Chapter 4 and 5 solutions

Chapter 4 and 5 solutions 4.4. 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 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 α = 0.05) and draw conclusions. Days Solution 1 3 4 1 13 18 39 16 4 17 44 3 5 4 1 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 1106.9 3 368.97 Model 703.50 351.75 40.7 0.0003 significant A 703.50 351.75 40.7 0.0003 Residual 51.83 6 8.64 Cor Total 186.5 11 The Model F value of 40.7 implies the model is significant. There is only a 0.03% chance that a "Model F Value" this large could occur due to noise. Std. Dev..94 R Squared 0.9314

Mean 18.75 Adj R Squared 0.9085 C.V. 15.68 Pred R Squared 0.755 PRESS 07.33 Adeq Precision 19.687 Treatment Means (Adjusted, If Necessary) Estimated Mean Standard Error 1 1 3.00 1.47 5.5 1.47 3 3 8.00 1.47 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t 1 vs.5 1.08 1.08 0.306 1 vs 3 15.00 1.08 7. 0.0004 vs 3 17.5 1.08 8.30 0.000 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. 4.7. Consider the hardness testing experiment described in Section 4.1. Suppose that the experiment was conducted as described and the following Rockwell C scale data (coded by subtracting 40 units) obtained: Coupon Tip 1 3 4 1 9.3 9.4 9.6 10.0 9.4 9.3 9.8 9.9 3 9. 9.4 9.5 9.7 4 9.7 9.6 10.0 10.

(a) Analyize the data from this experiment. There is a difference between the means of the four tips. Design Expert Output Response: Hardness ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Bock 0.8 3 0.7 Model 0.38 3 0.13 14.44 0.0009 significant A 0.38 3 0.13 14.44 0.0009 Residual 0.080 9 8.889E 003 Cor Total 1.9 15 The Model F value of 14.44 implies the model is significant. There is only a 0.09% chance that a "Model F Value" this large could occur due to noise. Std. Dev. 0.094 R Squared 0.880 Mean 9.63 Adj R Squared 0.7706 C.V. 0.98 Pred R Squared 0.4563 PRESS 0.5 Adeq Precision 15.635 Treatment Means (Adjusted, If Necessary) Estimated Mean Standard Error 1 1 9.57 0.047 9.60 0.047

3 3 9.45 0.047 4 4 9.88 0.047 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t 1 vs 0.05 1 0.067 0.38 0.7163 1 vs 3 0.13 1 0.067 1.87 0.0935 1 vs 4 0.30 1 0.067 4.50 0.0015 vs 3 0.15 1 0.067.5 0.0510 vs 4 0.7 1 0.067 4.1 0.006 3 vs 4 0.43 1 0.067 6.37 0.0001 (b) Use the Fisher LSD method to make comparisons among the four tips to determine specifically which tips differ in mean hardness readings. Based on the LSD bars in the Design Expert plot below, the mean of tip 4 differs from the means of tips 1,, and 3. The LSD method identifies a marginal difference between the means of tips and 3. 10. One Factor Plot 9.95 Hardness 9.7 9.45 9. 1 3 4 A: Tip

(c) Analyze the residuals from this experiment. The residual plots below do not identify any violations to the assumptions. Normal Plot of vs. Predicted 0.15 99 Normal % Probability 95 90 80 70 50 30 0 10 5 0.0875 0.05-0.0375 1-0.1-0.1-0.0375 0.05 0.0875 0.15 9. 9.47 9.71 9.96 10.0 Residual Predicted vs. Tip 0.15 0.0875 0.05-0.0375-0.1 1 3 4 Tip 4.8. A consumer products company relies on direct mail marketing pieces as a major component of its advertising campaigns. The company has three different designs for a new brochure and want to evaluate their effectiveness, as there are substantial differences in costs between the three designs. The company decides to test the three designs by mailing 5,000 samples of each to potential customers in four different regions of the country. Since there are known regional differences in the customer base, regions are considered as blocks. The number of responses to each mailing is shown below.

Region Design NE NW SE SW 1 50 350 19 375 400 55 390 580 3 75 340 00 310 (a) Analyze the data from this experiment. The residuals of the analsysis below identify concerns with the normality and equality of variance assumptions. As a result, a square root transformation was applied as shown in the second ANOVA table. The residuals of both analysis are presented for comparison in part (c) of this problem. The analysis concludes that there is a difference between the mean number of responses for the three designs. Design Expert Output Response: Number of responses ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Block 49035.67 3 16345. Model 90755.17 45377.58 50.15 0.000 significant A 90755.17 45377.58 50.15 0.000 Residual 548.83 6 904.81 Cor Total 1.45E+005 11 The Model F value of 50.15 implies the model is significant. There is only a 0.0% chance that a "Model F Value" this large could occur due to noise.

Std. Dev. 30.08 R Squared 0.9436 Mean 351.17 Adj R Squared 0.947 C.V. 8.57 Pred R Squared 0.774 PRESS 1715.33 Adeq Precision 16.197 Treatment Means (Adjusted, If Necessary) Estimated Mean Standard Error 1 1 98.50 15.04 473.75 15.04 3 3 81.5 15.04 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t 1 vs 175.5 1 1.7 8.4 0.000 1 vs 3 17.5 1 1.7 0.81 0.4483 vs 3 19.50 1 1.7 9.05 0.0001 Design Expert Output for Transformed Data Response: Number of responsestransform:square root Constant: 0 ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Block 35.89 3 11.96 Model 60.73 30.37 60.47 0.0001 significant A 60.73 30.37 60.47 0.0001 Residual 3.01 6 0.50 Cor Total 99.64 11

The Model F value of 60.47 implies the model is significant. There is only a 0.01% chance that a "Model F Value" this large could occur due to noise. Std. Dev. 0.71 R Squared 0.957 Mean 18.5 Adj R Squared 0.9370 C.V. 3.83 Pred R Squared 0.8109 PRESS 1.05 Adeq Precision 18.191 Treatment Means (Adjusted, If Necessary) Estimated Mean Standard Error 1 1 17.17 0.35 1.69 0.35 3 3 16.69 0.35 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t 1 vs 4.5 1 0.50 9.01 0.0001 1 vs 3 0.48 1 0.50 0.95 0.3769 vs 3 4.99 1 0.50 9.96 < 0.0001 (b) Use the Fisher LSD method to make comparisons among the three designs to determine specifically which designs differ in mean response rate. Based on the LSD bars in the Design Expert plot below, designs 1 and 3 do not differ; however, design is different than designs 1 and 3.

One Factor Plot 4.083 Sqrt(Number of responses) 1.598 19.113 16.67 14.14 1 3 A: Design (c) Analyze the residuals from this experiment. The first set of residual plots presented below represent the untransformed data. Concerns with normality as well as inequality of variance are presented. The second set of residual plots represent transformed data and do not identify significant violations of the assumptions. The residuals vs. design plot indicates a slight inequality of variance; however, not a strong violation and an improvement over the non transformed data. Normal Plot of vs. Predicted 36.5833 99 Normal % Probability 95 90 80 70 50 30 0 10 5 17 -.58333 -.1667 1-41.75-41.75 -.1667 -.58333 17 36.5833 199.75 85.88 37.00 458.13 544.5 Residual Predicted

vs. Design 36.5833 17 -.58333 -.1667-41.75 1 3 Design The following are the square root transformed data residual plots. Normal Plot of vs. Predicted 0.94069 99 Normal % Probability 95 90 80 70 50 30 0 10 5 0.4769 0.010514-0.45563 1-0.91041-0.91041-0.45563 0.010514 0.4769 0.94069 14.41 16.68 18.96 1.4 3.5 Residual Predicted

vs. Design 0.94069 0.4769 0.010514-0.45563-0.91041 1 3 Design 4.1. An aluminum master alloy manufacturer produces grain refiners in ingot form. The 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 furnaces as 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 as follows. Furnace Stirring Rate 1 3 4 5 8 4 5 6 10 14 5 6 9 15 14 6 9 0 17 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 165.19 3 55.06 Model.19 3 7.40 0.85 0.4995 not significant A.19 3 7.40 0.85 0.4995 Residual 78.06 9 8.67 Cor Total 65.44 15 The "Model F value" of 0.85 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 0.13 Mean 7.69 Adj R Squared 0.038 C.V. 38.31 Pred R Squared 1.4610 PRESS 46.7 Adeq Precision 5.390 Treatment Means (Adjusted, If Necessary) Estimated Mean Standard Error 1 5 5.75 1.47 10 8.50 1.47 3 15 7.75 1.47 4 0 8.75 1.47 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t 1 vs.75 1.08 1.3 0.193 1 vs 3.00 1.08 0.96 0.360 1 vs 4 3.00 1.08 1.44 0.1836

vs 3 0.75 1.08 0.36 0.770 vs 4 0.5 1.08 0.1 0.9071 3 vs 4 1.00 1.08 0.48 0.645 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. Normal plot of residuals 99 Normal % probability 95 90 80 70 50 30 0 10 5 1-3.815 -.065-0.315 1.4375 3.1875 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?

vs. Stirring Rate vs. Furnace 3.1875 3.1875 1.4375 1.4375-0.315-0.315 -.065 -.065-3.815-3.815 1 3 4 1 3 4 Stirring Rate Furnace 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 is no effect due to the stirring rate. 4.0. 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 run requires approximately 1 1/ 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 α = 0.05) and draw conclusions. Day Batch 1 3 4 5 1 A=8 B=7 D=1 C=7 E=3 C=11 E= A=7 D=3 B=8 3 B=4 A=9 C=10 E=1 D=5

4 D=6 C=8 E=6 B=6 A=10 5 E=4 D= B=3 A=8 C=8 The Minitab output below identifies the ingredients as having a significant effect on reaction time. Minitab Output General Linear Model Factor Type Levels Values Batch random 5 1 3 4 5 Day random 5 1 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 4 141.440 141.440 35.360 11.31 0.000 Batch 4 15.440 15.440 3.860 1.3 0.348 Day 4 1.40 1.40 3.060 0.98 0.455 Error 1 37.50 37.50 3.17 Total 4 06.640 5.3. 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 00 15 30 150 90.4 90.7 90.

90. 90.6 90.4 160 90.1 90.5 89.9 90.3 90.6 90.1 170 90.5 90.8 90.4 90.7 90.9 90.1 (a) Analyze the data and draw conclusions. Use α = 0.05. Both pressure (A) and temperature (B) are significant, the interaction is not. Response:Surface Finish 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 1.14 8 0.14 8.00 0.006 significant A 0.77 0.38 1.59 0.0004 B 0.30 0.15 8.47 0.0085 AB 0.069 4 0.017 0.97 0.4700 Residual 0.16 9 0.018 Lack of Fit 0.000 0 Pure Error 0.16 9 0.018 Cor Total 1.30 17 The Model F value of 8.00 implies the model is significant. There is only a 0.6% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms.

Values greater than 0.1000 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 residual plots show no serious deviations from the assumptions. vs. Predicted Normal plot of residuals 0.15 99 0.075 4.636E-014-0.075 Normal % probability 95 90 80 70 50 30 0 10 5 1-0.15 90.00 90.1 90.43 90.64 90.85-0.15-0.075-4.636E-014 0.075 0.15 Predicted Residual vs. Temperature vs. Pressure 0.15 0.15 0.075 0.075 3 4.636E-014 4.636E-014-0.075-0.075-0.15-0.15 1 3 1 3 Temperature Pressure

(c) Under what conditions would you operate this process? DESIGN-EXPERT Plot Yield 91.0008 Interaction Graph Temperature X = A: Pressure Y = B: Temperature 90.719 Design Points B1 150 B 160 B3 170 Yield 90.45 90.1371 89.849 00 15 30 Pressure Set pressure at 15 and Temperature at the high level, 170 degrees C, as this gives 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. Response:Surface Finish 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 1.13 5 0.3 16.18 < 0.0001 significant A 0.10 1 0.10 7. 0.0198 B 0.067 1 0.067 4.83 0.0483

A 0.67 1 0.67 47.74 < 0.0001 B 0.3 1 0.3 16.7 0.0015 AB 0.061 1 0.061 4.38 0.058 Residual 0.17 1 0.014 Lack of Fit 7.639E 003 3.546E 003 0.14 0.9314 not significant Pure Error 0.16 9 0.018 Cor Total 1.30 17 The Model F value of 16.18 implies the model is significant. There is only a 0.01% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B, A, B are significant model terms. Values greater than 0.1000 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. 0.1 R-Squared 0.8708 Mean 90.41 Adj R-Squared 0.8170 C.V. 0.13 Pred R-Squared 0.6794 PRESS 0.4 Adeq Precision 11.968 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 90.5 1 0.06 90.39 90.66 A Pressure 0.09 1 0.034 0.17 0.017 1.00 B Temperature 0.075 1 0.034 6.594E 004 0.15 1.00 A 0.41 1 0.059 0.54 0.8 1.00 B 0.4 1 0.059 0.11 0.37 1.00 AB 0.087 1 0.04 0.18 3.548E 003 1.00

Final Equation in Terms of Coded Factors: Yield = +90.5 0.09 * A +0.075 * B 0.41 * A +0.4 * B 0.087 * A * B Final Equation in Terms of Actual Factors: Yield = +48.54630 +0.86759 * Pressure 0.6404 * Temperature 1.81481E 003 * Pressure +.41667E 003 * Temperature 5.83333E 004 * Pressure * Temperature

170.00 90.8 Yield 90.7 B: Temperature 165.00 160.00 90.6 90. 90.5 90.5 90.4 90.3 90.1 90.4 90.3 Yield 91 90.8 90.6 90.4 90. 90 155.00 90.6 150.00 00.00 07.50 15.00.50 30.00 A: Pressure 170.00 00.00 165.00 07.50 160.00 15.00.50 A: Pres s ure 30.00 B: Temperature 155.00 150.00 5.4. 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) 0.15 0.18 0.0 0.5 74 79 8 99 0.0 64 68 88 104 60 73 9 96 9 98 99 104 0.5 86 104 108 110 88 88 95 99 99 104 108 114 0.30 98 99 110 111

10 95 99 107 (a) Analyze the data and draw conclusions. Use α = 0.05. The depth (A) and feed rate (B) are significant, as is the interaction (AB). Response: Surface Finish 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 584.67 11 531.15 18.49 < 0.0001 significant A 15.11 3 708.37 4.66 < 0.0001 B 3160.50 1580.5 55.0 < 0.0001 AB 557.06 6 9.84 3.3 0.0180 Residual 689.33 4 8.7 Lack of Fit 0.000 0 Pure Error 689.33 4 8.7 Cor Total 653.00 35 The Model F value of 18.49 implies the model is significant. There is only a 0.01% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 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. vs. Predicted Normal plot of residuals 8 99 3.83333-0.333333-4.5 Normal % probability 95 90 80 70 50 30 0 10 5 1-8.66667 66.00 77.17 88.33 99.50 110.67-8.66667-4.5-0.333333 3.83333 8 Predicted Residual vs. Feed Rate vs. Depth of Cut 8 8 3.83333 3.83333-0.333333-0.333333-4.5-4.5-8.66667-8.66667 1 3 1 3 4 Feed Rate Depth of Cut (c) Obtain point estimates of the mean surface finish at each feed rate. Feed Rate Average 0.0 81.58 0.5 97.58

0.30 103.83 DESIGN-EXPERT Plot Surface Finish X = B: Feed Rate 114 One Factor Plot Warning! Factor involved in an interaction. Actual Factor A: Depth of Cut = Average 100.5 Surface Finish 87 73.5 60 0.0 0.5 0.30 Feed Rate (d) Find P values for the tests in part (a). The P values are given in the computer output in part (a). 5.5. For the data in Problem 5.3, compute a 95 percent interval estimate of the mean difference in response for feed rates of 0.0 and 0.5 in/min. We wish to find a confidence interval on μ 1 μ, where μ 1 is the mean surface finish for 0.0 in/min and μ is the mean surface finish for 0.5 in/min. y y t MS E ( n 1) μ 1 μ y1.. y.. + tα, ab( n 1) n 1.... α, ab ) MS n E (8.7) ( 81.5833 97.5833) ± (.064) = 16 ± 9.03 3 Therefore, the 95% confidence interval for μ μ is 16.000 ± 9.03. 1

5.6. An article in Industrial Quality Control (1956, 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 1 3 80 300 90 1 90 310 85 85 95 90 30 60 0 35 40 5 40 35 30 (a) Is there any indication that either factor influences brightness? Use α = 0.05. Both factors, phosphor type (A) and Glass type (B) influence brightness. Response: Current in microamps 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 15516.67 5 3103.33 58.80 < 0.0001 significant A 933.33 466.67 8.84 0.0044 B 14450.00 1 14450.00 73.79 < 0.0001

AB 133.33 66.67 1.6 0.3178 Residual 633.33 1 5.78 Lack of Fit 0.000 0 Pure Error 633.33 1 5.78 Cor Total 16150.00 17 The Model F value of 58.80 implies the model is significant. There is only a 0.01% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms. (b) Do the two factors interact? Use α = 0.05. There is no interaction effect. (c) Analyze the residuals from this experiment. 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.

vs. Predicted Normal plot of residuals 15 99 8.75.5-3.75 Normal % probability 95 90 80 70 50 30 0 10 5 1-10 5.00 44.17 63.33 8.50 301.67-10 -3.75.5 8.75 15 Predicted Residual vs. Glass Type vs. Phosphor Type 15 15 8.75 8.75.5.5-3.75-3.75-10 -10 1 1 3 Glass Type Phosphor Type 5.7. Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences, Wiley 1977) describe an experiment to investigate the warping of copper plates. The two factors studied 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) 40 60 80 100

50 17,0 16,1 4, 8,7 75 1,9 18,13 17,1 7,31 100 16,1 18,1 5,3 30,3 15 1,17 3,1 3, 9,31 (a) Is there any indication that either factor affects the amount of warping? Is there any interaction between the factors? Use α = 0.05. Both factors, copper content (A) and temperature (B) affect warping, the interaction does not. Response: Warping 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 968. 15 64.55 9.5 < 0.0001 significant A 698.34 3 3.78 34.33 < 0.0001 B 156.09 3 5.03 7.67 0.001 AB 113.78 9 1.64 1.86 0.137 Residual 108.50 16 6.78 Lack of Fit 0.000 0 Pure Error 108.50 16 6.78 Cor Total 1076.7 31 The Model F value of 9.5 implies the model is significant. There is only a 0.01% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 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 vs. Predicted 3.5 99 Normal % probability 95 90 80 70 50 30 0 10 5 1.75 1.06581E-014-1.75 1-3.5-3.5-1.75-1.06581E-014 1.75 3.5 10.50 15.38 0.5 5.13 30.00 Residual Predicted vs. Copper Content vs. Temperature 3.5 3.5 1.75 1.06581E-014-1.75 1.75 1.06581E-014-1.75-3.5-3.5 1 3 4 1 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 40 40 100 B Temperature Average 50 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 15.5 0.9 13.55 17.45.76 9.64 1.36 Factor Name Level Low Level High Level A Copper Content 60 40 100 B Temperature Average 50 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 18.875 0.9 16.9 0.83.76 13.0 4.73 Factor Name Level Low Level High Level A Copper Content 80 40 100 B Temperature Average 50 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 1 0.9 19.05.95.76 15.14 6.86 Factor Name Level Low Level High Level A Copper Content 100 40 100 B Temperature Average 50 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high

Warping 8.5 0.9 6.30 30.0.76.39 34.11 Use a copper content of 40 for the lowest warping. MS 6.7815 S = E = = 0.9 b 8 Scaled t Distribution Cu=40 Cu=60 Cu=80 Cu=100 15.0 18.0 1.0 4.0 7.0 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 40. This is the same as for part (c).

DESIGN-EXPERT Plot Warping X = A: Copper Content Y = B: Temperature Design Points 3.760 6.5051 Interaction Graph Temperature B1 50 B 75 B3 100 B4 15 Warping 0.5 3 13.9949 7.73979 40 60 80 100 Copper Content 5.9. 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 α = 0.05. (A) Feed Rate (B) Drill Speed 0.015 0.030 0.045 0.060 15.70.45.60.75.78.49.7.86 00.83.85.86.94.86.80.87.88

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 0.8 7 0.040 15.53 0.0005 significant A 0.15 1 0.15 57.01 < 0.0001 B 0.09 3 0.031 11.86 0.006 AB 0.04 3 0.014 5.37 0.056 Residual 0.01 8.600E 003 Lack of Fit 0.000 0 Pure Error 0.01 8.600E 003 Cor Total 0.30 15 The Model F value of 15.53 implies the model is significant. There is only a 0.05% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 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.

DESIGN-EXPERT Plot Force.96879 Interaction Graph Drill Speed X = B: Feed Rate Y = A: Drill Speed Design Points A1 15 A 00.894 Force.69.5506.4111 0.015 0.030 0.045 0.060 Feed Rate 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, we can fit polynomial effects of both speed and feed rate, exactly as in Example 5.5 in the text. The Design Expert output with only the significant terms retained, including the response surface plots, now follows. Response: Force 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 0.3 3 0.075 11.49 0.0008 significant A 0.15 1 0.15.70 0.0005 B 0.019 1 0.019.94 0.1119 B 0.058 1 0.058 8.8 0.0117 Residual 0.078 1 6.530E 003 Lack of Fit 0.058 4 0.014 5.53 0.0196 significant Pure Error 0.01 8.600E 003 Cor Total 0.30 15

The Model F value of 11.49 implies the model is significant. There is only a 0.08% chance that a "Model F Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms. Values greater than 0.1000 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. 0.081 R-Squared 0.7417 Mean.77 Adj R-Squared 0.677 C.V..9 Pred R-Squared 0.5517 PRESS 0.14 Adeq Precision 9.69 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.69 1 0.03.6.76 A Drill Speed 0.096 1 0.00 0.05 0.14 1.00 B Feed Rate 0.047 1 0.07 0.013 0.11 1.00 B 0.13 1 0.045 0.036 0.3 1.00 Final Equation in Terms of Coded Factors: Force = +.69 +0.096 * A +0.047 * B +0.13 * B Final Equation in Terms of Actual Factors:

Force = +.48917 +3.06667E 003 * Drill Speed 15.76667 * Feed Rate +66.66667 * Feed Rate Force 0.06.9.85.8 0.05 3 B: Feed Rate.75 0.04.7.6.65 Force.9.8.7.6.5 0.03.8.85 0.0 15.00 143.75 16.50 181.5 00.00 A: Drill Speed 0.06 0.05 0.04 B: Feed Rate 0.03 0.0 15.00 143.75 00.00 181.5 16.50 A: Drill Speed