Chapter 5 Introduction to Factorial Designs Solutions

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Chapter 5 Introduction to Factorial Designs Solutions 5.1. The following output was obtained from a computer program that performed a two-factor ANOVA on a factorial experiment. Two-way ANOVA: y versus A, B Source DF SS MS F P A 1.3??? B? 8.554 4.771 4.59? Interaction????? Error 1 15.37 8.7773 Total 17 31.551 (a) Fill in the blanks in the ANOVA table. You can use bounds on the P-values. Two-way ANOVA: y versus A, B Source DF SS MS F P A 1.3.3.4.8513 B 8.554 4.771 4.59.331 Interaction 45.348.674.58.1167 Error 1 15.37 8.7773 Total 17 31.551 (b) How many levels were used for factor B? 3 levels. (c) How many replicates of the experiment were performed? 3 replicates. (d) What conclusions would you draw about this experiment? Only factor B is significant; factor A and the two-factor interaction are not significant. 5-1

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.. The following output was obtained from a computer program that performed a two-factor ANOVA on a factorial experiment. Two-way ANOVA: y versus A, B Source DF SS MS F P A 1?.?? B? 18.378??? Interaction 3 8.479??.93 Error 8 158.797? Total 15 347.653 (e) Fill in the blanks in the ANOVA table. You can use bounds on the P-values. Two-way ANOVA: y versus A, B Source DF SS MS F P A 1...1.998 B 3 18.378 6.16 3.97.93 Interaction 3 8.479.863.1439.93 Error 8 158.797 19.8496 Total 15 347.653 (a) How many levels were used for factor B? 4 levels. (b) How many replicates of the experiment were performed? replicates. (c) What conclusions would you draw about this experiment? Factor B is moderately significant with a P-value of.93. Factor A and the two-factor interaction are not significant. 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 15 3 15 9.4 9.7 9. 9. 9.6 9.4 16 9.1 9.5 89.9 9.3 9.6 9.1 17 9.5 9.8 9.4 9.7 9.9 9.1 (d) Analyze the data and draw conclusions. Use α =.5. 5-

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Both pressure (A) and temperature (B) are significant, the interaction is not. 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 1.14 8.14 8..6 significant A.77.38 1.59.4 B.3.15 8.47.85 AB.69 4.17.97.47 Residual.16 9.18 Lack of Fit. Pure Error.16 9.18 Cor Total 1.3 17 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.1 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.15 99.75-4.636E-14 -.75 Normal % probability 95 9 8 7 5 3 1 5 1 -.15 9. 9.1 9.43 9.64 9.85 -.15 -.75-4.636E-14.75.15 Predicted Residual 5-3

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY.15 vs. Temperature.15 vs. Pressure.75.75 3-4.636E-14-4.636E-14 -.75 -.75 -.15 -.15 1 3 1 3 Temperature Pressure (c) Under what conditions would you operate this process? DESIGN-EXPERT Plot Yield 91.8 Interaction Graph Tem perature X = A: Pressure Y = B: Temperature Design Points B1 15 B 16 B3 17 Yie ld 9.719 9.45 9.1371 89.849 15 3 P res s ure Set pressure at 15 and Temperature at the high level, 17 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. 5-4

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 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 1.13 5.3 16.18 <.1 significant A.1 1.1 7..198 B.67 1.67 4.83.483 A.67 1.67 47.74 <.1 B.3 1.3 16.7.15 AB.61 1.61 4.38.58 Residual.17 1.14 Lack of Fit 7.639E-3 3.546E-3.14.9314 not significant Pure Error.16 9.18 Cor Total 1.3 17 The Model F-value of 16.18 implies the model is significant. There is only a.1% 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.1 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..1 R-Squared.878 Mean 9.41 Adj R-Squared.817 C.V..13 Pred R-Squared.6794 PRESS.4 Adeq Precision 11.968 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 9.5 1.6 9.39 9.66 A-Pressure -.9 1.34 -.17 -.17 1. B-Temperature.75 1.34 6.594E-4.15 1. A -.41 1.59 -.54 -.8 1. B.4 1.59.11.37 1. AB -.87 1.4 -.18 3.548E-3 1. Final Equation in Terms of Coded Factors: Yield = +9.5 -.9 * A +.75 * B -.41 * A +.4 * B -.87 * A * B Final Equation in Terms of Actual Factors: Yield = +48.5463 +.86759 * Pressure -.644 * Temperature -1.81481E-3 * Pressure +.41667E-3 * Temperature -5.83333E-4 * Pressure * Temperature 5-5

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Yield 17. 9.8 9.7 B: Temperature 165. 9.6 9. 9.5 9.5 16. 9.4 9.4 9.3 9.1 9.3 Yield 91 9.8 9.6 9.4 9. 9 155. 9.6 15.. 7.5 15..5 3. A: Pressure 17.. 165. 7.5 16. 15..5 155.B: Temperature A: Pressure 3. 15. 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).15.18..5 74 79 8 99. 64 68 88 14 6 73 9 96 9 98 99 14.5 86 14 18 11 88 88 95 99 99 14 18 114.3 98 99 11 111 1 95 99 17 (a) Analyze the data and draw conclusions. Use α =.5. The depth (A) and feed rate (B) are significant, as is the interaction (AB). 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 11 531.15 18.49 <.1 significant A-Depth 15.11 3 78.37 4.66 <.1 B-Feed 316.5 158.5 55. <.1 AB 557.6 6 9.84 3.3.18 Residual 689.33 4 8.7 Lack of Fit. Pure Error 689.33 4 8.7 Cor Total 653. 35 5-6

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY The Model F-value of 18.49 implies the model is significant. There is only a.1% 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 vs. Predicted Normal plot of residuals 99 3.83333 -.333333-4.5 Normal % probability 95 9 8 7 5 3 1 5-8.66667 1 66. 77.17 88.33 99.5 11.67-8.66667-4.5 -.333333 3.83333 8 Predicted Residual 8 vs. Feed Rate 8 vs. Depth of Cut 3.83333 3.83333 -.333333 -.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. 81.58.5 97.58.3 13.83 5-7

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 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 1.5 Surface Finish 87 7 3.5 6.. 5.3 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 5.4, compute a 95 percent interval estimate of the mean difference in response for feed rates of. and.5 in/min. We wish to find a confidence interval on, where µ 1 is the mean surface finish for. in/min and µ is the mean surface finish for.5 in/min. MS E MS E y1.. y.. tα, ab( n 1) µ 1 µ y1.. y.. + tα, ab( n 1) ) n n (8.7) ( 81.5833 97.5833) ± (.64) = 16 ± 9.3 3 Therefore, the 95% confidence interval for µ 1 µ is -16. ± 9.3. 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 8 3 9 1 9 31 85 85 95 9 3 6 35 4 5 4 35 3 5-8

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (a) Is there any indication that either factor influences brightness? Use α =.5. Both factors, phosphor type (A) and Glass type (B) influence brightness. 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 15516.67 5 313.33 58.8 <.1 significant A 933.33 466.67 8.84.44 B 1445. 1 1445. 73.79 <.1 AB 133.33 66.67 1.6.3178 Residual 633.33 1 5.78 Lack of Fit. Pure Error 633.33 1 5.78 Cor Total 1615. 17 The Model F-value of 58.8 implies the model is significant. There is only a.1% 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 α =.5. 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 9 8 7 5 3 1 5 1-1 5. 44.17 63.33 8.5 31.67-1 -3.75.5 8.75 15 Predicted Residual 5-9

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY vs. Glass Type vs. Phosphor Type 15 15 8.75 8.75.5.5-3.75-3.75-1 -1 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) 4 6 8 1 5 17, 16,1 4, 8,7 75 1,9 18,13 17,1 7,31 1 16,1 18,1 5,3 3,3 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 =.5. Both factors, copper content (A) and temperature (B) affect warping, the interaction does not. 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. 15 64.55 9.5 <.1 significant A 698.34 3 3.78 34.33 <.1 B 156.9 3 5.3 7.67.1 AB 113.78 9 1.64 1.86.137 Residual 18.5 16 6.78 Lack of Fit. Pure Error 18.5 16 6.78 Cor Total 176.7 31 The Model F-value of 9.5 implies the model is significant. There is only a.1% 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. 5-1

Solutions from Montgomery, D. C. (1) 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 vs. Predicted 3.5 99 Normal % probability 95 9 8 7 5 3 1 5 1.75-1.6581E-14-1.75 1-3.5-3.5-1.75-1.6581E-14 1.75 3.5 1.5 15.38.5 5.13 3. Residual Predicted vs. Copper Content vs. Temperature 3.5 3.5 1.75-1.6581E-14-1.75 1.75-1.6581E-14-1.75-3.5-3.5 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? Factor Name Level Low Level High Level A Copper Content 4 4 1 B Temperature Average 5 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 15.5.9 13.55 17.45.76 9.64 1.36 Factor Name Level Low Level High Level A Copper Content 6 4 1 B Temperature Average 5 15 5-11

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 18.875.9 16.9.83.76 13. 4.73 Factor Name Level Low Level High Level A Copper Content 8 4 1 B Temperature Average 5 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 1.9 19.5.95.76 15.14 6.86 Factor Name Level Low Level High Level A Copper Content 1 4 1 B Temperature Average 5 15 Prediction SE Mean 95% CI low 95% CI high SE Pred 95% PI low 95% PI high Warping 8.5.9 6.3 3..76.39 34.11 Use a copper content of 4 for the lowest warping. MS 6.7815 S = E = =.9 b 8 Scaled t Distribution Cu=4 Cu=6 Cu=8 Cu=1 15. 18. 1. 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). 5-1

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Warping 3.76 X = A: Copper Content Y = B: Temperature 6.551 Design Points Interaction Graph Tem perature B1 5 B 75 B3 1 B4 15 Warping.5 3 13.9949 7.73979 4 6 8 1 Copper Content 5.8. 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: Operator Machine 1 3 4 1 19 11 18 11 11 115 19 18 11 11 111 114 11 111 19 11 3 116 11 114 1 114 115 119 117 (a) Analyze the data and draw conclusions. Use α =.5. Only the Operator (A) effect is significant. 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 17.46 11 19.77 5.1.41 significant A 16.33 8.17 1.14.1 B 1.46 3 4.15 1.1.3888 AB 44.67 6 7.44 1.96.157 Residual 45.5 1 3.79 Lack of Fit. Pure Error 45.5 1 3.79 Cor Total 6.96 3 The Model F-value of 5.1 implies the model is significant. There is only a.41% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms aresignificant. 5-13

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 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. Normal Plot of vs. Predicted 3 99 Normal % Probability 95 9 8 7 5 3 1 5 1.5-1.5 1-3 -.5-1.5 1.5.5 18.5 111. 113.5 116. 118.5 Residual Predicted vs. Operator vs. Machine 3 3 1.5 1.5 3 3-1.5-1.5-3 -3 1 3 1 3 4 Operator Machine 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 =.5. (A) Feed Rate (B) Drill Speed.15.3.45.6 15.7.45.6.75 5-14

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY.78.49.7.86.83.85.86.94.86.8.87.88 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 15.53.5 significant A.15 1.15 57.1 <.1 B.9 3.31 11.86.6 AB.4 3.14 5.37.56 Residual.1 8.6E-3 Lack of Fit. Pure Error.1 8.6E-3 Cor Total.3 15 The Model F-value of 15.53 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. DESIGN-EXPERT Plot Force.96879 Interaction Graph D rill S p e e d X = B: Feed Rate Y = A: Drill Speed Design Points A1 15 A Force.894.6 9.556.4111.15.3.45.6 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 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 3.75 11.49.8 significant A.15 1.15.7.5 B.19 1.19.94.1119 5-15

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY B.58 1.58 8.8.117 Residual.78 1 6.53E-3 Lack of Fit.58 4.14 5.53.196 significant Pure Error.1 8.6E-3 Cor Total.3 15 The Model F-value of 11.49 implies the model is significant. There is only a.8% 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.1 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..81 R-Squared.7417 Mean.77 Adj R-Squared.677 C.V..9 Pred R-Squared.5517 PRESS.14 Adeq Precision 9.69 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept.69 1.3.6.76 A-Drill Speed.96 1..5.14 1. B-Feed Rate.47 1.7 -.13.11 1. B.13 1.45.36.3 1. Final Equation in Terms of Coded Factors: Force = +.69 +.96 * A +.47 * B +.13 * B Final Equation in Terms of Actual Factors: Force = +.48917 +3.6667E-3 * Drill Speed -15.76667 * Feed Rate +66.66667 * Feed Rate Force.6.9.85.8.5 3 B: Feed Rate.75.7.4.6.65 Force.9.8.7.6.5.3.8.85. 15. 143.75 16.5 181.5. A: Drill Speed.6.5.4 B: Feed Rate.3. 15. 143.75. 181.5 16.5 A: Drill Speed 5-16

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.1. 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 1 15 15 58 19 139 1 568 187 138 57 185 1386 55 17 138 53 135 131 579 1 199 546 145 867 3 575 153 94 599 166 889 (a) Use α =.5 in the analysis. Is there a significant interaction effect? Does glass type or temperature affect the response? What conclusions can you draw? 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.41E+6 8 3.15E+5 84.77 <.1 significant A 1.59E+5 7543.6 6.37 <.1 B 1.78E+6 1 1.78E+6 4869.13 <.1 B 1.96E+5 1 1.96E+5 51.39 <.1 AB.6E+5 1.131E+5 39.39 <.1 AB 64373.93 3186.96 88.6 <.1 Pure Error 6579.33 18 365.5 Cor Total.418E+6 6 The Model F-value of 84.77 implies the model is significant. There is only a.1% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 19.1 R-Squared.9973 Mean 94.19 Adj R-Squared.9961 C.V..3 Pred R-Squared.9939 PRESS 1483.5 Adeq Precision 75.466 From the analysis of variance, both factors, Glass Type (A) and Temperature (B) are significant, as well as the interaction (AB). B and AB interation terms are also significant. The interaction and pure quadratic tems can be clearly seen in the plot shown below. For glass types 1 and the temperature is fairly linear, for glass type 3, there is a quadratic effect. 5-17

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Light Output X = B: Temperature Y = A: Glass Ty pe A1 1 A A3 3 14.4 1184.3 Interaction Graph A: Glass Type Light Output 966.199 748.99 53 1. 11.5 15. 137.5 15. B: Tem perature (b) Fit an appropriate model relating light output to glass type and temperature. The model, both coded and uncoded are shown in the Design Expert output below. Final Equation in Terms of Coded Factors: Light Output = +159. +8.33 * A[1] -4. * A[] +314.44 * B -178. * B +9. * A[1]B +65.56 * A[]B +7. * A[1]B +76. * A[]B Final Equation in Terms of Actual Factors: Glass Type 1 Light Output = -3646. +59.46667 * Temperature -.178 * Temperature Glass Type Light Output = -3415. +56. * Temperature -.163 * Temperature Glass Type 3 Light Output = -7845.33333 +136.13333 * Temperature -.51947 * Temperature (c) Analyze the residuals from this experiment. Comment on the adequacy of the models you have considered. 5-18

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY The only concern from the residuals below is the inequality of variance observed in the residuals versus glass type plot shown below. Normal Plot of vs. Predicted 35 99 Normal % Probability 95 9 8 7 5 3 1 5 17.5-17.5 1-35 -35-17.5 17.5 35 553. 761.5 969.5 1177.75 1386. Residual Predicted vs. Glass Type vs. Temperature 35 35 17.5 17.5-17.5-17.5-35 -35 1 3 1 18 117 15 133 14 15 Glass Type Temperature 5.11. Consider the data in Problem 5.3. Fit an appropriate model to the response data. Use this model to provide guidance concerning operating conditions for the process. See the alternative analysis shown in Problem 5.3 part (c). 5.1. Use Tukey s test to determine which levels of the pressure factor are significantly different for the data in Problem 5.3. Because the AB interaction is not significant, the sum of squares for the interaction is included as lack of fit in the residual error sum of squares for a SSE of.3 and degrees of freedom of 13. The sample size is also assumed to be increased from to 6. 5-19

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY The three pressure averages, arranged in ascending order are and T y.3. = 9.18 y.1. = 9.37 y.. = 9.68 MS.18 = q ( 3,9) E = 3.95 =. n 6.5.5 Comparing the differences with T.5, we have y.. y.3. =.5 > T.5 =. y y =.3 > T =....1..5 y y =.18 < T =..1..3..5 Therefore, the difference in yield between a pressure of 15 and 3 psig is statistically significant as is the difference in yield between 15 and psig. However, the difference in yield between and 3 psig is not statistically significant. 5.13. 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 57 163 565 1 565 18 51 583 143 59 58 988 56 547 16 538 51 14 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 = µ + τ + β + ε. Both factors, furnace position (A) and temperature (B) are significant. Other than the residual representing standard order 14 being marginally low, the residual plots show nothing unusual. 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 3.175E+5 718.4 <.1 significant A 716.6 1 716.6 16..13 B 9.453E+5 4.77E+5 169.6 <.1 Residual 6188.78 14 44.6 Lack of Fit 818.11 49.6.91.471 not significant Pure Error 537.67 1 447.56 Cor Total 9.587E+5 17 i j ijk 5-

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY The Model F-value of 718.4 implies the model is significant. There is only a.1% 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. Normal Plot of vs. Predicted 6.5556 99 Normal % Probability 95 9 8 7 5 3 1 5 6.55556-13.4444-33.4444 1-53.4444-53.4444-33.4444-13.4444 6.55556 6.5556 53.56 656.15 788.75 91.35 153.94 Residual Predicted vs. Position vs. Temperature 6.5556 6.5556 6.55556 6.55556-13.4444-13.4444-33.4444-33.4444-53.4444-53.4444 1 1 3 Position Temperature 5-1

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.14. Derive the expected mean squares for a two-factor analysis of variance with one observation per cell, assuming that both factors are fixed. E E E ( MS ) A ( MS ) ( MS ) AB B = σ a i ( a 1 ) i= 1 Degrees of Freedom τ = σ + b a-1 b β = σ + a b-1 + j ( b 1 ) j= 1 a b ( τβ ) ij ( a 1 )( b 1 ) i= 1 j= 1 ( a 1)( b 1) ab 1 5.15. Consider the following data from a two-factor factorial experiment. Analyze the data and draw conclusions. Perform a test for nonadditivity. Use α =.5. Column Factor Row Factor 1 3 4 1 36 39 36 3 18 3 3 37 33 34 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 11.88 5.36.6 significant A 58.5 9.5 6.4.1 B 8.9 3 9.64.1.147 Residual 8.83 6 4.81 Cor Total 638.5 11 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= 1 j= 1 4114 = = 3. 5451 = SS Error a b yij yi. y. j y.. SS A + SS abss SS Re sidual ( 357) 357 58. 5 + 8. 91667 + ( 4)( 3)( 58. 5)( 8. 91667) SS N A B B y.. + ab ( 4)( 3) = 8. 8333 3. 5451 = 5. 979 Source of Sum of Degrees of Mean 5-

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Variation Squares Freedom Square F Row 58.5 9.5 57.378 Column 8.91667 3 9.63889 1.954 Nonadditivity 3.5451 1 3.5451.6999 Error 5.979 5 5.58558 Total 638.5 11 5.16. 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) Pressure (lb/in) 5 6 7 1 9.6 11.8 9. 13 9.69 1.1 9.57 14 8.43 11.1 9.3 15 9.98 1.44 9.8 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 1.5.9.114 not significant A.58 3.19.54.677 B 4.66.33 6.49.316 Residual.15 6.36 Cor Total 7.39 11 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a 11.4 % 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. SS SS N N = a b i= 1 j= 1 117.93 13789.5797 ( 117.93).586917 + 4.65765 + = Error Re sidual N y.. yij yi. y. j y.. SS A + SSB + ab abss SS A ( 4)( 3)(.586917)( 4.65765) SSN =.48948 SS = SS SS =.1538833.48948 = 1. 6644 B ( 4)( 3) 5-3

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Source of Sum of Degrees Mean of Variation Squares Freedom Square F Row.586917 3.1935639.5815 Column 4.65765.3885 6.996 Nonadditivit.48948 1.48948 1.474 y Error 1.6644 5.3388 Total 7.395 11 5.17. Consider the three-factor model yijk i j k ( τβ ) + ( βγ ) εijk = µ + τ + β + γ + + ij jk i = 1,,...,a j = 1,,...,b k = 1,,...,c 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-1 a τ i σ + bc a 1 B b-1 C c-1 AB BC (a-1)(b-1) (b-1)(c-1) σ + ac σ + c σ + ab σ + a a ( ) i= 1 b j ( b 1 ) b i= j= j= 1 c β γ k = ( c 1 k 1 ) ( τβ ) ij ( a )( b ) 1 1 1 1 b c ( βγ ) jk ( b 1 )( c 1 ) j= 1 k = 1 Error (AC + ABC) b(a-1)(c-1) σ Total abc-1 5-4

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.18. 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 65 196.6 197.7 199.8 198.4 199.6.6 196. 196. 199.4 198.6.4.9 4 198.5 196. 198.4 197.5 198.7 199.6 197. 196.9 197.6 198.1 198. 199. 8 197.5 195.6 197.4 197.6 197. 198.5 196.6 196. 198.1 198.4 197.8 199.8 Percentage Block Time 3. Time 4. of Hardwood Cooking Pressure Hours Cooking Pressure Hours Concentration 4 5 65 4 5 65 1 196.6 197.7 199.8 198.4 199.6.6 196. 196. 199.4 198.6.4.9 4 1 198.5 196. 198.4 197.5 198.7 199.6 197. 196.9 197.6 198.1 198. 199. 8 1 197.5 195.6 197.4 197.6 197. 198.5 196.6 196. 198.1 198.4 197.8 199.8 (a) Analyze the data and draw conclusions. Use α =.5. 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 17 3.51 9.61 <.1 significant A 7.76 3.88 1.6.9 B.5 1.5 55.4 <.1 C 19.37 9.69 6.5 <.1 AB.8 1.4.85.843 AC 6.9 4 1.5 4.17.146 BC.19 1.1 3..75 ABC 1.97 4.49 1.35.93 Residual 6.58 18.37 Lack of Fit. Pure Error 6.58 18.37 Cor Total 66.31 35 The Model F-value of 9.61 implies the model is significant. There is only a.1% 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 1% level. 5-5

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (b) Prepare appropriate residual plots and comment on the model s adequacy. vs. Pressure vs. Cooking Time.85.85.45.45 -.45 -.45 -.85 -.85 1 3 1 Pressure Cooking Time vs. Predicted vs. Hardwood.85.85.45.45 -.45 -.45 -.85 -.85 195.9 197.11 198.33 199.54.75 1 3 Predicted Hardwood There is nothing unusual about the residual plots. 5-6

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (c) Under what set of conditions would you run the process? Why? DESIGN-EXPERT Plot strength.9 Interaction Graph P res s ure DESIGN-EXPERT Plot strength.9 Interaction Graph Hardwood X = B: Cooking Time Y = C: Pressure 199.575 C1 4 C 5 C3 65 Actual Factor A: Hardwood = Average 198.5 strength X = B: Cooking Time Y = A: Hardwood A1 A 4 A3 8 Actual Factor C: Pressure = Average strength 199.575 198.5 196.95 196.95 195.6 195.6 3 4 3 4 Cooking Tim e Cooking Tim e DESIGN-EXPERT Plot strength.9 Interaction Graph Hardwood X = C: Pressure Y = A: Hardwood 199.575 A1 A 4 A3 8 Actual Factor B: Cooking Time = Average 198.5 strength 196.95 195.6 4 5 65 P res s ure 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 a linear term for the two-level factor and linear and quadratic components for the three-level factors. The Minitab output, with the ABC interaction removed due to insignificance, now follows. Also included is the Design Expert output; however, if the student choses to use Design Expert, sequential sum of squares must be selected to assure that the sum of squares for the model equals the total of the sum of squares for each factor included in the model. 5-7

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Minitab Output General Linear Model: Strength versus Factor Type Levels Values Analysis of Variance for Strength, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Hardwood 1 6.967 4.999 4.999 13.3.1 Time 1.5 1.3198 1.3198 3.49.74 Pressure 1 15.565 1.514 1.514 3.97.58 Hardwood*Hardwood 1.8571.7951.7951 7.4.1 Pressure*Pressure 1 3.8134 1.83 1.83 4.83.38 Hardwood*Time 1.7779 1.5779 1.5779 4.18.53 Hardwood*Pressure 1.1179 3.4564 3.4564 9.15.6 Time*Pressure 1.19.193.193 5.81.4 Hardwood*Hardwood*Time 1 1.338 1.338 1.338 3.45.76 Hardwood*Hardwood* Pressure 1.1885.1885.1885 5.79.5 Hardwood*Pressure* Pressure 1 1.6489 1.6489 1.6489 4.36.48 Time*Pressure*Pressure 1.176.176.176 5.76.5 Error 3 8.6891 8.6891.3778 Total 35 66.389 Term Coef SE Coef T P Constant 36.9 9.38 8.6. Hardwood 1.78.949 3.64.1 Time -14.961 8.4-1.87.74 Pressure -.57.113-1.99.58 Hardwood*Hardwood -.659.4 -.7.1 Pressure*Pressure.34.17..38 Hardwood*Time -1.175.5749 -.4.53 Hardwood*Pressure -.533.6788-3..6 Time*Pressure.745.39.41.4 Hardwood*Hardwood*Time.178.553 1.86.76 Hardwood*Hardwood*Pressure.648.69.41.5 Hardwood*Pressure*Pressure.1.6.9.48 Time*Pressure*Pressure -.7.9 -.4.5 Unusual Observations for Strength Obs Strength Fit SE Fit Residual St Resid 6 198.5 197.461.364 1.39.1R R denotes an observation with a large standardized residual. 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 57.6 1 4.8 1.71 <.1 significant A 6.91 1 6.91 18.8.3 B.5 1.5 53.6 <.1 C 15.56 1 15.56 41.19 <.1 A.86 1.86.7.1456 C 3.81 1 3.81 1.9.4 AB.78 1.78.6.1648 AC.1 1.1 5.61.67 BC.19 1.19.5.845 AB 1.3 1 1.3 3.45.761 AC.19 1.19 5.79.45 AC 1.65 1 1.65 4.36.479 BC.18 1.18 5.76.49 Residual 8.69 3.38 Lack of Fit.11 5.4 1.15.3691 not significant Pure Error 6.58 18.37 Cor Total 66.31 35 5-8

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY The Model F-value of 1.71 implies the model is significant. There is only a.1% 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.1 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..61 R-Squared.869 Mean 198.6 Adj R-Squared.86 C.V..31 Pred R-Squared.6794 PRESS 1.6 Adeq Precision 15.4 Coefficient Standard 95% CI 95% CI Factor Estimate DF Error Low High VIF Intercept 197.1 1.6 196.67 197.74 A-Hardwood -.98 1.3-1.45 -.5 3.36 B-Cooking Time.78 1.6.4 1.31 6.35 C-Pressure.19 1.5 -.33.71 4.4 A.4 1.5 -.93.94 1.4 C.79 1.3.31 1.6 1.3 AB -. 1.13 -.48.39 1.6 AC -.46 1.15 -.78 -.14 1.8 BC.6 1.13 -..3 1. AB.46 1.5 -.53.98 3.96 AC.73 1.3.1 1.36 3.97 AC.57 1.7 5.65E-3 1.14 3.3 BC -.55 1.3-1. -.75 3.3 Final Equation in Terms of Coded Factors: Strength = +197.1 -.98 * A +.78 * B +.19 * C +.4 * A +.79 * C -. * A * B -.46 * A * C +.6 * B * C +.46 * A * B +.73 * A * C +.57 * A * C -.55 * B * C Final Equation in Terms of Actual Factors: Strength = +36.9176 +1.7773 * Hardwood -14.96111 * Cooking Time -.569 * Pressure -.6587 * Hardwood +.34333E-4 * Pressure -1.175 * Hardwood * Cooking Time -.533 * Hardwood * Pressure +.745 * Cooking Time * Pressure +.178 * Hardwood * Cooking Time +6.486E-4 * Hardwood * Pressure +1.143E-5 * Hardwood * Pressure -7.E-5 * Cooking Time * Pressure 5-9

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY Strength 65. 198.5 C: Pressure.5 6. 198 199.5 55. 199 198.5 5. Strength 1.5 1.5 199.5 199 198.5 198 197.5 197 45. 65. 6. 4.. 3.5 5. 6.5 8. A: Hardwood. 3.5 A: Hardwood 5. 6.5 55. 5. C: Pressure 45. 8. 4. Cooking Time: B = 4. 5.19. 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 C 35 C Operator Operator Cycle Time 1 3 1 3 3 7 31 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 31 8 35 6 6 36 8 6 4 35 7 9 37 6 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. 5-3

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 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 139.33 17 7.9.4 <.1 significant A 436. 18. 66.51 <.1 B 61.33 13.67 39.86 <.1 C 5.7 1 5.7 15.8.4 AB 355.67 4 88.9 7.13 <.1 AC 78.81 39.41 1..1 BC 11.6 5.63 1.7.1939 ABC 46.19 4 11.55 3.5.159 Residual 118. 36 3.8 Lack of Fit. Pure Error 118. 36 3.8 Cor Total 1357.33 53 The Model F-value of.4 implies the model is significant. There is only a.1% 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 39 Interaction Graph Tem perature DESIGN-EXPERT Plot Score 39 Interaction Graph Operator X = A: Cycle Time Y = C: Temperature C1 3 C 35 Actual Factor B: Operator = Average S core 35 31 X = A: Cycle Time Y = B: Operator B1 1 B B3 3 Actual Factor C: Temperature = Average S core 35 31 7 7 3 3 4 5 6 4 5 6 C ycle Tim e C ycle Tim e 5-31

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY vs. Operator vs. Cycle Time 3 3 1.5 1.5-4.636E-14 3-4.636E-14 3 3-1.5-1.5-3 -3 1 3 1 3 Operator Cycle Time vs. Predicted vs. Temperature 3 3 1.5 1.5-4.636E-14-4.636E-14 4 3-1.5-1.5-3 -3 4. 7.5 3.5 33.75 37. 1 Predicted Temperature 5.. In Problem 5.3, 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.1, how many replicates should be run? nad Φ = bσ n = ( 3)(. 5) ( 3)(. 1) = 1. 5n n Φ Φ υ 1 = ( b 1) υ = ab( n 1) β 5 5 (3)(3)(1).14 replications will be enough to detect the given difference. 5-3

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.1. 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 1 Day Pressure Pressure Temperature 5 6 7 5 6 7 Low 86.3 84. 85.8 86.1 85. 87.3 Medium 88.5 87.3 89. 89.4 89.9 9.3 High 89.1 9. 91.3 91.7 93. 93.7 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 13.1 1 13.1 Model 19.81 8 13.73 5.84 <.1 significant A 5.51.75 5.18.36 B 99.85 49.93 93.98 <.1 AB 4.45 4 1.11.1.1733 Residual 4.5 8.53 Cor Total 17.7 17 The Model F-value of 5.84 implies the model is significant. There is only a.1% 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 5.7. Analyze the data, assuming that replicates are blocks. 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 11.8 1 11.8 Model 968. 15 64.55 9.96 <.1 significant A 698.34 3 3.78 35.9 <.1 B 156.9 3 5.3 8.3. AB 113.78 9 1.64 1.95.114 Residual 97. 15 6.48 Cor Total 176.7 31 The Model F-value of 9.96 implies the model is significant. There is only a.1% 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 5.7. 5-33

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 5.3. Consider the data in Problem 5.8. Analyze the data, assuming that replicates are blocks. 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 1.4 Model 17.46 11 19.77 5..64 significant A 16.33 8.17.9. B 1.46 3 4.15 1.5.487 AB 44.67 6 7.44 1.88.1716 Residual 43.46 11 3.95 Cor Total 6.96 3 The Model F-value of 5. implies the model is significant. There is only a.64% 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 5.8. 5.4. An article in the Journal of Testing and Evaluation (Vol. 16, no., pp. 58-515) 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 1.9 1.47.5 1.93.48.3 1.75.1.3.6.65 3. 3.1 1.68 3.18 3.4.6 3.96 3.98.38 3.64 3.4.4 11. 9.96.1.71 11. 1.1.81 9.6 9.36.8 11.3 1.4 5-34

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (a) Analyze the data from this experiment (use α =.5). 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 376.11 8 47.1 34. <.1 significant A 9.89 14.95 5.4 <.1 B 64.5 3.13 159.9 <.1 AB 11.97 4 5.49 16.89 <.1 Residual 5.4 7. Lack of Fit. Pure Error 5.4 7. Cor Total 381.53 35 The Model F-value of 34. implies the model is significant. There is only a.1% 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. vs. Predicted Normal plot of residuals.71 99.15 -.41 -.97 Normal % probability 95 9 8 7 5 3 1 5 1-1.53 1.91 4.8 6.5 8.4 1.59-1.53 -.97 -.41.15.71 Predicted Residual 5-35

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY vs. Environment vs. Frequency.71.71.15.15 -.41 -.41 -.97 -.97-1.53-1.53 1 3 1 3 Environment Frequency (c) Repeat the analyses from parts (a) and (b) using ln(y) as the response. Comment on the results. 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 13.46 8 1.68 179.57 <.1 significant A 7.57 3.79 44.9 <.1 B.36 1.18 15.85 <.1 AB 3.53 4.88 94.17 <.1 Residual.5 7 9.367E-3 Lack of Fit. Pure Error.5 7 9.367E-3 Cor Total 13.71 35 The Model F-value of 179.57 implies the model is significant. There is only a.1% 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 based on the transformed data look better. 5-36

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY.16534 vs. Predicted Normal plot of residuals 99.8783.414 -.8988 Normal % probability 95 9 8 7 5 3 1 5 1 -.16484.65 1.7 1.5 1.93.36 -.16484 -.8988.414.8783.16534 Predicted Residual vs. Environment vs. Frequency.16534.16534.8783.8783.414.414 -.8988 -.8988 -.16484 -.16484 1 3 1 3 Environment Frequency 5.5. An article in the IEEE Transactions on Electron Devices (Nov. 1986, pp. 1754) 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 1 1 x 1 4.6 1.15 11.1 4.4 1. 1.58 x 1 3. 9.38 1.81 3.5 1. 1.6 (a) Is there evidence (with α =.5) indicating that either polysilicon doping level or anneal temperature affect base current? 5-37

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY 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 11.74 5.55 35.91 <.1 significant A.98 1.98 15.6.79 B 111.19 55.59 865.16 <.1 AB.58.9 4.48.645 Residual.39 6.64 Lack of Fit. Pure Error.39 6.64 Cor Total 113.13 11 The Model F-value of 35.91 implies the model is significant. There is only a.1% 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 1% level. (b) Prepare graphical displays to assist in interpretation of this experiment. 11.151 Interaction Graph Doping 9.888 Base Current 7.75 5.5618 A- A+ 3.3986 9 95 1 Anneal 5-38

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (c) Analyze the residuals and comment on model adequacy. vs. Predicted Normal plot of residuals.3 99.16-8.88178E-16 -.16 Normal % probability 95 9 8 7 5 3 1 5 1 -.3 3.35 5.1 7.7 8.93 1.8 -.3 -.16-8.88178E-16.16.3 Predicted Residual vs. Doping vs. Anneal.3.3.16.16-8.88178E-16-8.88178E-16 -.16 -.16 -.3 -.3 1 1 3 Doping Anneal There is a funnel shape in the plot of residuals versus predicted, indicating some inequality of variance. 5-39

Solutions from Montgomery, D. C. (1) Design and Analysis of Experiments, Wiley, NY (d) Is the model y = β + β1x1 + βx + βx + β1x1x + ε supported by this experiment (x 1 = doping level, x = temperature)? Estimate the parameters in this model and plot the response surface. 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 11.73 4 8.18 493.73 <.1 significant A.98 1.98 17.18.43 B 93.16 1 93.16 163.9 <.1 B 18.3 1 18.3 315.81 <.1 AB.56 1.56 9.84.164 Residual.4 7.57 Lack of Fit.14 1.14..6569 not significant Pure Error.39 6.64 Cor Total 113.13 11 The Model F-value of 493.73 implies the model is significant. There is only a.1% 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 1.1 9.66 1. A-Doping -.9 1.69 -.45 -.1 1. B-Anneal 3.41 1.84 3.1 3.61 1. B -.6 1.15 -.95 -.5 1. AB.7 1.84.65.46 1. All of the coefficients in the assumed model are significant. The quadratic effect is easily observable in the response surface plot. Base Current 1. 11 Anneal 975. 1 95. 9 8 95. 7 6 5 4 9. 1.E+ 1.5E+ 1.5E+ 1.75E+.E+ Doping Base Current 1 11 1 9 8 7 6 5 4 3 1. 1.E+ 1.5E+ 975. 95. 1.5E+ 1.75E+ 95. Anneal Doping.E+ 9. 5-4