( ) ijk d n ABCD ABC ( ) ijl c n ABCD ABD ( ) jkl a n BCD ( ) ikl b n ABCD ACD ( ) ijkl n ABCD ( ijkl) h No exact tests exist on main effects or two-factor interactions involving the fixed factor A. To test the fixed factor A use AF : MS MS MS MS MS MS MS MS A ABC ABD ACD AB AC AD ABCD Random main effects could be tested by, for example: DF : MS MS D ABC MS MS ABCD ABD For testing two-factor interactions involving A use: AB: F MS MS AB ABC MS MS ABCD ABD The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B random H L C random H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3 ** B.3.3.4.97 x -5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY C.3.3.36.76 x D.3.3.4.97 x A*B 3.3 3.3..796 x A*C 3.3 3.3..667 x A*D 3.3 3.3..796 x B*C 3.3 3.3..5 B*D 3.3 3.3..5 C*D 3.3 3.3..5 A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5 A*C*D 3.3 3.3..5 B*C*D 3.3 3.3.5.6 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A * (6) + (5) + 4(3) + 4() + 4() + 8(7) + 8(6) + 8(5) + 6Q[] B -.875 * (6) + 4(4) + 8(9) + 8(8) + 6() 3 C -.5 * (6) + 4(4) + 8() + 8(8) + 6(3) 4 D -.875 * (6) + 4(4) + 8() + 8(9) + 6(4) 5 A*B -3.5 * (6) + (5) + 4() + 4() + 8(5) 6 A*C. * (6) + (5) + 4(3) + 4() + 8(6) 7 A*D -3.5 * (6) + (5) + 4(3) + 4() + 8(7) 8 B*C. 4 (6) + 4(4) + 8(8) 9 B*D. 4 (6) + 4(4) + 8(9) C*D. 4 (6) + 4(4) + 8() A*B*C. 5 (6) + (5) + 4() A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D. 5 (6) + (5) + 4(3) 4 B*C*D -.35 6 (6) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS A.56 * (5) + (6) + (7) - () - () - (3) + (5) B.33 3.3 (8) + (9) - (4) 3 C.33 3.3 (8) + () - (4) 4 D.33 3.3 (9) + () - (4) 5 A*B.98 8.3 () + () - (5) 6 A*C.33 3.3 () + (3) - (5) 7 A*D.98 8.3 () + (3) - (5) (d) A and B are fixed and C and D are random. F F R R R a b c d n Factor i j k l h E(MS) i b c d n ACD AD AC A j a c d n BCD BC BD B k a b d n CD C l a b c n CD D ( ) ij c d n ( ) ik b d n ACD AC ( ) il b c n ACD AD ABCD ABC ABD AB -6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ( ) jk a d n BCD BC ( ) jl a c n BCD BD ( ) kl a b n CD ( ) ijk d n ABCD ABC ( ) ijl c n ABCD ABD ( ) jkl a n BCD ( ) ikl b n ACD ( ) ijkl n ABCD ( ijkl) h There are no exact tests on the fixed factors A and B, or their two-factor interaction AB. The appropriate test statistics are: AF : BF : AB: F MS MS MS MS MS MS A AC B BC AB MS MS MS MS ABC ACD AD BCD MS MS BD ABCD ABD The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B fixed H L C random H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3.96.64 x B.3.3.4.97 x C.3.3.36.656 D.3.3.4.874 A*B 3.3 3.3..796 x A*C 3.3 3.3..5 A*D 3.3 3.3..5 B*C 3.3 3.3..5 B*D 3.3 3.3..5 C*D 3.3 3.3.5.6 A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5 A*C*D 3.3 3.3.5.6 B*C*D 3.3 3.3.5.6 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A * (6) + 4(3) + 8(7) + 8(6) + 6Q[] B * (6) + 4(4) + 8(9) + 8(8) + 6Q[] 3 C -.5 (6) + 8() + 6(3) -7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 D -.875 (6) + 8() + 6(4) 5 A*B * (6) + (5) + 4() + 4() + 8Q[5] 6 A*C. 3 (6) + 4(3) + 8(6) 7 A*D. 3 (6) + 4(3) + 8(7) 8 B*C. 4 (6) + 4(4) + 8(8) 9 B*D. 4 (6) + 4(4) + 8(9) C*D -.563 6 (6) + 8() A*B*C. 5 (6) + (5) + 4() A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D -.35 6 (6) + 4(3) 4 B*C*D -.35 6 (6) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS A.33 3.3 (6) + (7) - (3) B.33 3.3 (8) + (9) - (4) 5 A*B.98 8.3 () + () - (5) (e) A, B and C are fixed and D is random. F F F R R a b c d n Factor i j k l h E(MS) i b c d n AD A j a c d n BD B k a b d n CD C l a b c n D ( ) ij c d n ABD AB ( ) ik b d n ACD AC ( ) il b c n AD ( ) jk a d n BCD BC ( ) jl a c n BD ( ) kl a b n CD ( ) ijk d n ABCD ABC ( ) ijl c n ABD ( ) jkl a n BCD ( ) ikl b n ACD ( ) ijkl n ABCD ( ijkl) h There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B fixed H L C fixed H L D random H L Analysis of Variance for y Source DF SS MS F P -8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A 6.3 6.3.96.395 B.3.3.4.874 C.3.3.36.656 D.3.3..9 A*B 3.3 3.3..795 A*C 3.3 3.3..5 A*D 3.3 3.3.5.6 B*C 3.3 3.3..5 B*D 3.3 3.3.5.6 C*D 3.3 3.3.5.6 A*B*C 3.3 3.3..5 A*B*D 8.3 8.3.7.5 A*C*D 3.3 3.3.5.6 B*C*D 3.3 3.3.5.6 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A 7 (6) + 8(7) + 6Q[] B 9 (6) + 8(9) + 6Q[] 3 C (6) + 8() + 6Q[3] 4 D -.7656 6 (6) + 6(4) 5 A*B (6) + 4() + 8Q[5] 6 A*C 3 (6) + 4(3) + 8Q[6] 7 A*D -.563 6 (6) + 8(7) 8 B*C 4 (6) + 4(4) + 8Q[8] 9 B*D -.563 6 (6) + 8(9) C*D -.563 6 (6) + 8() A*B*C 5 (6) + (5) + 4Q[] A*B*D 3.9375 6 (6) + 4() 3 A*C*D -.35 6 (6) + 4(3) 4 B*C*D -.35 6 (6) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) - Reconsider cases (c), (d) and (e) of Problem -9. Obtain the expected mean squares assuming the unrestricted model. You may use a computer package such as Minitab. Compare your results with those for the restricted model. A is fixed and B, C, and D are random. Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B random H L C random H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3 ** B.3.3 ** C.3.3.36.843 x D.3.3 ** A*B 3.3 3.3..796 x A*C 3.3 3.3..667 x A*D 3.3 3.3..796 x B*C 3.3 3.3..667 x B*D 3.3 3.3..796 x C*D 3.3 3.3..667 x A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A*C*D 3.3 3.3..5 B*C*D 3.3 3.3..5 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) A * (6) + (5) + 4(3) + 4() + 4() + 8(7) + 8(6) + 8(5) + Q[] B.375 * (6) + (5) + 4(4) + 4() + 4() + 8(9) + 8(8) + 8(5) + 6() 3 C -.5 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(8) + 8(6) + 6(3) 4 D.375 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(9) + 8(7) + 6(4) 5 A*B -3.5 * (6) + (5) + 4() + 4() + 8(5) 6 A*C. * (6) + (5) + 4(3) + 4() + 8(6) 7 A*D -3.5 * (6) + (5) + 4(3) + 4() + 8(7) 8 B*C. * (6) + (5) + 4(4) + 4() + 8(8) 9 B*D -3.5 * (6) + (5) + 4(4) + 4() + 8(9) C*D. * (6) + (5) + 4(4) + 4(3) + 8() A*B*C. 5 (6) + (5) + 4() A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D. 5 (6) + (5) + 4(3) 4 B*C*D. 5 (6) + (5) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS A.56 * (5) + (6) + (7) - () - () - (3) + (5) B.56 * (5) + (8) + (9) - () - () - (4) + (5) 3 C.4 3.3 (6) + (8) + () - () - (3) - (4) + (5) 4 D.56 * (7) + (9) + () - () - (3) - (4) + (5) 5 A*B.98 8.3 () + () - (5) 6 A*C.33 3.3 () + (3) - (5) 7 A*D.98 8.3 () + (3) - (5) 8 B*C.33 3.3 () + (4) - (5) 9 B*D.98 8.3 () + (4) - (5) C*D.33 3.3 (3) + (4) - (5) A and B are fixed and C and D are random. Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B fixed H L C random H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3.96.64 x B.3.3.4.97 x C.3.3.36.843 x D.3.3 ** A*B 3.3 3.3..796 x A*C 3.3 3.3..667 x A*D 3.3 3.3..796 x -3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B*C 3.3 3.3..667 x B*D 3.3 3.3..796 x C*D 3.3 3.3..667 x A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5 A*C*D 3.3 3.3..5 B*C*D 3.3 3.3..5 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) A * (6) + (5) + 4(3) + 4() + 4() + 8(7) + 8(6) + Q[,5] B * (6) + (5) + 4(4) + 4() + 4() + 8(9) + 8(8) + Q[,5] 3 C -.5 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(8) + 8(6) + 6(3) 4 D.375 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(9) + 8(7) + 6(4) 5 A*B * (6) + (5) + 4() + 4() + Q[5] 6 A*C. * (6) + (5) + 4(3) + 4() + 8(6) 7 A*D -3.5 * (6) + (5) + 4(3) + 4() + 8(7) 8 B*C. * (6) + (5) + 4(4) + 4() + 8(8) 9 B*D -3.5 * (6) + (5) + 4(4) + 4() + 8(9) C*D. * (6) + (5) + 4(4) + 4(3) + 8() A*B*C. 5 (6) + (5) + 4() A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D. 5 (6) + (5) + 4(3) 4 B*C*D. 5 (6) + (5) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS A.33 3.3 (6) + (7) - (3) B.33 3.3 (8) + (9) - (4) 3 C.4 3.3 (6) + (8) + () - () - (3) - (4) + (5) 4 D.56 * (7) + (9) + () - () - (3) - (4) + (5) 5 A*B.98 8.3 () + () - (5) 6 A*C.33 3.3 () + (3) - (5) 7 A*D.98 8.3 () + (3) - (5) 8 B*C.33 3.3 () + (4) - (5) 9 B*D.98 8.3 () + (4) - (5) C*D.33 3.3 (3) + (4) - (5) (e) A, B and C are fixed and D is random. Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed H L B fixed H L C fixed H L D random H L Analysis of Variance for y Source DF SS MS F P A 6.3 6.3.96.395-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B.3.3.4.874 C.3.3.36.656 D.3.3 ** A*B 3.3 3.3..795 A*C 3.3 3.3..5 A*D 3.3 3.3..796 x B*C 3.3 3.3..5 B*D 3.3 3.3..796 x C*D 3.3 3.3..667 x A*B*C 3.3 3.3..5 A*B*D 8.3 8.3 9..5 A*C*D 3.3 3.3..5 B*C*D 3.3 3.3..5 A*B*C*D 3.3 3.3.5.6 Error 6 98..38 Total 3 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) A 7 (6) + (5) + 4(3) + 4() + 8(7) + Q[,5,6,] B 9 (6) + (5) + 4(4) + 4() + 8(9) + Q[,5,8,] 3 C (6) + (5) + 4(4) + 4(3) + 8() + Q[3,6,8,] 4 D.375 * (6) + (5) + 4(4) + 4(3) + 4() + 8() + 8(9) + 8(7) + 6(4) 5 A*B (6) + (5) + 4() + Q[5,] 6 A*C 3 (6) + (5) + 4(3) + Q[6,] 7 A*D -3.5 * (6) + (5) + 4(3) + 4() + 8(7) 8 B*C 4 (6) + (5) + 4(4) + Q[8,] 9 B*D -3.5 * (6) + (5) + 4(4) + 4() + 8(9) C*D. * (6) + (5) + 4(4) + 4(3) + 8() A*B*C 5 (6) + (5) + Q[] A*B*D 6.5 5 (6) + (5) + 4() 3 A*C*D. 5 (6) + (5) + 4(3) 4 B*C*D. 5 (6) + (5) + 4(4) 5 A*B*C*D -4.65 6 (6) + (5) 6 Error.375 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 4 D.56 * (7) + (9) + () - () - (3) - (4) + (5) 7 A*D.98 8.3 () + (3) - (5) 9 B*D.98 8.3 () + (4) - (5) C*D.33 3.3 (3) + (4) - (5) - In Problem 5-7, assume that the three operators were selected at random. Analyze the data under these conditions and draw conclusions. Estimate the variance components. Minitab Output ANOVA: Score versus Cycle Time, Operator, Temperature Factor Type Levels Values Cycle Ti fixed 3 4 5 6 Operator random 3 3 Temperat fixed 3 35 Analysis of Variance for Score Source DF SS MS F P Cycle Ti 436. 8..45. Operator 6.333 3.667 39.86. Temperat 5.74 5.74 8.89.96-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Cycle Ti*Operator 4 355.667 88.97 7.3. Cycle Ti*Temperat 78.85 39.47 3.4.37 Operator*Temperat.59 5.63.7.94 Cycle Ti*Operator*Temperat 4 46.85.546 3.5.6 Error 36 8. 3.78 Total 53 357.333 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Cycle Ti 4 (8) + 6(4) + 8Q[] Operator 7.77 8 (8) + 8() 3 Temperat 6 (8) + 9(6) + 7Q[3] 4 Cycle Ti*Operator 4.73 8 (8) + 6(4) 5 Cycle Ti*Temperat 7 (8) + 3(7) + 9Q[5] 6 Operator*Temperat.63 8 (8) + 9(6) 7 Cycle Ti*Operator*Temperat.756 8 (8) + 3(7) 8 Error 3.778 (8) The following calculations agree with the Minitab results: MS MS MS ABC BC MS E ˆ 3. 7778 MS n MS an AC E MS bn MSC MS abn E E E ˆ ˆ ˆ ˆ. 54696 3. 77778. 756 3 88. 9667 3. 77778 4. 735 3 5. 6963 3. 77778. 63 3 3 3. 66667 3. 77778 7. 776 3 3 - Consider the three-factor model yijk i j k ijk ij jk Assuming that all the factors are random, develop the analysis of variance table, including the expected mean squares. Propose appropriate test statistics for all effects. Source DF E(MS) A a- c bc B b- c a ac C c- a ab AB (a-)(b-) c BC (b-)(c-) a Error (AC + ABC) b(a-)(c-) Total abc- There are exact tests for all effects except B. To test B, use the statistic F MS MS B AB MS MS E BC -33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -3 The three-factor model for a single replicate is ( ) ( ) ( ) ( ) y ijk i j k ij jk ik ijk ijk If all the factors are random, can any effects be tested? If the three-factor interaction and the ( ) ij interaction do not exist, can all the remaining effects be tested. The expected mean squares are found by referring to Table -9, deleting the line for the error term ( ijk ) l and setting n=. The three-factor interaction now cannot be tested; however, exact tests exist for the twofactor interactions and approximate F tests can be conducted for the main effects. For example, to test the main effect of A, use F MS MS A AB MS MS ABC AC If ( ) ijk and ( ) ij can be eliminated, the model becomes yijk i j k ijk ij jk ik ijk For this model, the analysis of variance is Source DF E(MS) A a- b bc B b- a ac C c- a b ab AC (a-)(c-) b BC (b-)(c-) a Error (AB + ABC) c(a-)(b-) Total abc- There are exact tests for all effect except C. To test the main effect of C, use the statistic: F MS MS C BC MS MS E AC -4 In Problem 5-6, assume that both machines and operators were chosen randomly. Determine the power of the test for detecting a machine effect such that, where is the variance component for the machine factor. Are two replicates sufficient? an n If, then an estimate of 379., and an estimate of n 745., from the analysis of variance table. Then -34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3 3. 79.. 49 7. 45 and the other OC curve parameters are 3 and 6. This results in 75. approximately, with 5., or 9. with.. Two replicates does not seem sufficient. -5 In the two-factor mixed model analysis of variance, show that Cov ii'. a Since (constant) we have i ij a i V Cov, a ij a a! a Cov ij, i' a! a! a aa Cov ij, i' j Cov, ij i' j a a ij, i' j a V ij, which implies that i ij i' j j for -6 Show that the method of analysis of variance always produces unbiased point estimates of the variance component in any random or mixed model. Let g be the vector of mean squares from the analysis of variance, chosen so that E(g) does not contain any fixed effects. Let be the vector of variance components such that E( g) A, where A is a matrix of constants. Now in the analysis of variance method of variance component estimation, we equate observed and expected mean squares, i.e. Since - A always exists then, E g = As ˆ s A - - - s = EA g A Eg= A As s Thus is an unbiased estimator of. This and other properties of the analysis of variance method are discussed by Searle (97a). - g -7 Invoking the usual normality assumptions, find an expression for the probability that a negative estimate of a variance component will be obtained by the analysis of variance method. Using this result, write a statement giving the probability that in a one-factor analysis of variance. Comment on the usefulness of this probability statement. -35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Suppose MS MS, where MS i for i=, are two mean squares and c is a constant. The c probability that ˆ (negative) is MS MS E MS E MS EMS P ˆ P MS MS P P P Fu, v MS MS EMS EMS EMS where u is the number of degrees of freedom for MS and v is the number of degrees of freedom for MS. For the one-way model, this equation reduces to P PF P F nk a, Na a, N a n where k. Using arbitrary values for some of the parameters in this equation will give an experimenter some idea of the probability of obtaining a negative estimate of ˆ. -8 Analyze the data in Problem -9, assuming that the operators are fixed, using both the unrestricted and restricted forms of the mixed models. Compare the results obtained from the two models. The restricted model is as follows: Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 3 4 5 6 7 8 9 Operator fixed Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.7. 7.33. Operator.47.47.69.47 Part*Operator 9 5.47.6.4.97 Error 4 6..5 Total 59 64.85 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Part.5836 4 (4) + 6() Operator 3 (4) + 3(3) + 3Q[] 3 Part*Operator -.994 4 (4) + 3(3) 4 Error.5 (4) The second approach is the unrestricted mixed model. Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 3 4 5 6 7 8 9 Operator fixed -36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.7. 8.8. Operator.47.47.69.47 Part*Operator 9 5.47.6.4.97 Error 4 6..5 Total 59 64.85 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) Part.7333 3 (4) + 3(3) + 6() Operator 3 (4) + 3(3) + Q[] 3 Part*Operator -.994 4 (4) + 3(3) 4 Error.5 (4) Source Sum of Squares DF Mean Square A.46667 a-=.46667 E(MS) F-test F i i n bn a a MS A F.69 MS AB B 99.6667 b-=9.85 n an AB 5.46667 (a-)(b-)=9.685 n Error 6. 4.5 Total 64.85 nabc-=59 MS B F 8.8 MS AB MS AB F.4 MS E In the unrestricted model, the F-test for B is different. The F-test for B in the unrestricted model should generally be more conservative, since MS AB will generally be larger than MS E. However, this is not the case with this particular experiment. -9 Consider the two-factor mixed model. Show that the standard error of the fixed factor mean (e.g. A) is / bn MS AB. The standard error is often used in Duncan s Multiple Range test. Duncan s Multiple Range Test requires the variance of the difference in two means, say V y i y.. where rows are fixed and columns are random. Now, assuming all model parameters to be independent, we have the following: b b yi.. ym.. i m ij mj ijk j b b j m.. bn b n j k bn b n j k mjk and V yi.. ym.. n b bn b b b bn bn bn bn -37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Since MS AB estimates n, we would use MS AB bn as the standard error to test the difference. However, the table of ranges for Duncan s Multiple Range test already include the constant. -3 Consider the variance components in the random model from Problem -9. (a) Find an exact 95 percent confidence interval on. femse fems, f E, f E. 5 4. 5 4 59. 34 4. 43.. 456 (b) Find approximate 95 percent confidence intervals on the other variance components using the Satterthwaite method. and are negative, and the Satterthwaithe method does not apply. The confidence interval on is r MS MS B MS B B MS an AB MS MS b a b AB AB r ˆ E. 85. 6859. 7333 9 3. 85. 6859. 85. 6859 r O, r, r 9. 86. 7333 8. 86. 7333 8 7. 5575. 895. 7957 6. 34759 8. 86-3 Use the experiment described in Problem 5-6 and assume that both factor are random. Find an exact 95 percent confidence interval on. Construct approximate 95 percent confidence interval on the other variance components using the Satterthwaite method. MS E 37967. femse fems, f E, f E 3. 7967 3. 7967 3. 34 E 4. 4-38

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Satterthwaite Method: r MS MS MS AB AB AB a b MS E n MS E MS df 9494.. 349 r E.. 7 44444 3 7967 8639. 7. 44444 3. 7967 7. 44444 3. 7967 3 E r, r, r. 94. 8639. 94. 8639 7. 9598. 9998. 564 49577.. 94, this variance component does not have a confidence interval using Satterthwaite s Method. r MS MS MS A A A MS bn MS AB MS a a b AB AB r ˆ 8. 6667 7. 44444 9. 98 4 8. 6667 7. 44444 8. 6667 7. 44444 r, r, r 3 (. 648)( 9. 98) (. 648)( 9. 98) 65395.. 35. 8348 465. 45637. 648-3 Consider the three-factor experiment in Problem 5-7 and assume that operators were selected at random. Find an approximate 95 percent confidence interval on the operator variance component. MSC MSE abn MSC MSE r MSC MS c df E E r ˆ 3. 66667 3. 77778 7. 776 3 3 3. 66667 3. 7778 3. 66667 3. 7778 36 r, r, r. 985. 9857. 776. 9857. 776 9. 5467. 454 46948. 498. 6653-39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -33 Rework Problem -3 using the modified large-sample approach described in Section -7.. Compare the two sets of confidence intervals obtained and discuss. G G F. 5, 9,. H F. 95, 9i,. 95, 9 9 ij V V V L L L MS MS an B AB O 88. 4689.. 77. 37 ˆ O F, f, f G F, f, f H 3. 8. 4689 3. 8 G i j c MS F, fi, f j i j. 85. 6859. 7333 3. 8 3. 77. 36366. 4689. 85. 77. 685. 36366. 85. 685. 8375 B H 6 c MS AB G c c MS B MS 6 AB L 7333. 8375. 875. V L 6 6-34 Rework Problem -3 using the modified large-sample method described in Section -7.. Compare this confidence interval with he one obtained previously and discuss. G F H F G V V V ij L L L. 95, 36, 3,.,. 95, 36. 5 6 36 MSC MS abn. 6538 E ˆ.. 54493. 6478 3. 66667 3. 77778 7. 776 3 3 F, f, f G F, f, f H. 88. 6538. 88 G i c j MS B F, fi, f j i j. 88. 54493. 7454. 6538 3. 66667. 54493 3. 7778. 7454 3. 666673. 7778. 95 H 8 c MS AB G c c MS B MS L 7. 776. 95. 4999 V L AB 8 8 8-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 3 Nested and Split-Plot Designs Solutions In this chapter we have not shown residual plots and other diagnostics to conserve space. A complete analysis would, of course, include these model adequacy checking procedures. 3- A rocket propellant manufacturer is studying the burning rate of propellant from three production processes. Four batches of propellant are randomly selected from the output of each process and three determinations of burning rate are made on each batch. The results follow. Analyze the data and draw conclusions. Process Process Process 3 Batch 3 4 3 4 3 4 5 9 5 5 9 3 8 35 4 35 38 5 3 8 7 6 7 4 7 5 54 9 6 4 3 4 7 5 4 5 33 Minitab Output ANOVA: Burn Rate versus Process, Batch Factor Type Levels Values Process fixed 3 3 Batch(Process) random 4 3 4 Analysis of Variance for Burn Rat Source DF SS MS F P Process 676.6 338.3.46.8 Batch(Process) 9 77.58 3.84.. Error 4 454. 8.9 Total 35 37.64 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Process (3) + 3() + Q[] Batch(Process) 7.64 3 (3) + 3() 3 Error 8.9 (3) There is no significant effect on mean burning rate among the different processes; however, different batches from the same process have significantly different burning rates. 3- The surface finish of metal parts made on four machines is being studied. An experiment is conducted in which each machine is run by three different operators and two specimens from each operator are collected and tested. Because of the location of the machines, different operators are used on each machine, and the operators are chosen at random. The data are shown in the following table. Analyze the data and draw conclusions. Machine Machine Machine 3 Machine 4 Operator 3 3 3 3 79 94 46 9 85 76 88 53 46 36 4 6 6 74 57 99 79 68 75 56 57 53 56 47 Minitab Output ANOVA: Finish versus Machine, Operator 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Type Levels Values Machine fixed 4 3 4 Operator(Machine) random 3 3 Analysis of Variance for Finish Source DF SS MS F P Machine 3 367.67 5.89 3.4.73 Operator(Machine) 8 87.67 35. 4.7.3 Error 4. 84.5 Total 3 7449.33 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Machine (3) + () + 6Q[] Operator(Machine) 33.85 3 (3) + () 3 Error 84.5 (3) There is a slight effect on surface finish due to the different processes; however, the different operators running the same machine have significantly different surface finish. 3-3 A manufacturing engineer is studying the dimensional variability of a particular component that is produced on three machines. Each machine has two spindles, and four components are randomly selected from each spindle. These results follow. Analyze the data, assuming that machines and spindles are fixed factors. Machine Machine Machine 3 Spindle 8 4 4 6 9 9 5 5 3 5 8 4 3 4 Minitab Output ANOVA: Variability versus Machine, Spindle Factor Type Levels Values Machine fixed 3 3 Spindle(Machine) fixed Analysis of Variance for Variabil Source DF SS MS F P Machine 55.75 7.875 8.93. Spindle(Machine) 3 43.75 4.583 9.9. Error 8 6.5.47 Total 3 6. There is a significant effects on dimensional variability due to the machine and spindle factors. 3-4 To simplify production scheduling, an industrial engineer is studying the possibility of assigning one time standard to a particular class of jobs, believing that differences between jobs is negligible. To see if this simplification is possible, six jobs are randomly selected. Each job is given to a different group of three operators. Each operator completes the job twice at different times during the week, and the following results were obtained. What are your conclusions about the use of a common time standard for all jobs in this class? What value would you use for the standard? Job Operator Operator Operator 3 58.3 59.4 59. 59.6 58.9 57.8 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output ANOVA: Time versus Job, Operator 54.6 54.9 57.7 56.8 54.8 56.3 3 6.5 6.6 6. 58.9 6.5 59.5 4 6. 58.7 57.5 58.9 6. 58.5 5 56.3 58. 58.3 56.9 57.7 56.9 6 63.7 6. 6.3 6.3 6.6 6.8 Factor Type Levels Values Job random 6 3 4 5 6 Operator(Job) random 3 3 Analysis of Variance for Time Source DF SS MS F P Job 5 48. 9.6 7.89. Operator(Job).743.6.69.738 Error 8 7.575.53 Total 35 88.43 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Job 4.76 (3) + () + 6() Operator(Job) -.35 3 (3) + () 3 Error.539 (3) The jobs differ significantly; the use of a common time standard would likely not be a good idea. 3-5 Consider the three-stage nested design shown in Figure 3-5 to investigate alloy hardness. Using the data that follow, analyze the design, assuming that alloy chemistry and heats are fixed factors and ingots are random. Alloy Chemistry Heats 3 3 Ingots 4 7 95 69 65 78 3 83 75 6 35 63 3 67 47 54 45 39 6 64 77 4 Minitab Output ANOVA: Hardness versus Alloy, Heat, Ingot Factor Type Levels Values Alloy fixed Heat(Alloy) fixed 3 3 Ingot(Alloy Heat) random Analysis of Variance for Hardness Source DF SS MS F P Alloy 35.4 35.4.85.39 Heat(Alloy) 4 6453.8 63.5 4.35.55 Ingot(Alloy Heat) 6 6.3 37..8.3 Error 4.5 78.5 Total 3 37. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Alloy 3 (4) + (3) + Q[] Heat(Alloy) 3 (4) + (3) + 4Q[] 3 Ingot(Alloy Heat) 96.9 4 (4) + (3) 4 Error 78.46 (4) 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Alloy hardness differs significantly due to the different heats within each alloy. 3-6 Reanalyze the experiment in Problem 3-5 using the unrestricted form of the mixed model. Comment on any differences you observe between the restricted and unrestricted model results. You may use a computer software package. Minitab Output ANOVA: Hardness versus Alloy, Heat, Ingot Factor Type Levels Values Alloy fixed Heat(Alloy) fixed 3 3 Ingot(Alloy Heat) random Analysis of Variance for Hardness Source DF SS MS F P Alloy 35.4 35.4.85.39 Heat(Alloy) 4 6453.8 63.5 4.35.55 Ingot(Alloy Heat) 6 6.3 37..8.3 Error 4.5 78.5 Total 3 37. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) Alloy 3 (4) + (3) + Q[,] Heat(Alloy) 3 (4) + (3) + Q[] 3 Ingot(Alloy Heat) 96.9 4 (4) + (3) 4 Error 78.46 (4) 3-7 Derive the expected means squares for a balanced three-stage nested design, assuming that A is fixed and that B and C are random. Obtain formulas for estimating the variance components. F R R R a b c n Factor i j k l E(MS) i b c n n ji cn () c n n cn k ij ( ) n n ( ijk) l bcn a i MS E ˆ MSC B MS n E ˆ MS BA MSCB cn The expected mean squares can be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C Factor Type Levels Values A fixed - B(A) random - C(A B) random - Analysis of Variance for y Source DF SS MS F P 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A B(A).5 8.5.5 4.5.6.83.35.76 C(A B) 4 49..5.3.68 Error 8 46. 5.75 Total 5 3.75 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) A (4) + (3) + 4() + 8Q[] B(A) -. 3 (4) + (3) + 4() 3 C(A B) 3.5 4 (4) + (3) 4 Error 5.75 (4) 3-8 Repeat Problem 3-7 assuming the unrestricted form of the mixed model. You may use a computer software package. Comment on any differences you observe between the restricted and unrestricted model analysis and conclusions. Minitab Output ANOVA: y versus A, B, C Factor Type Levels Values A fixed - B(A) random - C(A B) random - Analysis of Variance for y Source DF SS MS F P A.5.5.6.83 B(A) 8.5 4.5.35.76 C(A B) 4 49..5.3.68 Error 8 46. 5.75 Total 5 3.75 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) A (4) + (3) + 4() + Q[] B(A) -. 3 (4) + (3) + 4() 3 C(A B) 3.5 4 (4) + (3) 4 Error 5.75 (4) In this case there is no difference in results between the restricted and unrestricted models. 3-9 Derive the expected means squares for a balanced three-stage nested design if all three factors are random. Obtain formulas for estimating the variance components. Assume the restricted form of the mixed model. R R R R a b c n Factor i j k l E(MS) i b c n n cn bcn ji () c n n cn k ij ( ) n n ( ijk) l MS E MS C( B) n MS E MS MS B( A) C( B) cn MS A MS bcn B( A) The expected mean squares can be generated in Minitab as follows: 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output ANOVA: y versus A, B, C Factor Type Levels Values A random - B(A) random - C(A B) random - Analysis of Variance for y Source DF SS MS F P A.5.5.6.83 B(A) 8.5 4.5.35.76 C(A B) 4 49..5.3.68 Error 8 46. 5.75 Total 5 3.75 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) A -.5 (4) + (3) + 4() + 8() B(A) -. 3 (4) + (3) + 4() 3 C(A B) 3.5 4 (4) + (3) 4 Error 5.75 (4) 3- Verify the expected mean squares given in Table 3-. F F R a b n Factor i j l E(MS) i b n bn a j i n n ab ijkl i j i R R R a b n Factor i j l E(MS) i b n n bn j i n n ijkl F R R a b n Factor i j l E(MS) i b n n j i n n ijkl bn a i 3- Unbalanced designs. Consider an unbalanced two-stage nested design with b j levels of B under the ith level of A and n ij replicates in the ijth cell. 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (a) Write down the least squares normal equations for this situation. Solve the normal equations. The least squares normal equations are: n.. ˆ a i n ˆ i. i b i a b i i j nij ˆ j i y i ni. ˆ ni. ˆ i n ˆ ij j i yi.., for i,,..., a j j i nij ˆ nij ˆ ˆ i nij j i yij., for i,,..., a and j,,..., b i There are +a+b equations in +a+b unknowns. However, there are a+linear dependencies in these equations, and consequently, a+ side conditions are needed to solve them. Any convenient set of a+ linearly independent equations can be used. The easiest set is, i, for i=,,,a. Using these conditions we get, i,... ji () y ij. as the solution to the normal equations. See Searle (97) for a full discussion. (b) Construct the analysis of variance table for the unbalanced two-stage nested design. The analysis of variance table is Source SS DF A a yi.. y... n n a- B Error Total a a i i. b i i j b n i ij i j k y n y ij. ij ijk a.. y n i.. i i. a b i i j a b n y... i ij yijk n i j k.. (c) Analyze the following data, using the results in part (b). Factor A Factor B 3 6-3 5 4 7 4 8 9 3-3 6 y n ij. ij b.-a n..-b n..- Note that a=, b =, b =3, b.=b +b =5, n =3, n =, n =4, n =3 and n 3 =3 Source SS DF MS 3-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A.3.3 B 53.78 3 5.6 Error 35.4 3.54 Total 89.33 4 The analysis can also be performed in Minitab as follows. The adjusted sum of squares is utilized by Minitab s general linear model routine. Minitab Output General Linear Model: y versus A, B Factor Type Levels Values A fixed B(A) fixed 5 3 Analysis of Variance for y, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P A.33.898.898.5.65 B(A) 3 53.783 53.783 5.6 4.47. Error 35.47 35.47 3.54 Total 4 89.333 3- Variance components in the unbalanced two-stage nested design. Consider the model yijk i j i kij i,,...,a j,,...,b k,,...,n ij where A and B are random factors. Show that E E MS E MS A B A c MS E N N a c where a bi nij N n i j i. c b a a bi a nij n i j i. i c a c i i. c j a n b i n ij N 3-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY See Variance Component Estimation in the -way Nested Classification, by S.R. Searle, Annals of Mathematical Statistics, Vol. 3, pp. 6-66, 96. A good discussion of variance component estimation from unbalanced data is in Searle (97a). 3-3 A process engineer is testing the yield of a product manufactured on three machines. Each machine can be operated at two power settings. Furthermore, a machine has three stations on which the product is formed. An experiment is conducted in which each machine is tested at both power settings, and three observations on yield are taken from each station. The runs are made in random order, and the results follow. Analyze this experiment, assuming all three factors are fixed. Machine Machine Machine 3 Station 3 3 3 Power Setting 34. 33.7 36. 3. 33. 3.8 3.9 33.8 33.6 3.3 34.9 36.8 33.5 34.7 35. 33. 33.4 3.8 3.6 35. 37. 34. 33.9 34.3 33. 3.8 3.7 Power Setting 4.3 8. 5.7 4. 4. 6. 4. 3. 4.7 6.3 9.3 6. 5. 5. 7. 6. 7.4. 7. 8.6 4.9 6.3 7.9 3.9 5.3 8. 4.8 y The linear model is ijkl i j kj ijk l ij ik( j ) Minitab Output ANOVA: Yield versus Machine, Power, Station Factor Type Levels Values Machine fixed 3 3 Power fixed Station(Machine) fixed 3 3 Analysis of Variance for Yield Source DF SS MS F P Machine.43.57 6.46.4 Power 853.63 853.63 5.8. Station(Machine) 6 3.583 5.43 3.3. Machine*Power.66.38.9.89 Power*Station(Machine) 6 8.94 4.84.95.9 Error 36 58.893.636 Total 53 995.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Machine 6 (6) + 8Q[] Power 6 (6) + 7Q[] 3 Station(Machine) 6 (6) + 6Q[3] 4 Machine*Power 6 (6) + 9Q[4] 5 Power*Station(Machine) 6 (6) + 3Q[5] 6 Error.636 (6) 3-4 Suppose that in Problem 3-3 a large number of power settings could have been used and that the two selected for the experiment were chosen randomly. Obtain the expected mean squares for this situation and modify the previous analysis appropriately. R F F R 3 3 3 Factor i j k l E(MS) i 3 3 3 7 j 3 3 9 9 j 3-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ( ) ij 3 3 k j 9 ( ) 3 ( ) ik ( j ) 3 ( ijk) l 3 k( j ) The analysis of variance and the expected mean squares can be completed in Minitab as follows: Minitab Output ANOVA: Yield versus Machine, Power, Station Factor Type Levels Values Machine fixed 3 3 Power random Station(Machine) fixed 3 3 Analysis of Variance for Yield 3 Source DF SS MS F P Machine.43.57 34.33.8 Power 853.63 853.63 5.8. Station(Machine) 6 3.583 5.43.3.445 Machine*Power.66.38.9.89 Power*Station(Machine) 6 8.94 4.84.95.9 Error 36 58.893.636 Total 53 995.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Machine 4 (6) + 9(4) + 8Q[] Power 3.5554 6 (6) + 7() 3 Station(Machine) 5 (6) + 3(5) + 6Q[3] 4 Machine*Power -.476 6 (6) + 9(4) 5 Power*Station(Machine).65 6 (6) + 3(5) 6 Error.6359 (6) 3-5 Reanalyze the experiment in Problem 3-4 assuming the unrestricted form of the mixed model. You may use a computer software program to do this. Comment on any differences between the restricted and unrestricted model analysis and conclusions. ANOVA: Yield versus Machine, Power, Station Factor Type Levels Values Machine fixed 3 3 Power random Station(Machine) fixed 3 3 Analysis of Variance for Yield Source DF SS MS F P Machine.43.57 34.33.8 Power 853.63 853.63 77.86. Station(Machine) 6 3.583 5.43.3.445 Machine*Power.66.38.6.939 Power*Station(Machine) 6 8.94 4.84.95.9 Error 36 58.893.636 Total 53 995.88 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) Machine 4 (6) + 3(5) + 9(4) + Q[,3] Power 3.646 4 (6) + 3(5) + 9(4) + 7() 3 Station(Machine) 5 (6) + 3(5) + Q[3] 4 Machine*Power -.57 5 (6) + 3(5) + 9(4) 5 Power*Station(Machine).65 6 (6) + 3(5) 6 Error.6359 (6) 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY There are differences between several of the expected mean squares. However, the conclusions that could be drawn do not differ in any meaningful way from the restricted model analysis. 3-6 A structural engineer is studying the strength of aluminum alloy purchased from three vendors. Each vendor submits the alloy in standard-sized bars of.,.5, or. inches. The processing of different sizes of bar stock from a common ingot involves different forging techniques, and so this factor may be important. Furthermore, the bar stock if forged from ingots made in different heats. Each vendor submits two tests specimens of each size bar stock from the three heats. The resulting strength data follow. Analyze the data, assuming that vendors and bar size are fixed and heats are random. Vendor Vendor Vendor 3 Heat 3 3 3 Bar Size: inch.3.346.35.3.346.35.47.75.34.59.4.6.63.39.3.96.68.35 / inch.36.39.5.74.384.346.73.6.39.3.36.39.68.375.357.64.65.364 inch.87.346.73.47.36.336.3.8.39.9.38.5.5.38.34.6.7.33 y kj ( ik j ijk l ijkl i j ij ) Minitab Output ANOVA: Strength versus Vendor, Bar Size, Heat Factor Type Levels Values Vendor fixed 3 3 Heat(Vendor) random 3 3 Bar Size fixed 3..5. Analysis of Variance for Strength Source DF SS MS F P Vendor.88486.4443.6.776 Heat(Vendor) 6.93.676 4.3. Bar Size.563.63.37.9 Vendor*Bar Size 4.3754.5939.65.64 Bar Size*Heat(Vendor).33.99.7.37 Error 7.935.44 Total 53.35934 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Vendor (6) + 6() + 8Q[] Heat(Vendor).7 6 (6) + 6() 3 Bar Size 5 (6) + (5) + 8Q[3] 4 Vendor*Bar Size 5 (6) + (5) + 6Q[4] 5 Bar Size*Heat(Vendor).6 6 (6) + (5) 6 Error.4 (6) 3-7 Reanalyze the experiment in Problem 3-6 assuming the unrestricted form of the mixed model. You may use a computer software program to do this. Comment on any differences between the restricted and unrestricted model analysis and conclusions. Minitab Output ANOVA: Strength versus Vendor, Bar Size, Heat Factor Type Levels Values Vendor fixed 3 3 Heat(Vendor) random 3 3 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Bar Size fixed 3..5. Analysis of Variance for Strength Source DF SS MS F P Vendor.88486.4443.6.776 Heat(Vendor) 6.93.676 8.7. Bar Size.563.63.37.9 Vendor*Bar Size 4.3754.5939.65.64 Bar Size*Heat(Vendor).33.99.7.37 Error 7.935.44 Total 53.35934 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) Vendor (6) + (5) + 6() + Q[,4] Heat(Vendor).63 5 (6) + (5) + 6() 3 Bar Size 5 (6) + (5) + Q[3,4] 4 Vendor*Bar Size 5 (6) + (5) + Q[4] 5 Bar Size*Heat(Vendor).6 6 (6) + (5) 6 Error.4 (6) There are some differences in the expected mean squares. However, the conclusions do not differ from those of the restricted model analysis. 3-8 Suppose that in Problem 3-6 the bar stock may be purchased in many sizes and that the three sizes are actually used in experiment were selected randomly. Obtain the expected mean squares for this situation and modify the previous analysis appropriately. Use the restricted form of the mixed model. Minitab Output ANOVA: Strength versus Vendor, Bar Size, Heat Factor Type Levels Values Vendor fixed 3 3 Heat(Vendor) random 3 3 Bar Size random 3..5. Analysis of Variance for Strength Source DF SS MS F P Vendor.88486.4443.7.77 x Heat(Vendor) 6.93.676 8.7. Bar Size.563.63.37.9 Vendor*Bar Size 4.3754.5939.65.64 Bar Size*Heat(Vendor).33.99.7.37 Error 7.935.44 Total 53.35934 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Vendor * (6) + (5) + 6(4) + 6() + 8Q[] Heat(Vendor).63 5 (6) + (5) + 6() 3 Bar Size. 5 (6) + (5) + 8(3) 4 Vendor*Bar Size -.5 5 (6) + (5) + 6(4) 5 Bar Size*Heat(Vendor).6 6 (6) + (5) 6 Error.4 (6) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS Vendor 5.75.6376 () + (4) - (5) Notice that a Satterthwaite type test is used for vendor. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 3-9 Steel in normalized by heating above the critical temperature, soaking, and then air cooling. This process increases the strength of the steel, refines the grain, and homogenizes the structure. An experiment is performed to determine the effect of temperature and heat treatment time on the strength of normalized steel. Two temperatures and three times are selected. The experiment is performed by heating the oven to a randomly selected temperature and inserting three specimens. After minutes one specimen is removed, after minutes the second specimen is removed, and after 3 minutes the final specimen is removed. Then the temperature is changed to the other level and the process is repeated. Four shifts are required to collect the data, which are shown below. Analyze the data and draw conclusions, assume both factors are fixed. Temperature (F) Shift Time(minutes) 5 6 63 89 54 9 3 6 6 5 8 5 7 3 59 69 3 48 73 74 8 3 7 69 4 54 88 48 9 3 59 64 This is a split-plot design. Shifts correspond to blocks, temperature is the whole plot treatment, and time is the subtreatments (in the subplot or split-plot part of the design). The expected mean squares and analysis of variance are shown below. R F F R 4 3 Factor i j k l E(MS) i (blocks) 3 6 j (temp) 4 3 3 ij 3 3 / j k (time) 4 8 / k ik jk 4 / 3 ijk ijkl (not estimable) jk The following Minitab Output has been modified to display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, in general, correct. Notice that the Error term in the analysis of variance is actually the three factor interaction. Minitab Output ANOVA: Strength versus Shift, Temperature, Time Factor Type Levels Values Shift random 4 3 4 Temperat fixed 5 6 Time fixed 3 3 Analysis of Variance for Strength 3-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Standard Split Plot Source DF SS MS F P F P Shift 3 45.46 48.49.9.39 Temperat 34.38 34.38 9.. 9.. Shift*Temperat 3 4.46 8.5.97. Time 59.5 79.63..4..4 Shift*Time 6 478.4 79.74.96.7 Temperat*Time 795.5 397.63 9.76.3 9.76.3 Error 6 44.4 4.74 Total 3 443.63 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Shift.9 7 (7) + 6() Temperat 3 (7) + 3(3) + Q[] 3 Shift*Temperat 3.39 7 (7) + 3(3) 4 Time 5 (7) + (5) + 8Q[4] 5 Shift*Time 9.5 7 (7) + (5) 6 Temperat*Time 7 (7) + 4Q[6] 7 Error 4.736 (7) 3- An experiment is designed to study pigment dispersion in paint. Four different mixes of a particular pigment are studied. The procedure consists of preparing a particular mix and then applying that mix to a panel by three application methods (brushing, spraying, and rolling). The response measured is the percentage reflectance of the pigment. Three days are required to run the experiment, and the data obtained follow. Analyze the data and draw conclusions, assuming that mixes and application methods are fixed. Mix Day App Method 3 4 64.5 66.3 74. 66.5 68.3 69.5 73.8 7. 3 7.3 73. 78. 7.3 65. 65. 73.8 64.8 69. 7.3 74.5 68.3 3 7. 7.8 79. 7.5 3 66. 66.5 7.3 67.7 69. 69. 75.4 68.6 3 7.8 74. 8. 7.4 This is a split plot design. Days correspond to blocks, mix is the whole plot treatment, and method is the subtreatment (in the subplot or split plot part of the design). The expected mean squares are: R F F R 3 4 3 Factor i j k l E(MS) i (blocks) 4 3 j (temp) 3 3 9 3 ij 3 3 / j 3 k (time) 3 4 4 / k ik 4 4 jk 3 / 6 ijk ijkl (not estimable) 3 jk The following Minitab Output has been modified to display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, 3-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY in general, correct. Notice that the Error term in the analysis of variance is actually the three factor interaction. Minitab Output ANOVA: Reflectance versus Day, Mix, Method Factor Type Levels Values Day random 3 3 Mix fixed 4 3 4 Method fixed 3 3 Analysis of Variance for Reflecta Standard Split Plot Source DF SS MS F P F P Day.4..39.85 Mix 3 37.479.493 35.77. 35.75. Day*Mix 6 4.59.755.3.45 Method.95.47 6.4. 6.6. Day*Method 4.963.49.67.65 Mix*Method 6.36.673.8.5.8.5 Error 8.786.73 Total 35 556.93 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Day.46 7 (7) + () Mix 3 (7) + 3(3) + 9Q[] 3 Day*Mix.759 7 (7) + 3(3) 4 Method 5 (7) + 4(5) + Q[4] 5 Day*Method -.63 7 (7) + 4(5) 6 Mix*Method 7 (7) + 3Q[6] 7 Error.733 (7) 3- Repeat Problem 3-, assuming that the mixes are random and the application methods are fixed. The expected mean squares are: R R F R 3 4 3 Factor i j k l E(MS) i (blocks) 4 3 j (temp) 3 3 ij 3 3 3 9 3 k (time) 3 4 ik 4 jk 3 ijk 4 / k 4 3 ijkl (not estimable) The F-tests are the same as those in Problem 3-. The following Minitab Output has been edited to display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, in general, correct. Again, the Error term in the analysis of variance is actually the three factor interaction. Minitab Output ANOVA: Reflectance versus Day, Mix, Method 3-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Factor Type Levels Values Day random 3 3 Mix random 4 3 4 Method fixed 3 3 Analysis of Variance for Reflecta Standard Split Plot Source DF SS MS F P F P Day.4..35.38 Mix 3 37.479.493 35.77. 35.75. Day*Mix 6 4.59.755.3.45 Method.95.47 77.58. x 6.6. Day*Method 4.963.49.67.65 Mix*Method 6.36.673.8.5.8.5 Error 8.786.73 Total 35 556.93 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Day. 3 (7) + 3(3) + () Mix.34 3 (7) + 3(3) + 9() 3 Day*Mix.76 7 (7) + 3(3) 4 Method * (7) + 3(6) + 4(5) + Q[4] 5 Day*Method -.63 7 (7) + 4(5) 6 Mix*Method.335 7 (7) + 3(6) 7 Error.73 (7) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 4 Method 3.59.43 (5) + (6) - (7) 3- Consider the split-split-plot design described in example 3-3. Suppose that this experiment is conducted as described and that the data shown below are obtained. Analyze and draw conclusions. Technician 3 Blocks Dose Strengths 3 3 3 Wall Thickness 95 7 8 96 7 8 95 7 4 8 5 99 84 8 6 3 85 7 95 83 5 5 84 3 4 8 85 6 97 85 9 7 87 5 95 78 7 4 9 69 6 84 9 79 76 4 3 3 86 6 99 8 8 8 9 4 9 84 86 9 8 86 3 3 96 7 7 94 66 9 73 98 5 8 6 84 97 75 3 6 88 4 87 9 8 4 4 3 9 7 9 7 9 4 9 68 9 98 68 6 98 7 84 8 3 78 5 3 85 5 85 5 8 4 4 88 8 8 85 6 95 Using the computer output, the F-ratios were calculated by hand using the expected mean squares found in Table 3-8. The following Minitab Output has been edited to display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, in general, correct. Notice that the Error term in the analysis of variance is actually the four factor interaction. Minitab Output 3-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ANOVA: Time versus Day, Tech, Dose, Thick Factor Type Levels Values Day random 4 3 4 Tech fixed 3 3 Dose fixed 3 3 Thick fixed 4 3 4 Analysis of Variance for Time Standard Split Plot Source DF SS MS F P F P Day 3 48.4 6.4 3.38.9 Tech 48.35 4.7 4.6.6 4.6.6 Day*Tech 6 6.5 6.86 5.6. Dose 57.6 85.3 55.44. 55.3. Day*Dose 6. 8.69 3.9.4 Tech*Dose 4 5.94 3.49 3.3.48 3.3.48 Day*Tech*Dose 3.89 9.49.99.56 Thick 3 386.9 68.97 36.47. 36.48. Day*Thick 9 33. 34.79 7.8. Tech*Thick 6 6.49.8.6.84.6.84 Day*Tech*Thick 8 67.57 9.3.95.44 Dose*Thick 6 4.8 67.5 7.3. 7.5. Day*Dose*Thick 8 7.44 3.9.8.668 Tech*Dose*Thick 5.89 7.6 3.59. 3.59. Error 36 7.6 4.78 Total 43 6644.66 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Day.355 5 (5) + 36() Tech 3 (5) + (3) + 48Q[] 3 Day*Tech.84 5 (5) + (3) 4 Dose 5 (5) + (5) + 48Q[4] 5 Day*Dose.588 5 (5) + (5) 6 Tech*Dose 7 (5) + 4(7) + 6Q[6] 7 Day*Tech*Dose.779 5 (5) + 4(7) 8 Thick 9 (5) + 9(9) + 36Q[8] 9 Day*Thick 3.3346 5 (5) + 9(9) Tech*Thick (5) + 3() + Q[] Day*Tech*Thick.5 5 (5) + 3() Dose*Thick 3 (5) + 3(3) + Q[] 3 Day*Dose*Thick -.886 5 (5) + 3(3) 4 Tech*Dose*Thick 5 (5) + 4Q[4] 5 Error 4.7793 (5) 3-3 Rework Problem 3-, assuming that the dosage strengths are chosen at random. Use the restricted form of the mixed model. The following Minitab Output has been edited to display the results of the split-plot analysis. Minitab will calculate the sums of squares correctly, but the expected mean squares and the statistical tests are not, in general, correct. Again, the Error term in the analysis of variance is actually the four factor interaction. Minitab Output ANOVA: Time versus Day, Tech, Dose, Thick Factor Type Levels Values Day random 4 3 4 Tech fixed 3 3 Dose random 3 3 Thick fixed 4 3 4 Analysis of Variance for Time Standard Split Plot Source DF SS MS F P F P Day 3 48.4 6.4.86.59 Tech 48.35 4.7.54.55 4.6.6 Day*Tech 6 6.5 6.86.83.59 Dose 57.6 85.3 55.44. 55.3. 3-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Day*Dose 6. 8.69 3.9.4 Tech*Dose 4 5.94 3.49 3.3.48 3.3.48 Day*Tech*Dose 3.89 9.49.99.56 Thick 3 386.9 68.97.96. x 36.48. Day*Thick 9 33. 34.79 8.89. Tech*Thick 6 6.49.8.97.475 x.6.84 Day*Tech*Thick 8 67.57 9.3.95.44 Dose*Thick 6 4.8 67.5 7.3. 7.5. Day*Dose*Thick 8 7.44 3.9.8.668 Tech*Dose*Thick 5.89 7.6 3.59. 3.59. Error 36 7.6 4.78 Total 43 6644.66 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) Day -.7 5 (5) + (5) + 36() Tech * (5) + 4(7) + 6(6) + (3) + 48Q[] 3 Day*Tech.447 7 (5) + 4(7) + (3) 4 Dose 3.88 5 (5) + (5) + 48(4) 5 Day*Dose.59 5 (5) + (5) 6 Tech*Dose.375 7 (5) + 4(7) + 6(6) 7 Day*Tech*Dose.78 5 (5) + 4(7) 8 Thick * (5) + 3(3) + () + 9(9) + 36Q[8] 9 Day*Thick 3.43 3 (5) + 3(3) + 9(9) Tech*Thick * (5) + 4(4) + 3() + Q[] Day*Tech*Thick.5 5 (5) + 3() Dose*Thick 5.6 3 (5) + 3(3) + () 3 Day*Dose*Thick -.89 5 (5) + 3(3) 4 Tech*Dose*Thick 3.95 5 (5) + 4(4) 5 Error 4.779 (5) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS Tech 6.35 48.85 (3) + (6) - (7) 8 Thick.84 97.9 (9) + () - (3) Tech*Thick 5.69.69 () + (4) - (5) The expected mean squares can also be shown as follows: R F R F R 4 3 3 4 Factor i j k h l E(MS) i 3 3 4 36 j 4 3 4 4 6 ( 48 / ) j ij 3 4 k 4 3 4 ik 3 4 jk 4 4 ijk 4 4 3 48 4 6 4 h 4 3 3 9 36 3 ih 3 3 jh ijh 3 kh 4 3 3 / h 3 9 4 3 3 6 4 / jh 3 3 3-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ikh 3 jkh 4 ijkh 3 4 ijkl There are no exact tests on technicians j, dosage strengths k, wall thickness h, or the technician x wall thickness interaction. The approximate F-tests are as follows: jh H : j = p q MS F MS B AB MS MS ABC BC 4. 74 9. 49. 9 6. 859 3. 486 MS MS 4. 74 9. 49 MS 6 Do not reject H : j = B B ABC ABC MS 4. 74 9. 49 MS MS 6. 859 3. 486 MS AB AB BC BC MS 4 6. 859 6 3. 486 4. 35 9. 48 H : k = MS F MS p q C CD MS MS ACD AD 85. 8 3. 94. 39 67. 46 34. 79 MS MS 85. 8 3. 94 MS C C 6 Reject H : k = H : h = p ACD ACD MS 8 85. 8 3. 94 8 MS MS 67. 46 34. 79 MS CD CD MS F MS D MS 9 CD MS AD AD MS ACD AD 67. 46 6 34. 79 9.. 736 68. 97 3. 94. 499 67. 46 34. 79 MS MS 68. 97 3. 94 MS D D 3 ACD ACD MS 8 68. 97 3 3. 94 8 3. 9 3-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY q MS MS 67. 46 34. 79 MS CD CD 6 MS 9 AD AD 67. 46 6 34. 79 9. 736 Reject H : h = jh H : = MS F MS BD BCD MS MS ABCD ABD. 8 4. 779. 977 7. 57 9. 39 jh F<, Do not reject H : = 3-4 Suppose that in Problem 3- four technicians had been used. Assuming that all the factors are fixed, how many blocks should be run to obtain an adequate number of degrees of freedom on the test for differences among technicians? The number of degrees of freedom for the test is (a-)(4-)=3(a-), where a is the number of blocks used. Number of Blocks (a) DF for test 3 3 6 4 9 5 At least three blocks should be run, but four would give a better test. 3-5 Consider the experiment described in Example 3-3. Demonstrate how the order in which the treatments combinations are run would be determined if this experiment were run as (a) a split-split-plot, (b) a split-plot, (c) a factorial design in a randomized block, and (d) a completely randomized factorial design. (a) Randomization for the split-split plot design is described in Example 3-3. (b) In the split-plot, within a block, the technicians would be the main treatment and within a blocktechnician plot, the combinations of dosage strength and wall thickness would be run in random order. The design would be a two-factor factorial in a split-plot. (c) To run the design in a randomized block, the 36 combinations of technician, dosage strength, and wall thickness would be ran in random order within each block. The design would be a three factor factorial in a randomized block. (d) The blocks would be considered as replicates, and all 44 observations would be 4 replicates of a three factor factorial. 3-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 4 Other Design and Analysis Topics Solutions 4- Reconsider the experiment in Problem 5-. Use the Box-Cox procedure to determine if a transformation on the response is appropriate (or useful) in the analysis of the data from this experiment. DESIGN-EXPERT Plot Crack Growth Lambda Current = Best =. Low C.I. = -.44 High C.I. =.56 Recommend transform: Log (Lambda = ) Ln(ResidualSS) Box-Cox Plot for Power Transforms 5.6 4.49 3.36.3. -3 - - 3 Lambda With the value of lambda near zero, and since the confidence interval does not include one, a natural log transformation would be appropriate. 4- In example 6-3 we selected a log transformation for the drill advance rate response. Use the Box- Cox procedure to demonstrate that this is an appropriate data transformation. 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Advance Rate Lambda Current = Best = -.3 Low C.I. = -.79 High C.I. =.3 Recommend transform: Log (Lambda = ) Ln(ResidualSS) Box-Cox Plot for Power Transforms 6.85 5.4 3.95.5.5-3 - - 3 Lambda Because the value of lambda is very close to zero, and the confidence interval does not include one, the natural log was the correct transformation chosen for this analysis. 4-3 Reconsider the smelting process experiment in Problem 8-3, where a 6-3 fractional factorial design was used to study the weight of packing material stuck to carbon anodes after baking. Each of the eight runs in the design was replicated three times and both the average weight and the range of the weights at each test combination were treated as response variables. Is there any indication that that a transformation is required for either response? DESIGN-EXPERT Plot Weight Box-Cox Plot for Power Transforms DESIGN-EXPERT Plot Range Box-Cox Plot for Power Transforms Lambda Current = Best =.33 Low C.I. = -.7 High C.I. = 4.9.5.6 Lambda Cu rre n t = Best =.58 Low C.I. = -.74 High C.I. =.9 3.6. Recommend transform: None (Lambda = ) Ln(ResidualSS) 9.47 Recommend transform: None (Lambda = ) Ln(ResidualSS).7 8.68.3 7.89 9.9-3 - - 3-3 - - 3 Lambda Lambda There is no indication that a transformation is required for either response. 4-4 In Problem 8-4 a replicated fractional factorial design was used to study substrate camber in semiconductor manufacturing. Both the mean and standard deviation of the camber measurements were used as response variables. Is there any indication that a transformation is required for either response? 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Camber Avg Box-Cox Plot for Power Transforms DESIGN-EXPERT Plot Cam ber StDev Box-Cox Plot for Power Transforms Lambda Current = Best = -.3 Low C.I. = -.79 High C.I. =.74..35 Lambda Cu rre n t = Best =.57 Low C.I. = -.3 High C.I. =.6 3..35 Recommend transform: Log (Lambda = ) Ln(ResidualSS).49 Recommend transform: None (Lambda = ) Ln(ResidualSS) 9.7 9.6 8.5 8.76 6.4-3 - - 3-3 - - 3 Lambda Lambda The Box-Cox plot for the Camber Average suggests a natural log transformation should be applied. This decision is based on the confidence interval for lambda not including one and the point estimate of lambda being very close to zero. With a lambda of approximately.5, a square root transformation could be considered for the Camber Standard Deviation; however, the confidence interval indicates that no transformation is needed. 4-5 Reconsider the photoresist experiment in Problem 8-5. Use the variance of the resist thickness at each test combination as the response variable. Is there any indication that a transformation is required? DESIGN-EXPERT Plot Thick StDev Lambda Current = Best = -.4 Low C.I. = -.77 High C.I. =.76 Recommend transform: Log (Lambda = ) Ln(ResidualSS) Box-Cox Plot for Power Transforms 9.93 9.8 8.6 7.97 7.3-3 - - 3 Lambda With the point estimate of lambda near zero, and the confidence interval for lambda not inclusive of one, a natural log transformation would be appropriate. 4-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4-6 In the grill defects experiment described in Problem 8-9 a variation of the square root transformation was employed in the analysis of the data. Use the Box-Cox method to determine if this is the appropriate transformation. DESIGN-EXPERT Plot c Lambda Current = Best = -.6 Low C.I. = -.69 High C.I. =.74 Recommend transform: Log (Lambda = ) k =.56 (used to make response values positive) Ln(ResidualSS) Box-Cox Plot for Power Transforms.75.35 7.95 5.55 3.5-3 - - 3 Lambda Because the confidence interval for the minimum lambda does not include one, the decision to use a transformation is correct. Because the lambda point estimate is close to zero, the natural log transformation would be appropriate. This is a stronger transformation than the square root. 4-7 In the central composite design of Problem -4, two responses were obtained, the mean and variance of an oxide thickness. Use the Box-Cox method to investigate the potential usefulness of transformation for both of these responses. Is the log transformation suggested in part (c) of that problem appropriate? DESIGN-EXPERT Plot Mean Thick Box-Cox Plot for Power Transforms DESIGN-EXPERT Plot Var Thick Box-Cox Plot for Power Transforms Lambda Current = Best = -. Low C.I. = -3.58 High C.I. = 3.8 8.73 8.57 Lambda Cu rre n t = Best = -.47 Low C.I. = -.85 High C.I. =.5.63.3 Recommend transform: None (Lambda = ) Ln(ResidualSS) 8.4 Recommend transform: None (Lambda = ) Ln(ResidualSS).97 8.5.65 8.9.3-3 - - 3-3 - - 3 Lambda Lambda The Box-Cox plot for the Mean Thickness model suggests that a natural log transformation could be applied; however, the confidence interval for lambda includes one. Therefore, a transformation would 4-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY have a minimal effect. The natural log transformation applied to the Variance of Thickness model appears to be acceptable; however, again the confidence interval for lambda includes one. 4-8 In the 3 3 factorial design of Problem -33 one of the responses is a standard deviation. Use the Box-Cox method to investigate the usefulness of transformations for this response. Would your answer change if we used the variance of the response? DESIGN-EXPERT Plot Std. Dev. Lambda Current = Best =.9 Low C.I. =. High C.I. =.6 Box-Cox Plot for Power Transforms.3 9. Recommend transform: Square Root (Lambda =.5) k =.58 (used to make response values positive) Ln(ResidualSS) 6.3 3.3 9.94-3 - - 3 Lambda Because the confidence interval for lambda does not include one, a transformation should be applied. The natural log transformation should not be considered due to zero not being included in the confidence interval. The square root transformation appears to be acceptable. However, notice that the value of zero is very close to the lower confidence limit, and the minimizing value of lambda is between and.5. It is likely that either the natural log or the square root transformation would work reasonably well. 4-9 Problem -34 suggests using the ln(s ) as the response (refer to part b). Does the Box-Cox method indicate that a transformation is appropriate? 4-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Variance Lambda Current = Best = -.7 Low C.I. = -.53 High C.I. = -.7 Recommend transform: Inverse (Lambda = -) Ln(ResidualSS) Box-Cox Plot for Power Transforms 7.56 4.3.7 7.8 3.85-3 - - 3 Lambda Because the confidence interval for lambda does not include one, a transformation should be applied. The confidence interval does not include zero; therefore, the natural log transformation is inappropriate. With the point estimate of lambda at.7, the reciprocal transformation is appropriate. 4- A soft drink distributor is studying the effectiveness of delivery methods. Three different types of hand trucks have been developed, and an experiment is performed in the company s methods engineering laboratory. The variable of interest is the delivery time in minutes (y); however, delivery time is also strongly related to the case volume delivered (x). Each hand truck is used four times and the data that follow are obtained. Analyze the data and draw the appropriate conclusions. Use =.5. Hand Truck Type 3 3 y x y x y x 7 4 5 6 4 38 44 4 35 3 6 33 35 46 4 53 5 4 4 6 5 8 From the analysis performed in Minitab, hand truck does not have a statistically significant effect on delivery time. Volume, as expected, does have a significant effect. Minitab Output General Linear Model: Time versus Truck Factor Type Levels Values Truck fixed 3 3 Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Volume 3.7 7.55 7.55 3.. Truck.65.65 5.8..375 Error 8 4.95 4.95 5.4 Total 85.67 Term Coef SE Coef T P Constant -4.747.638 -.8. Volume.736.7699 5.4. 4-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4- Compute the adjusted treatment means and the standard errors of the adjusted treatment means for the data in Problem 4-. S S S adj y y ˆ x x i. i. 45 39 398 73 4 4 3 5 398. 73 35 4 4 33 34 398. 73 3 4 4 adj. 34 39 y.. adj 5 y.. adj 86 adj.y. adj.y. adj.y3. y 3.. S adj.yi. MS 5. 4 4 5. 4 4 5. 4 4 E n i... x x i. E xx 34. 75 33. 7 884. 5 3. 5 33. 7 884. 5 33. 5 33. 7 884. 5 The solutions can also be obtained with Minitab as follows: Minitab Output Least Squares Means for Time Truck Mean SE Mean 34.39.5 35.5.54 3 3.86.45... 5. 54. 45 4- The sums of squares and products for a single-factor analysis of covariance follow. Complete the analysis and draw appropriate conclusions. Use =.5. Source of Degrees of Sums of Squares and Products Variation Freedom x xy x Treatment 3 5 65 Error 6 55 Total 5 75 Sums of Squares & Products Adjusted Source df x xy y y df MS F Treatment 3 5 65 - - Error 6 55 3 8.8 Total 5 75 559.67 4 4-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Adjusted Treat. 44.67 3 8.56.89 Treatments differ only at %. 4-3 Find the standard errors of the adjusted treatment means in Example 4-4. From Example 4-4 y. 4. 38, adj y. 44., adj y 3. 37. 78 S S S adj.y. adj.y. adj.y3.. 54 5. 54 5. 54 5 5. 4. 3 95. 6 6. 4. 3 95. 6. 4. 3 95. 6. 73. 7439. 787 4-4 Four different formulations of an industrial glue are being tested. The tensile strength of the glue when it is applied to join parts is also related to the application thickness. Five observations on strength (y) in pounds and thickness (x) in. inches are obtained for each formulation. The data are shown in the following table. Analyze these data and draw appropriate conclusions. Glue Formulation 3 3 4 4 y x y x y x y x 46.5 3 48.7 46.3 5 44.7 6 45.9 4 49. 47. 4 43. 5 49.8 5. 48.9 5. 46. 48.5 48. 48. 44.3 4 45. 4 5.3 48.6 From the analysis performed in Minitab, glue formulation does not have a statistically significant effect on strength. As expected, glue thickness does affect strength. Minitab Output General Linear Model: Strength versus Glue Factor Type Levels Values Glue fixed 4 3 4 Analysis of Variance for Strength, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Thick 68.85 59.566 59.566 4.6. Glue 3.77.77.59.4.74 Error 5.96.96.397 Total 9 9.585 Term Coef SE Coef T P Constant 6.89.944 3.9. Thick -.99.547-6.53. Unusual Observations for Strength 4-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Obs Strength Fit SE Fit Residual St Resid 3 49.8 47.599.558.7.7R R denotes an observation with a large standardized residual. Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term Thick (3) + Q[] Glue (3) + Q[] 3 Error (3) Error Terms for Tests, using Adjusted SS Source Error DF Error MS Synthesis of Error MS Thick 5..397 (3) Glue 5..397 (3) Variance Components, using Adjusted SS Source Estimated Value Error.397 4-5 Compute the adjusted treatment means and their standard errors using the data in Problem 4-4. adj yi. yi. ˆ xi. x.. adj y. 46. 5. 993.. 45 47. 8 adj y. 48. 3. 99. 8. 45 47. 64 adj y 3. 48. 6. 99.. 45 47. 9 adj 47. 8. 99. 8. 45 47 43 S S S S y 4.. adj.y. adj.y. adj.y3. adj.y4. S adj.yi. MS. 4 5. 4 5. 4 5. 4 5 E n x x i. E xx 3.. 45 58. 4. 8. 45 58. 4.. 45 58. 4. 8. 45 58. 4... 536. 5386. 536. 539 The adjusted treatment means can also be generated in Minitab as follows: Minitab Output Least Squares Means for Strength Glue Mean SE Mean 47.8.5355 47.64.538 3 47.9.53 4-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 47.43.534 4-6 An engineer is studying the effect of cutting speed on the rate of metal removal in a machining operation. However, the rate of metal removal is also related to the hardness of the test specimen. Five observations are taken at each cutting speed. The amount of metal removed (y) and the hardness of the specimen (x) are shown in the following table. Analyze the data using and analysis of covariance. Use =.5. Cutting Speed (rpm) 4 4 y x y x y x 68 65 8 75 9 4 94 4 8 3 98 5 65 73 4 77 5 74 5 9 4 88 36 85 33 8 3 As shown in the analysis performed in Minitab, there is no difference in the rate of removal between the three cutting speeds. As expected, the hardness does have an impact on rate of removal. Minitab Output General Linear Model: Removal versus Speed Factor Type Levels Values Speed fixed 3 4 Analysis of Variance for Removal, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Hardness 375.7 39.3 39.3 347.96. Speed.4.4..4.87 Error 95.5 95.5 8.7 Total 4 373.6 Term Coef SE Coef T P Constant -4.656 6.97-6.3. Hardness.9346.58 8.65. Speed.478.85.44.668.36.76.3.974 Unusual Observations for Removal Obs Removal Fit SE Fit Residual St Resid 8 65. 7.49.558-5.49 -.R R denotes an observation with a large standardized residual. Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term Hardness (3) + Q[] Speed (3) + Q[] 3 Error (3) Error Terms for Tests, using Adjusted SS Source Error DF Error MS Synthesis of Error MS Hardness. 8.7 (3) Speed. 8.7 (3) Variance Components, using Adjusted SS Source Estimated Value 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Error 8.677 Means for Covariates Covariate Mean StDev Hardness 37. 5.94 Least Squares Means for Removal Speed Mean SE Mean 86.88.35 86.44.38 4 85.89.38 4-7 Show that in a single factor analysis of covariance with a single covariate a (-) percent confidence interval on the i th adjusted treatment mean is y i. ˆ xi. x.. t,a n MS E n x x Using this formula, calculate a 95 percent confidence interval on the adjusted mean of machine in Example 4-4. The (-) percent interval on the i th adjusted treatment mean would be since y ˆ x x ˆ xi. x.. t,an S adjy i. yi. i. i... is an estimator of the i th adjusted treatment mean. The standard error of the adjusted treatment mean is found as follows: adj.y V y ˆ x x i. E xx.. V y x x V ˆ V i. i. i... i. i... Since the y i. and are independent. From regression analysis, we havev ˆ E xx. Therefore, V adj.y i. n x x x x i. E.. xx n i. E xx.. Replacing by its estimator MS E, yields y Vˆ adj.y S adj.y i. i. MS E MS x n E n i. x E xx.. or x x Substitution of this result into i. i..., a( n) adjy i. i. E xx.. ˆ x x t S will produce the desired confidence interval. A 95% confidence interval on the mean of machine would be found as follows: 4-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY y ˆ i. xi. x 4. 38 adj.yi. 73 38 t. 5. 73 38.. 73 38. 59 adj.yi... 4. 4. S. 4., Therefore, 38. 79 4. 96, where denotes the true adjusted mean of treatment one. 4-8 Show that in a single-factor analysis of covariance with a single covariate, the standard error of the difference between any two adjusted treatment means is adj.y The variance of this statistic is S Adjyi. Adjy adj.y MS j. E n y ˆ x x x x i. E xx.. ˆ x y x i. j. i. i... j. j.. y ˆ i. adj. y j. yi. y j. xi. x j. adj ˆ y y x x V y V y x x V ˆ i. j. i. j. i. j. i. j x x x x V. n n i. E j. xx n Replacing by its estimator MS E,, and taking the square root yields the standard error i. E xx j... S Adjyi. Adjy j. MS E n x x i. E xx.. 4-9 Discuss how the operating characteristic curves for the analysis of variance can be used in the analysis of covariance. To use the operating characteristic curves, fixed effects case, we would use as the parameter, a n The test has a- degrees of freedom in the numerator and a(n-)- degrees of freedom in the denominator. i 4-