Chapter 2 Simple Comparative Experiments Solutions


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1 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 ( ) = y o zo n 4 Since z.5 =.645, do not reject. (c) Find the Pvalue for the test in part (b). From the ztable: P (d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is y z y z n n 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? 
2 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY y o 8 8 zo n 6 4 Since z / = z.5 =.96, do not reject. (c) What is the Pvalue 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 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 zo 5.8. n (c) Find the Pvalue 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 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.. 
3 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Since y N(,9), a 95% twosided 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 halfinterval is.5. Since z / = z.5 =.96,. 963 n n n The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested, and the following results are obtained: Days (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 = [ (83) + (43) + (43) + (63) + (53) + (383) + (633) + (593) + (343) + ( 393) ] / (  ) s = 3438 / 9 = 38 s t o y o 3 s n since t.,9 =.8; do not reject H Minitab Output TTest of the Mean Test of mu =. vs mu >. Variable N Mean StDev SE Mean T P Shelf Life T Confidence Intervals 3
4 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Variable N Mean StDev SE Mean 99. % CI Shelf Life (.9, 5.9) (c) Find the Pvalue 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 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 ttest in problem 5 is not too serious unless the departure from normality is severe. Normal Probability Plot for Shelf Life ML Estimates Percent ML Estimates Mean 3 StDev Goodness of Fit AD* Data 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
5 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  ) = s y o to.67 s n 6 since t.5,5 =.753; do not reject H Minitab Output TTest of the Mean Test of mu = 5. vs mu > 5. Variable N Mean StDev SE Mean T P Hours T Confidence Intervals Variable N Mean StDev SE Mean 95. % CI Hours ( 88.9, 94.) (c) Find the Pvalue 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 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
6 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot for Hours ML Estimates Percent ML Estimates Mean 4.5 StDev Goodness of Fit AD* Data 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 (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 y n z o y y n n z.5 =.96; do not reject (c) What is the Pvalue 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
7 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 ( ) (9.6) 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 n n z. =.5; do not reject The 99 percent confidence interval is ( ) (.575) y y z y y z n n n n 6.4 ( ) (.575) 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 (a) Test the hypotheses that the two variances are equal. Use =.5. 7
8 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY F 599.,, 43. F H: H: F ,, S S S S Do not reject. F ,,. (b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use =.5. What is the Pvalue for this test? ( n ) S ( n ) S S p n n 8 S p 93. y y t S p 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 TTest and Confidence Interval Two sample T for Type vs Type N Mean StDev SE Mean Type Type % CI for mu Type  mu Type : ( 8.6, 9.) TTest 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 ttest. However, moderate departure from normality has little impact on the performance of the ttest. 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
9 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot for Type ML Estimates Percent ML Estimates Mean 7.4 StDev Goodness of Fit AD* Data 8 9 Normal Probability Plot for Type ML Estimates Percent ML Estimates Mean 7. StDev Goodness of Fit AD* Data 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) (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 TTest and Confidence Interval 9
10 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Two sample T for Uniformity Flow Rat N Mean StDev SE Mean % CI for mu (5)  mu (): ( .63,.4) TTest mu (5) = mu () (vs not =): T = .35 P =. DF = Both use Pooled StDev =.79 (b) What is the Pvalue for the test in part (a)? From the computer printout, P=. (c) Does the C F 6 flow rate affect the wafertowafer variability in etch uniformity? Use =.5. H: H: F555.,, F Do not reject; C F 6 flow rate does not affect wafertowafer 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 ttest in part (a). 5 Uniformity 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. 
11 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY. 5, 7, S. 7 F S 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 )( ) S p n n 8 9 S p y y. 5. t. 479 S p n n 8 9 t. 5, 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: (a) Construct a 95 percent confidence interval estimate of. n S n S, n ( ), n (b) Test the hypothesis that =.. Use =.5. What are your conclusions? H 59.,. H : : SS ,
12 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 ML Estimates Mean 5.88 StDev Goodness of Fit AD* Data 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^ (a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use =.5. 
13 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY H : H : or equivalently H H : : d d Minitab Output Paired TTest and Confidence Interval Paired T for Caliper  Caliper N Mean StDev SE Mean Caliper Caliper Difference % CI for mean difference: (.4,.54) TTest of mean difference = (vs not = ): TValue =.43 PValue =.674 (b) Find the Pvalue 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 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/ S/ S3/ S4/ S5/ S/ S/ S/ S/ Sum = 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 n i 3
14 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 TTest and Confidence Interval Paired T for Karlsruhe  Lehigh N Mean StDev SE Mean Karlsruh Lehigh Difference t, n t. 5, 9 95% CI for mean difference: (.7,.3777) TTest of mean difference = (vs not = ): TValue = 6.8 PValue =. d.74 t 6.8 Sd.35 n 9. 36, reject the null hypothesis. (b) What is the Pvalue 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 d d d S d d t,n n (d) Investigate the normality assumption for both samples. Normal Probability Plot.999 Probability Av erage:.34 StDev :.463 N: Karlsruhe AndersonDarling Normality Test ASquared:.86 PValue:.5374
15 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot Probability Lehigh Av erage:.66 StDev : N: 9 AndersonDarling Normality Test ASquared:.77 PValue:.8 (e) Investigate the normality assumption for the difference in ratios for the two methods. Normal Probability Plot Probability Difference Av erage: StDev :.3599 N: 9 AndersonDarling Normality Test ASquared:.38 PValue:.464 (f) Discuss the role of the normality assumption in the paired ttest. As in any ttest, the assumption of normality is of only moderate importance. In the paired ttest, 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
16 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 Av erage:.5 StDev :.757 N: Form 5 5 AndersonDarling Normality Test ASquared:.45 PValue:.7 Normal Probability Plot Probability Av erage: StDev : N: Form 95 5 AndersonDarling Normality Test ASquared:.443 PValue:.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 TTest and Confidence Interval Two sample T for Form vs Form N Mean StDev SE Mean Form.5..9 Form % CI for mu Form  mu Form : ( ., 5.9) TTest mu Form = mu Form (vs >): T =.8 P =.4 DF = Both use Pooled StDev =. (c) What is the Pvalue for the test in part (a)? P =.46
17 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 TwoSample TTest and CI: Form, Form Twosample T for Form vs Form N Mean StDev SE Mean Form.5..9 Form r Difference = mu Form  mu Form Estimate for difference: % lower bound for difference:.36 TTest of difference = 3 (vs >): TValue =.8 PValue =.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 (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 TTest and Confidence Interval Two sample T for Solution vs Solution N Mean StDev SE Mean Solution Solution % CI for mu Solution  mu Solution: ( .83, .43) TTest 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
18 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot Probability Av erage: 9.95 StDev :.4659 N: Solution.5 AndersonDarling Normality Test ASquared:. PValue:.743 Normal Probability Plot.999 Probability Av erage:.365 StDev :.369 N: Solution AndersonDarling Normality Test ASquared:.58 PValue:.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
19 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
20 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 n y S This agrees with the result in Table where Construct a data set for which the paired ttest statistic is very large, but for which the usual twosample or pooled ttest statistic is small. In general, describe how you created the data. Does this give you any insight regarding how the paired ttest works? A B delta
21 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Minitab Output Paired TTest and Confidence Interval Paired T for A  B N Mean StDev SE Mean A B Difference % CI for mean difference: (.74, .65) TTest of mean difference = (vs not = ): TValue = 7.7 PValue =. Two Sample TTest and Confidence Interval Two sample T for A vs B N Mean StDev SE Mean A B % CI for mu A  mu B: ( .4, 8.6) TTest 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 twosample ttest 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. 
22 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 ) (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 E significant A 4.897E E Residual.539E Lack of Fit. Pure Error.539E Cor Total 6.436E+5 5 The Model Fvalue of.73 implies the model is significant. There is only a.5% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs <. 3 vs The Fvalue is.73 with a corresponding Pvalue of.5. Mixing technique has an effect. (b) Construct a graphical display as described in Section to compare the mean tensile strengths for the four mixing techniques. What are your conclusions? S y i. MS n E
23 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (4) (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 ( ) Treatment vs. Treatment 4 = = 49. > Treatment vs. Treatment 3 = =.5 > Treatment vs. Treatment = = 85.5 > Treatment vs. Treatment 4 = = > Treatment vs. Treatment 3 = = 37.5 < Treatment 3 vs. Treatment 4 = = 67.5 > The Fisher LSD method is also presented in the DesignExpert 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
24 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 99 Norm al % probability Residual (e) Plot the residuals versus the predicted tensile strength. Comment on the plot. There is nothing unusual about this plot. Residuals vs. Predicted Residuals Predicted (f) Prepare a scatter plot of the results to aid the interpretation of the results of this experiment. DesignExpert 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. 33
25 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 33 One Factor Plot Tensile Strength 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 MS n, S E y i. 3, S y i. 4, S yi. Treatment vs. Treatment 4 = = 49. > (R 4 ) Treatment vs. Treatment 3 = =.5 > 8.9 (R 3 ) Treatment vs. Treatment = = 85.5 > (R ) Treatment vs. Treatment 4 = = > 8.9 (R 3 ) Treatment vs. Treatment 3 = = 37.5 < (R ) Treatment 3 vs. Treatment 4 = = 67.5 > (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)
26 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 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. 33 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 : Treatment : Treatment 3: Treatment 4: y y i. j. t,n a MS n E i j y y i. j. t,n a MS n 34 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 E 35
27 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 DesignExpert 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 not significant A Residual Lack of Fit. Pure Error Cor Total.5 7 The "Model Fvalue" of. implies the model is not significant relative to the noise. There is a 5.69 % chance that a "Model Fvalue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs vs (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. 36
28 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals. Residuals vs. Predicted 99 Normal % probability Residuals Residual Predicted (d) Construct a graphical display of the treatments as described in Section Does this graph adequately summarize the results of the analysis of variance in part (b). Yes. Scaled t Distribution (5) (75,5,) M ean D ensity 35 Rework Part (d) of Problem 34 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 37
29 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 DesignExpert output indicates that the Pvalue 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 33, which accommodates different sample sizes. Equation 33 simplifies to Equation 333 for a balanced design experiment. 36 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 (a) Is there a difference in conductivity due to coating type? Use =.5. Yes, there is a difference in means. Refer to the DesignExpert 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 significant A Residual Lack of Fit. Pure Error Cor Total The Model Fvalue of 4.3 implies the model is significant. There is only a.3% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs vs (b) Estimate the overall mean and the treatment effects. 38
30 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ˆ 7 / ˆ y. y ˆ y. y ˆ 3 y3. y ˆ y y (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 Treatment 4: Treatment  Treatment 4: (d) Test all pairs of means using the Fisher LSD method with =.5. Refer to the DesignExpert 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 to compare the means. Which coating produces the highest conductivity? S y i. MS E Coating types and produce the highest conductivity. n 4 Scaled t Distribution (4) (3) ()() Conductivity 39
31 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. 37 Reconsider the experiment in Problem 36. 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 nonconstant variance. Normal plot of residuals Residuals vs. Predicted Normal % probability Residuals Residual Predicted Residuals vs. Coating Type Residuals Coating Type 38 An article in the ACI Materials Journal (Vol. 84, 987. pp. 36) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3 x 6 cylinder was used, and the 3
32 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 (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 not significant A Residual Lack of Fit. Pure Error Cor Total The "Model Fvalue" of.87 implies the model is not significant relative to the noise. There is a.38 % chance that a "Model Fvalue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs vs (b) Find the Pvalue 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
33 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted Normal % probability Residuals Residual Predicted Residuals vs. Rodding Level Residuals Rodding Level (d) Construct a graphical display to compare the treatment means as describe in Section
34 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (, 5) (5) () Mean Compressive Strength 39 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 (%) (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 <. significant A <. Residual Lack of Fit. Pure Error Cor Total The Model Fvalue of 3.85 implies the model is significant. There is only a.% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) EstimatedStandard Mean Error
35 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs <. vs <. vs <. vs vs vs <. vs <. 3 vs vs <. 3 vs <. 4 vs vs vs (b) Find the Pvalue 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 Residuals Residual Predicted 34
36 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 4 Residuals vs. Orifice Diameter.85 Residuals Orifice Diameter (d) Find a 95 percent confidence interval on the mean percent radon released when the orifice diameter is Treatment 5 (Orifice =.4): (e) Construct a graphical display to compare the treatment means as describe in Section What conclusions can you draw? Scaled t Distribution (6) (5) (4) (3) () () 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. 35
37 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Circuit Type Response Time (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 significant A Residual Lack of Fit. Pure Error Cor Total The Model Fvalue of 6.8 implies the model is significant. There is only a.4% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs (b) Use Tukey s test to compare pairs of treatment means. Use =.. S y i. MS E n 5 q., 3, 5. 4 t vs. :.8.=.4 > 9.66 vs. 3:.88.4=.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 to compare the treatment means. What conclusions can you draw? How do they compare with the conclusions from part (a). The scaledt plot agrees with part (b). In this case, the large difference between the mean of treatment and the other two treatments is very obvious. 36
38 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (3) () () 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 SS F C C 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. 37
39 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 7.8 Residuals vs. Predicted 99 Normal % probability Residuals Residual Predicted Residuals vs. Circuit Type Residuals 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 (a) Is there any indication that the fluids differ? Use =.5. At =.5 there are no difference, but at since the Pvalue is just slightly above.5, there is probably a difference in means. Design Expert Output Response: Life in in h 38
40 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 not significant A Residual Lack of Fit. Pure Error Cor Total The Model Fvalue of 3.5 implies there is a 5.5% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs vs (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 Residuals Residual Predicted 39
41 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY.95 Residuals vs. Fluid Type.55 Residuals 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 (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 <. significant A <. Residual Lack of Fit. Pure Error Cor Total The Model Fvalue of.78 implies the model is significant. There is only a.% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs <. 3
42 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY vs vs <. vs vs vs (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 Normal % probability Residuals Residual Predicted Residuals vs. Circuit Design Residuals 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 Pvalue greater than.5, it is less than.. Unless there are other reasons for choosing Type 3, Type would be selected. 3
43 Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 33 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 (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 not significant A Residual Lack of Fit. Pure Error Cor Total.9 The Model Fvalue of 3.5 implies there is a 8.3% chance that a "Model FValue" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error Mean Standard t for H Treatment Difference DF Error Coeff= Prob > t vs vs vs vs vs vs (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. 3
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