Chapter 6 The 2 k Factorial Design Solutions

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Chapter 6 The k Factorial Design Solutions 6.. An engineer is interested in the effects of cutting speed (A), tool geometry (B), and cutting angle on the life (in hours) of a machine tool. Two levels of each factor are chosen, and three replicates of a 3 factorial design are run. The results are as follows: Treatment Replicate A B C Combination I II III - - - () 3 5 + - - a 3 43 9 - + - b 35 34 5 + + - ab 55 47 46 - - + c 44 45 38 + - + ac 4 37 36 - + + bc 6 5 54 + + + abc 39 4 47 (a) Estimate the factor effects. Which effects appear to be large? From the normal probability plot of effects below, factors B, C, and the AC interaction appear to be significant. DESIGN-EXPERT Plot Life Normal plot A: Cutting Speed B: Tool Geometry C: Cutting Angle Norm al % probability 9 8 7 5 3 5 A C B AC -8.8 3-3.7 9. 5 6. 9.33 E ffe ct (b) Use the analysis of variance to confirm your conclusions for part (a). The analysis of variance confirms the significance of factors B, C, and the AC interaction. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 6.67 7 3.38 7.64.4 significant A.67.67..8837 B 77.67 77.67 5.55. C 8.7 8.7 9.9.77 AB 6.67 6.67.55.468 AC 468.7 468.7 5.5. BC 48.7 48.7.6.45 ABC 8.7 8.7.93.3483 Pure Error 48.67 6 3.7 Cor Total.33 3 The Model F-value of 7.64 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. The reduced model ANOVA is shown below. Factor A was included to maintain hierarchy. Response: Life in hours ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 59.67 4 379.9.54 <. significant A.67.67..8836 B 77.67 77.67 5.44 <. C 8.7 8.7 9.5.67 AC 468.7 468.7 5.45.9 Residual 575.67 9 3.3 Lack of Fit 93. 3 3..3.467 not significant Pure Error 48.67 6 3.7 Cor Total.33 3 The Model F-value of.54 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Effects B, C and AC are significant at %. (c) Write down a regression model for predicting tool life (in hours) based on the results of this experiment. y = 4. 8333 +. 667x + 5. 6667x + 3. 467x + 4. 467x ijk A B C A x C Coefficient Standard % CI % CI Factor Estimate DF Error Low High VIF Intercept 4.83. 38.48 43.9 A-Cutting Speed.7. -.9.5. B-Tool Geometry 5.67. 3.3 8.. C-Cutting Angle 3.4..6 5.77. AC -4.4. -6.77 -.6. Final Equation in Terms of Coded Factors: Life = +4.83 +.7 * A +5.67 * B +3.4 * C -4.4 * A * C Final Equation in Terms of Actual Factors: 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Life = +4.83333 +.6667 * Cutting Speed +5.66667 * Tool Geometry +3.4667 * Cutting Angle -4.4667 * Cutting Speed * Cutting Angle The equation in part (c) and in the given in the computer output form a hierarchial model, that is, if an interaction is included in the model, then all of the main effects referenced in the interaction are also included in the model. (d) Analyze the residuals. Are there any obvious problems? Normal plot of residuals vs. Predicted.5 Normal % probability 9 8 7 5 3 5 6.7967.8333 -.65-7.33333-7.33333 -.65.8333 6.7967.5 7.7 33.9 4.67 47.4 54.7 Residual Predicted There is nothing unusual about the residual plots. (e) Based on the analysis of main effects and interaction plots, what levels of A, B, and C would you recommend using? Since B has a positive effect, set B at the high level to increase life. The AC interaction plot reveals that life would be maximized with C at the high level and A at the low level. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Life 6 Interaction Graph Cutting Angle DESIGN-EXPERT Plot Life 6 One Factor Plot X = A: Cutting Speed Y = C: Cutting Angle C- -. C+. Actual Factor B: Tool Geometry =. L ife 5.5 4 X = B: Tool Geometry Actual Factors A: Cutting Speed =. C: Cutting Angle =. L ife 5.5 4 3.5 3.5 -. -.5..5. -. -.5..5. Cutting Speed Tool Geom etry 6.. Reconsider part (c) of Problem 6.. Use the regression model to generate response surface and contour plots of the tool life response. Interpret these plots. Do they provide insight regarding the desirable operating conditions for this process? The response surface plot and the contour plot in terms of factors A and C with B at the high level are shown below. They show the curvature due to the AC interaction. These plots make it easy to see the region of greatest tool life. Life.3 3 5 46 5 C: Cutting Angle.5. 48 46 Life 56 54 5 5 48 46 44 4 4 38 -.5 44 4 4 -.3 3 -. -.5..5. A: Cutting Speed..5. C: Cutting Angle -.5 -. -. -.5..5. A: Cutting Speed 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.3. Find the standard error of the factor effects and approximate percent confidence limits for the factor effects in Problem 6.. Do the results of this analysis agree with the conclusions from the analysis of variance? SE( effect ) = S = 3. 7 = k 3 n ( 3) Variable Effect A.333 B.333 * AB -.667 C 6.833 * AC -8.833 * BC -.833 ABC -.67 The % confidence intervals for factors B, C and AC do not contain zero. This agrees with the analysis of variance approach.. 4 6.4. Plot the factor effects from Problem 6. on a graph relative to an appropriately scaled t distribution. Does this graphical display adequately identify the important factors? Compare the conclusions from this MSE 3. 7 plot with the results from the analysis of variance. S = = = 3. 7 n 3 Scaled t Distribution AC C B -... Factor Effects This method identifies the same factors as the analysis of variance. 6.5. A router is used to cut locating notches on a printed circuit board. The vibration level at the surface of the board as it is cut is considered to be a major source of dimensional variation in the notches. Two factors are thought to influence vibration: bit size (A) and cutting speed (B). Two bit sizes (/6 and /8 inch) and two speeds (4 and 9 rpm) are selected, and four boards are cut at each set of conditions shown below. The response variable is vibration measured as a resultant vector of three accelerometers (x, y, and z) on each test circuit board. 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Treatment Replicate A B Combination I II III IV - - () 8. 8.9.9 4.4 + - a 7. 4..4.5 - + b 5.9 4.5 5. 4. + + ab 4. 43.9 36.3 39.9 (a) Analyze the data from this experiment. Response: Vibration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 638. 3 546.4 9.36 <. significant A 7.3 7.3 85.5 <. B 7.6 7.6 38. <. AB 33.63 33.63 5.8 <. Residual 7.7 5.98 Lack of Fit. Pure Error 7.7 5.98 Cor Total 79.83 5 The Model F-value of 9.36 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. (b) Construct a normal probability plot of the residuals, and plot the residuals versus the predicted vibration level. Interpret these plots. Normal plot of residuals 3.65 vs. Predicted Normal % probability 9 8 7 5 3 5.75 -.75 -.75-3.975-3.975 -.75 -.75.75 3.65 4.9.6 7.6 33.94 4.7 Residual Predicted There is nothing unusual about the residual plots. (c) Draw the AB interaction plot. Interpret this plot. What levels of bit size and speed would you recommend for routine operation? To reduce the vibration, use the smaller bit. Once the small bit is specified, either speed will work equally well, because the slope of the curve relating vibration to speed for the small tip is approximately zero. The process is robust to speed changes if the small bit is used. 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Vibration 4 3.9 Interaction Graph Cutting Speed X = A: Bit Size Y = B: Cutting Speed Design Points 36.5 B- -. B+. Vibration 8.4.65.9 -. -.5..5. B it S ize 6.6. Reconsider the experiment described in Problem 6.. Suppose that the experimenter only performed the eight trials from replicate I. In addition, he ran four center points and obtained the following response values: 36, 4, 43, 45. (a) Estimate the factor effects. Which effects are large? DESIGN-EXPERT Plot Life Normal plot A: Cutting Speed B: Tool Geometry C: Cutting Angle Norm al % probability 9 8 7 5 3 5 AC C B -3.75-7. 3 -.5 6..75 E ffe ct Effects B, C, and AC appear to be large. (b) Perform an analysis of variance, including a check for pure quadratic curvature. What are your conclusions? SS ( y y ) ( 8)( 4)( 4. 875 4. ) nf nc F C PureQuadra tic = = = nf + nc 8 + 4. 47 6-7

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

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Plot of vs. Predicted 6.83333 Normal % Probability 9 8 7 5 3 5 3.383 -.46667-3.4797-6.9667-6.9667-3.4797 -.46667 3.383 6.83333.7 3.3 4.9 49.35 58.4 Residual Predicted (e) What conclusions would you draw about the appropriate operating conditions for this process? To maximize life run with B at the high level, A at the low level and C at the high level Cube Graph Life 58.38 45.88 B+ 34.88 49.88 B: Tool Geometry 45.63 33.3 C+ C: Cutting Angle B-.3 37.3 C- A- A+ A: Cutting Speed 6.7. An experiment was performed to improve the yield of a chemical process. Four factors were selected, and two replicates of a completely randomized experiment were run. The results are shown in the following table: Treatment Replicate Replicate Treatment Replicate Replicate Combination I II Combination I II () 9 93 d 98 a 74 78 ad 7 76 b 8 85 bd 87 83 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ab 83 8 abd 85 86 c 77 78 cd 9 ac 8 8 acd 79 75 bc 88 8 bcd 87 84 abc 73 7 abcd 8 8 (a) Estimate the factor effects. Term Effect SumSqr % Contribtn Model Intercept Error A -9.65 657.3 4.374 Error B -.35 3.78.84679 Error C -.6875 57.783 3.5538 Error D 3.9375 4.3 7.6 Error AB 4.65 3.3 8.67 Error AC.6875 3.785.3339 Error AD -.875 38.83.35 Error BC -.565.535.55533 Error BD -.875.85.784 Error CD.6875.78.38 Error ABC -5.875 5.8 3.8 Error ABD 4.6875 75.78.89 Error ACD -.9375 7.35.4336 Error BCD -.9375 7.35.4336 Error ABCD.4375 47.533.6 (b) Prepare an analysis of variance table, and determine which factors are important in explaining yield. Response: yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 54.97 5.33 3. <. significant A 657.3 657.3 85.8 <. B 3.78 3.78.8.984 C 57.78 57.78 7.55.43 D 4.3 4.3 6.. AB 3.3 3.3 7.4.7 AC 3.78 3.78.49.493 AD 38.8 38.8 5..3 BC.53.53.33.5733 BD.8.8.37.854 CD.78.78.98.38 ABC 5.8 5.8 8. <. ABD 75.78 75.78.96. ACD 7.3 7.3.9.35 BCD 7.3 7.3.9.35 ABCD 47.53 47.53 6..4 Residual.5 6 7.66 Lack of Fit. Pure Error.5 6 7.66 Cor Total 67.47 3 The Model F-value of 3. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, C, D, AB, AD, ABC, ABD, ABCD are significant model terms. F.,, 6 = 8. 53, and F. 5,, 6 = 6. therefore, factors A and D and interactions AB, ABC, and ABD are significant at %. Factor C and interactions AD and ABCD are significant at 5%. 6-

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

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals 6.96875 vs. Predicted Normal % probability 9 8 7 5 3 5 3.96875.96875 -.35-5.35-5.35 -.35.96875 3.96875 6.96875 7.9 78.3 84.69 9.8 97.47 Residual Predicted (e) Two three-factor interactions, ABC and ABD, apparently have large effects. Draw a cube plot in the factors A, B, and C with the average yields shown at each corner. Repeat using the factors A, B, and D. Do these two plots aid in data interpretation? Where would you recommend that the process be run with respect to the four variables? Cube Graph yield Cube Graph yield 86.53 76.34 86. 83.5 B+ 84.3 84. B+ 84.56 77.6 B: B 85.4 77.47 C+ B: B 94.75 74.75 D+ C: C D: D B- 93.8 74.97 C- A- A+ A: A B- 83.94 77.69 D- A- A+ A: A Run the process at A low B low, C low and D high. 6.8. A bacteriologist is interested in the effects of two different culture media and two different times on the growth of a particular virus. She performs six replicates of a design, making the runs in random order. Analyze the bacterial growth data that follow and draw appropriate conclusions. Analyze the residuals and comment on the model s adequacy. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Culture Medium Time M di 5 6 hr 3 8 4 5 6 9 7 37 39 3 34 8 hr 38 38 9 33 35 36 3 35 Response: Virus growth ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 69.46 3 3.49 45. <. significant A 9.38 9.38.84.96 B 59.4 59.4 5.5 <. AB 9.4 9.4 8..4 Residual.7 5. Lack of Fit. Pure Error.7 5. Cor Total 793.63 3 The Model F-value of 45. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B, AB are significant model terms. Normal plot of residuals 4.66667 vs. Predicted Normal % probability 9 8 7 5 3 5.66667.666667 -.33333-3.33333-3.33333 -.33333.666667.66667 4.66667 3.33 6.79 3.5 33.7 37.7 Residual Predicted Growth rate is affected by factor B (Time) and the AB interaction (Culture medium and Time). There is some very slight indication of inequality of variance shown by the small decreasing funnel shape in the plot of residuals versus predicted. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Virus growth X = A: Culture Medium Y = B: Time Design Points B-. B+ 8. Virus growth 39 34.5 9.5 Interaction Graph Tim e 4.75 Culture Medium 6.9. An industrial engineer employed by a beverage bottler is interested in the effects of two different types of 3-ounce bottles on the time to deliver -bottle cases of the product. The two bottle types are glass and plastic. Two workers are used to perform a task consisting of moving 4 cases of the product 5 feet on a standard type of hand truck and stacking the cases in a display. Four replicates of a factorial design are performed, and the times observed are listed in the following table. Analyze the data and draw the appropriate conclusions. Analyze the residuals and comment on the model s adequacy. Worker Bottle Type Glass 5. 4.89 6.65 6.4 4.98 5. 5.49 5.55 Plastic 4. 4.43 5.8 4.9 4.7 4.5 4.75 4.7 Response: Times ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4.86 3.6 3.4.4 significant A.. 6.8.7 B.54.54.4.7 AB.3.3.4.463 Residual.49. Lack of Fit. Pure Error.49. Cor Total 6.35 5 The Model F-value of 3.4 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B are significant model terms. There is some indication of non-constant variance in this experiment. 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals vs. Predicted.6675 Normal % probability 9 8 7 5 3 5.3775.875 -.5 -.4 -.4 -.5.875.3775.6675 4.47 4.85 5.3 5.6 5.98 Residual Predicted vs. Worker vs. Bottle Type.6675 3.3775.5.875 -.5 -.5 -.4-3 Worker Bottle Type 6.. In problem 6.9, the engineer was also interested in potential fatigue differences resulting from the two types of bottles. As a measure of the amount of effort required, he measured the elevation of heart rate (pulse) induced by the task. The results follow. Analyze the data and draw conclusions. Analyze the residuals and comment on the model s adequacy. Worker Bottle Type Glass 39 45 3 58 35 6 Plastic 44 35 3 4 6 5 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Pulse ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 784.9 3 98.6 6.3. significant A 66.56 66.56 45.37 <. B 5.6 5.6.8.8 AB 5.56 5.56.9.35 Residual 694.75 57.9 Lack of Fit. Pure Error 694.75 57.9 Cor Total 3478.94 5 The Model F-value of 6.3 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A are significant model terms. Normal Plot of vs. Predicted 3.75 Normal % Probability 9 8 7 5 3 5 6.6875 -.375-7.4375-4.5-4.5-7.4375 -.375 6.6875 3.75 3.5.9 8.88 36.56 44.5 Residual Predicted vs. Worker vs. Bottle Type 3.75 3.75 6.6875 6.6875 -.375 -.375-7.4375-7.4375-4.5-4.5 Worker Bottle Type There is an indication that one worker exhibits greater variability than the other. 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.. Calculate approximate percent confidence limits for the factor effects in Problem 6.. Do the results of this analysis agree with the analysis of variance performed in Problem 6.? SE k n ( effect ) = S = 57. 9 = ( 4) Variable Effect C.I. A -5.65 ±3.8(.96)= ±7.448 B -5.5 ±3.8(.96)= ±7.448 AB 3.65 ±3.8(.96)= ±7.448 3. 8 The % confidence interval for factor A does not contain zero. This agrees with the analysis of variance approach. 6.. An article in the AT&T Technical Journal (March/April 986, Vol. 65, pp. 39-5) describes the application of two-level factorial designs to integrated circuit manufacturing. A basic processing step is to grow an epitaxial layer on polished silicon wafers. The wafers mounted on a susceptor are positioned inside a bell jar, and chemical vapors are introduced. The susceptor is rotated and heat is applied until the epitaxial layer is thick enough. An experiment was run using two factors: arsenic flow rate (A) and deposition time (B). Four replicates were run, and the epitaxial layer thickness was measured (in mm). The data are shown below: Replicate Factor Levels A B I II III IV Low (-) High (+) - - 4.37 6.65 3.97 3.97 A 55% 59% + - 3.88 3.86 4.3 3.94 - + 4.8 4.757 4.843 4.878 B Short Long + + 4.888 4.9 4.45 4.93 ( min) (5 min) (a) Estimate the factor effects. Term Effect SumSqr % Contribtn Model Intercept Error A -.375.459 6.79865 Error B.586.37358 3.96 Error AB.85.36969 5.3574 Error Lack Of Fit Error Pure Error 3.8848 64.655 (b) Conduct an analysis of variance. Which factors are important? From the analysis of variance shown below, no factors appear to be important. Factor B is only marginally interesting with an F-value of 4.3. Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.9 3.7.9.45 not significant 6-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY A.4.4.6.833 B.37.37 4.3.6 AB.3.3..3386 Residual 3.83.3 Lack of Fit. Pure Error 3.83.3 Cor Total 5.9 5 The "Model F-value" of.9 implies the model is not significant relative to the noise. There is a 4.5 % chance that a "Model F-value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case there are no significant model terms. (c) Write down a regression equation that could be used to predict epitaxial layer thickness over the region of arsenic flow rate and deposition time used in this experiment. Final Equation in Terms of Coded Factors: Thickness = +4.5 -.6 * A +.9 * B +.4 * A * B Final Equation in Terms of Actual Factors: Thickness = +37.6656 -.439 * Flow Rate -.48735 * Dep Time +.85 * Flow Rate * Dep Time (d) Analyze the residuals. Are there any residuals that should cause concern? Observation # falls outside the groupings in the normal probability plot and the plot of residual versus predicted. Normal plot of residuals vs. Predicted.64475 Normal % probability 9 8 7 5 3 5.85.5575 -.4875 -.635 -.635 -.4875.5575.85.64475 3.9 4.5 4.37 4.6 4.8 Residual Predicted (e) Discuss how you might deal with the potential outlier found in part (d). One approach would be to replace the observation with the average of the observations from that experimental cell. Another approach would be to identify if there was a recording issue in the original data. 6-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY The first analysis below replaces the data point with the average of the other three. The second analysis assumes that the reading was incorrectly recorded and should have been 4.65. The analysis with the run associated with standard order replaced with the average of the remaining three runs in the cell, 3.97, is shown below. Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.97 3. 53.57 <. significant A 7.439E-3 7.439E-3.4.5375 B.96.96 6.9 <. AB.76E-4.76E-4..3 Pure Error..8 Cor Total 3.9 5 The Model F-value of 53.57 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. Final Equation in Terms of Coded Factors: Thickness = +4.38 -. * A +.43 * B +3.688E-3 * A * B Final Equation in Terms of Actual Factors: Thickness = +3.3665 -. * Flow Rate +.9 * Dep Time +7.375E-4 * Flow Rate * Dep Time Normal plot of residuals.43 vs. Predicted Normal % probability 9 8 7 5 3 5.375 -.55 -.4475 -.374 -.374 -.4475 -.55.375.43 3.9 4.5 4.37 4.6 4.8 Residual Predicted A new outlier is present and should be investigated. 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Analysis with the run associated with standard order replaced with the value 4.65: Response: Thickness ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.8 3.94 45.8 <. significant A.8.8.87.3693 B.8.8 34.47 <. AB 3.969E-3 3.969E-3.9.66 Pure Error.5. Cor Total 3.7 5 The Model F-value of 45.8 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case B are significant model terms. Final Equation in Terms of Coded Factors: Thickness = +4.39 -.34 * A +.4 * B +.6 * A * B Final Equation in Terms of Actual Factors: Thickness = +5.556 -.5688 * Flow Rate -.35 * Dep Time +3.5E-3 * Flow Rate * Dep Time Normal plot of residuals 3. vs. Predicted Normal % probability 9 8 7 5 3 5 Studentized.5. -.5-3. -3. -.96 -.9..6 3.9 4.5 4.37 4.6 4.8 Studentized Predicted Another outlier is present and should be investigated. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.3. Continuation of Problem 6.. Use the regression model in part (c) of Problem 6. to generate a response surface contour plot for epitaxial layer thickness. Suppose it is critically important to obtain layer thickness of 4.5 mm. What settings of arsenic flow rate and deposition time would you recommend? Arsenic flow rate may be set at any of the experimental levels, while the deposition time should be set at.4 minutes. DESIGN-EXPERT Plot Thickness X = A: Flow Rate Y = B: Dep Time Design Points Thickness 4 4 5. 4.674 3.75 DESIGN-EXPERT Plot Thickness X = A: Flow Rate Y = B: Dep Time Design Points 6.65 5.43 Interaction Graph Dep Tim e Dep Tim e.5 4.537 B-. B+ 5. Thickness 4.857.5 4.373 4.6 4.56 4.7 4 4. 55. 56. 57. 58. 59. 3.4864 55. 56. 57. 58. 59. Flow Rate Flow Rate 6.4. Continuation of Problem 6.3. How would your answer to Problem 6.3 change if arsenic flow rate was more difficult to control in the process than the deposition time? Running the process at a high level of Deposition Time there is no change in thickness as flow rate changes. 6.5. A nickel-titanium alloy is used to make components for jet turbine aircraft engines. Cracking is a potentially serious problem in the final part, as it can lead to non-recoverable failure. A test is run at the parts producer to determine the effects of four factors on cracks. The four factors are pouring temperature (A), titanium content (B), heat treatment method (C), and the amount of grain refiner used (D). Two replicated of a 4 design are run, and the length of crack (in µm) induced in a sample coupon subjected to a standard test is measured. The data are shown below: Treatment Replicate Replicate A B C D Combination I II - - - - () 7.37 6.376 + - - - a 4.77 5.9 - + - - b.635.89 + + - - ab 7.73 7.85 - - + - c.43.5 + - + - ac 4.368 4.98 - + + - bc 9.36 9.53 + + + - abc 3.44.93 - - - + d 8.56 8. 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY + - - + ad 6.867 7.5 - + - + bd 3.876 3.658 + + - + abd 9.84 9.639 - - + + cd.846.337 + - + + acd 6.5 5.94 - + + + bcd.9.935 + + + + abcd 5.653 5.53 (a) Estimate the factor effects. Which factors appear to be large? From the half normal plot of effects shown below, factors A, B, C, D, AB, AC, and ABC appear to be large. Term Effect SumSqr % Contribtn Model Intercept Model A 3.888 7.989.748 Model B 3.97588 6.46. Model C -3.5965 3.464 8.84 Model D.775 3.663 5.3583 Model AB.934 9.967 5.969 Model AC -4.775 8.496.4548 Error AD.765.4688.8845 Error BC.96.7378.884 Error BD.475.7865.3 Error CD -.76875.4778.8685 Model ABC 3.375 78.75 3.768 Error ABD.98.7683.3464 Error ACD..963.534 Error BCD.3565.53.7746 Error ABCD.45.5963.7893 DESIGN-EXPERT Plot Crack Length Half Normal plot A: Pour Temp B: Titanium Content C: Heat Treat Method D: Grain Ref iner 97 AC Half Normal % probability 9 85 8 7 6 BC D AB C ABC A B 4... 3. 4. Effect 6-

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (b) Conduct an analysis of variance. Do any of the factors affect cracking? Use α=.5. The Design Expert output below identifies factors A, B, C, D, AB, AC, and ABC as significant. Response: Crack Lengthin mm x ^- ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 57. 5 38.6 468. <. significant A 7.9 7.9 898.34 <. B 6.46 6.46 558.7 <. C 3.46 3.46 74.8 <. D 3.66 3.66 377.8 <. AB 9.93 9.93 368.74 <. AC 8.5 8.5 583.6 <. AD.47.47.58.4586 BC.74.74.9.3547 BD.8.8..6453 CD.47.47.58.4564 ABC 78.75 78.75 97.33 <. ABD.77.77..345 ACD.96E-3.96E-3.36.858 BCD...3.78 ABCD.596E-3.596E-3..89 Residual.3 6.8 Lack of Fit. Pure Error.3 6.8 Cor Total 57.5 3 The Model F-value of 468. implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C, D, AB, AC, ABC are significant model terms. (c) Write down a regression model that can be used to predict crack length as a function of the significant main effects and interactions you have identified in part (b). Final Equation in Terms of Coded Factors: Crack Length= +. +.5 *A +. *B -.8 *C +.98 *D +.97 *A*B -. *A*C +.57 * A * B * C (d) Analyze the residuals from this experiment. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals vs. Predicted.454875 Normal % probability 9 8 7 5 3 5.3688.5 -.687 -.433875 -.433875 -.687.5.3688.454875 4.9 8.6.93 5.8 9.66 Residual Predicted There is nothing unusual about the residuals. (e) Is there an indication that any of the factors affect the variability in cracking? By calculating the range of the two readings in each cell, we can also evaluate the effects of the factors on variation. The following is the normal probability plot of effects: DESIGN-EXPERT Plot Range Normal plot A: Pour Temp B: Titanium Content C: Heat Treat Method D: Grain Refiner Norm al % probability 9 8 7 5 3 5 AB CD -. -.. 5. 3. E ffe ct It appears that the AB and CD interactions could be significant. The following is the ANOVA for the range data: 6-4

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Range ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model.9.4.46.4 significant AB.3.3 9.98.75 CD.6.6.94.3 Residual.6 3.3 Cor Total.45 5 The Model F-value of.46 implies the model is significant. There is only a.4% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case AB, CD are significant model terms. Final Equation in Terms of Coded Factors: Range = +.37 +.89 * A * B +. * C * D (f) What recommendations would you make regarding process operations? Use interaction and/or main effect plots to assist in drawing conclusions. From the interaction plots, choose A at the high level and B at the low level. In each of these plots, D can be at either level. From the main effects plot of C, choose C at the high level. Based on the range analysis, with C at the high level, D should be set at the low level. From the analysis of the crack length data: DESIGN-EXPERT Plot Crack Length 9.84 Interaction Graph B: Titanium Content DESIGN-EXPERT Plot Crack Length 9.84 Interaction Graph C: Heat Treat Method X = A: Pour Temp Y = B: Titanium Content 5.8 B- -. B+. Actual Factors C: Heat Treat Method = D: Grain Refiner =..96 Crack Length X = A: Pour Temp Y = C: Heat Treat Method 5.8 C - C Actual Factors B: Titanium Content =. D: Grain Refiner =..96 Crack Length 8. 8. 4.98 4.98 -. -.5..5. -. -.5..5. A: Pour Tem p A: Pour Tem p 6-5

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Crack Length X = D: Grain Refiner 9.84 One Factor Plot DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content Z = C: Heat Treat Method Cube Graph Crack Length.8 4.7 Actual Factors 5.8 A: Pour Temp =. B: Titanium Content =. C: Heat Treat Method = Crack Length.96 8. Actual Factor D: Grain Refiner =. B: Titanium Content B+.8.8 8.64 5. C+ C: Heat Treat Me 4.98 -. -.5..5. D: Grain Refiner B- 7.73 5.96 C- A- A+ A: Pour Tem p From the analysis of the ranges: DESIGN-EXPERT Plot Range.66 Interaction Graph B: Titanium Content DESIGN-EXPERT Plot Range.66 Interaction Graph D: Grain Refiner X = A: Pour Temp Y = B: Titanium Content.55 B- -. B+. Actual Factors C: Heat Treat Method =. D: Grain Refiner =..384 Range X = C: Heat Treat Method Y = D: Grain Refiner.55 D- -. D+. Actual Factors A: Pour Temp =. B: Titanium Content =..384 Range.455.455.7.7 -. -.5..5. -. -.5..5. A: Pour Tem p C: Heat Treat Method 6.6. Continuation of Problem 6.5. One of the variables in the experiment described in Problem 6.5, heat treatment method (c), is a categorical variable. Assume that the remaining factors are continuous. (a) Write two regression models for predicting crack length, one for each level of the heat treatment method variable. What differences, if any, do you notice in these two equations? Final Equation in Terms of Actual Factors Heat Treat Method - Crack Length = +3.7869 +3.533 * Pour Temp +.934 * Titanium Content +.97888 * Grain Refiner 6-6

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY -.669 * Pour Temp * Titanium Content Heat Treat Method Crack Length = +.84 -.49444 * Pour Temp +.3594 * Titanium Content +.97888 * Grain Refiner +.5358 * Pour Temp * Titanium Content (b) Generate appropriate response surface contour plots for the two regression models in part (a). DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content. Crack Length 8 DESIGN-EXPERT Plot Crack Length X = A: Pour Temp Y = B: Titanium Content. Crack Length B: Titanium Content Actual Factors C: Heat Treat Method = - D: Grain Refiner =..5. 4 6 B: Titanium Content Actual Factors C: Heat Treat Method = D: Grain Refiner =..5. -.5 -.5 8 -. -. -.5..5. 6 -. -. -.5..5. A: Pour Tem p A: Pour Tem p (c) What set of conditions would you recommend for the factors A, B and D if you use heat treatment method C=+? High level of A, low level of B, and low level of D. (d) Repeat part (c) assuming that you wish to use heat treatment method C=-. Low level of A, low level of B, and low level of D. 6.7. An experimenter has run a single replicate of a 4 design. The following effect estimates have been calculated: A = 76. AB = -5.3 ABC = -.8 B = -67.5 AC =.69 ABD = -6.5 C = -7.84 AD = 9.78 ACD =. D = -8.73 BC =.78 BCD = -7.98 BD = 4.74 ABCD = -6.5 CD =.7 (a) Construct a normal probability plot of these effects. The plot from Minitab follows. 6-7

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Probability Plot.9.. A Probability.8.5..5. B AB. Average: -.57 StDev: 3.357 N: 5-5 5 Value Anderson-Darling Normality Test A-Squared:.78 P-Value:.33 (b) Identify a tentative model, based on the plot of the effects in part (a). yö = Intercept + 38. 475x 33. 76x 5. 66x A 6.8. The effect estimates from a 4 factorial design are as follows: ABCD = -.538, ABC = -.66, ABD = -.985, ACD = -.7566, BCD = -.484, CD = -.7, BD = -.793, AD =.5988, BC =.96, AC =.66, AB =.366, D = 4.6744, C = 5.458, B = 8.469, and A =.75. Are you comfortable with the conclusions that all main effects are active? B A x B Probability Plot of Effect Estimate Normal - % CI Percent 9 8 7 6 5 4 3 Mean.975 StDev 4.49 N 5 AD.8 P-Value <.5 5 - -5 5 Effect Estimate 5 The upper right 4 dots are the four main effects. Since they do not follow the rest of the data, the normal probability plot shows that they are active. 6.9. The effect estimates from a 4 factorial experiment are listed here. Are any of the effects significant? 6-8

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY ABCD = -.55 AD = -.6564 BCD = 4.454 AC =.9 ACD = -.493 AB = -.59 ABD = -5.84 D = -6.75 ABC = -5.7696 C = -8.45 CD = 4.677 B = -6.534 BD = -4.66 A = -.794 BC = -.798 Probability Plot of Effect Est Normal - % CI Percent 9 8 7 6 5 4 3 Mean -.859 StDev 4.379 N 5 AD. P-Value.794 5 - -5 - -5 Effect Est 5 No effects appear to be significant. 6.. Consider a variation of the bottle filling experiment from Example 5.3. Suppose that only two levels of carbonation are used so that the experiment is a 3 factorial design with two replicates. The data are shown below. Coded Factors Fill Height Deviation Run A B C Replicate Replicate - - - -3 - + - - 3 - + - - 4 + + - 3 5 - - + - 6 + - + 7 - + + 8 + + + 6 5 Factor Levels Low (-) High (+) A (%) 6-9

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY B (psi) 5 3 C (b/m) 5 (a) Analyze the data from this experiment. Which factors significantly affect fill height deviation? The half normal probability plot of effects shown below identifies the factors A, B, and C as being significant and the AB interaction as being marginally significant. The analysis of variance in the Design Expert output below confirms that factors A, B, and C are significant and the AB interaction is marginally significant. DESIGN-EXPERT Plot Fill Deviation Half Normal plot A: Carbonation B: Pressure C: Speed Half Norm al % probability 97 9 85 8 7 6 4 AB C B A..75.5.5 3. E ffe ct Response: Fill Deviation ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 7.75 4 7.69 6.84 <. significant A 36. 36. 54.6 <. B.5.5 3.7. C.5.5 8.59. AB.5.5 3.4.97 Residual 7.5.66 Lack of Fit.5 3.75..37 not significant Pure Error 5. 8.63 Cor Total 78. 5 The Model F-value of 6.84 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C are significant model terms. Std. Dev..8 R-Squared.97 Mean. Adj R-Squared.8733 C.V. 8.8 Pred R-Squared.833 PRESS 5.34 Adeq Precision 5.44 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots below do not identify any violations to the assumptions. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal plot of residuals vs. Predicted.5 Normal % probability 9 8 7 5 3 5.565 -.565 -.5 -.5 -.565.565.5 -.3 -.38.38 3.3 4.88 Residual Predicted vs. Run vs. Carbonation.5.5.565.565 -.565 -.565 -.5 -.5 4 7 3 6 Run Number Carbonation vs. Pressure vs. Speed.5.5.565.565 3 -.565 -.565 -.5 -.5 5 6 7 8 9 3 8 7 5 33 4 5 Pressure Speed 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY (c) Obtain a model for predicting fill height deviation in terms of the important process variables. Use this model to construct contour plots to assist in interpreting the results of the experiment. The model in both coded and actual factors are shown below. Final Equation in Terms of Coded Factors Fill Deviation = +. +.5 * A +.3 * B +.88 * C +.38 * A * B Final Equation in Terms of Actual Factors Fill Deviation = +9.65 -.65 * Carbonation -. * Pressure +.35 * Speed +.5 * Carbonation * Pressure The following contour plots identify the fill deviation with respect to carbonation and pressure. The plot on the left sets the speed at b/m while the plot on the right sets the speed at 5 b/m. Assuming a faster bottle speed is better, settings in pressure and carbonation that produce a fill deviation near zero can be found in the lower left hand corner of the contour plot on the right. B: Pressure 3. 8.75 7.5 Fill Deviation Fill Deviation 3.5.5.5 -.5 - B: Pressure 3. 4.5 4 3.5 8.75 3.5 7.5.5 6.5 -.5 6.5.5-5...5..5. 5...5..5. A: Carbonation Speed set at b/m A: Carbonation Speed set at 5 b/m (d) In part (a), you probably noticed that there was an interaction term that was borderline significant. If you did not include the interaction term in your model, include it now and repeat the analysis. What difference did this make? If you elected to include the interaction term in part (a), remove it and repeat the analysis. What difference does this make? The following analysis of variance, residual plots, and contour plots represent the model without the interaction. As in the original analysis, the residual plots do not identify any concerns with the assumptions. The contour plots did not change significantly either. The interaction effect is small relative to the main effects. 6-3

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Response: Fill Deviation 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 68.5 3.83 8.84 <. significant A 36. 36. 45.47 <. B.5.5 5.58.3 C.5.5 5.47. Residual 9.5.79 Lack of Fit 4.5 4.3.8. not significant Pure Error 5. 8.63 Cor Total 78. 5 The Model F-value of 8.84 implies the model is significant. There is only a.% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, C are significant model terms. Std. Dev..89 R-Squared.878 Mean. Adj R-Squared.8478 C.V. 88.98 Pred R-Squared.7835 PRESS 6.89 Adeq Precision 5.735 Final Equation in Terms of Coded Factors Fill Deviation = +. +.5 * A +.3 * B +.88 * C Final Equation in Terms of Actual Factors Fill Deviation = -35.75 +.5 * Carbonation +.45 * Pressure +.35 * Speed Normal plot of residuals vs. Predicted.5 Normal % probability 9 8 7 5 3 5.85.5 -.565 -.5 -.5 -.565.5.85.5 -.5 -.75..75 4.5 Residual Predicted 6-33

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY vs. Run vs. Carbonation.5.5.85.85.5.5 -.565 -.565 3 -.5 -.5 4 7 3 6 Run Number Carbonation vs. Pressure vs. Speed.5.5.85.85.5.5 -.565 -.565 -.5 -.5 5 6 7 8 9 3 8 7 5 33 4 5 Pressure Speed Fill Deviation 3..5 Fill Deviation 3. 4 3.5 B: Pressure 8.75 7.5 -.5.5.5 B: Pressure 8.75 7.5.5.5 3 6.5 -.5-6.5.5-5...5..5. -.5 5...5..5. A: Carbonation Speed set at b/m A: Carbonation Speed set at 5 b/m 6-34

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.. I am always interested in improving my golf scores. Since a typical golfer uses the putter for about 35-45% of his or her strokes, it seems logical that in improving one s putting score is a logical and perhaps simple way to improve a golf score ( The man who can putt is a match for any man. Willie Parks, 864-, two-time winner of the British Open). An experiment was conducted to study the effects of four factors on putting accuracy. The design factors are length of putt, type of putter, breaking putt vs. straight putt, and level versus downhill putt. The response variable is distance from the ball to the center of the cup after the ball comes to rest. One golfer performs the experiment, a 4 factorial design with seven replicates was used, and all putts were made in random order. The results are as follows. Design Factors Distance from cup (replicates) Length of putt (ft) Type of putter Break of putt Slope of putt 3 4 5 6 7 Mallet Straight Level. 8. 4..5 9. 6. 8.5 3 Mallet Straight Level. 6.5 4.5 7.5.5 7.5 33. Cavity-back Straight Level 4. 6.. 4.5. 4. 5. 3 Cavity-back Straight Level.. 34.. 5.5.5. Mallet Breaking Level.. 8.5 9.5 6. 5.. 3 Mallet Breaking Level 5..5 8.. 9.5 9.. Cavity-back Breaking Level 6.5 8.5 7.5 6.... 3 Cavity-back Breaking Level 6.5 4.5. 3.5 8. 8. 8. Mallet Straight Downhill 4.5 8. 4.5.. 7.5 6. 3 Mallet Straight Downhill 9.5 8. 6. 5.5. 7. 36. Cavity-back Straight Downhill 5. 6. 8.5..5 9. 3. 3 Cavity-back Straight Downhill 4.5 39. 6.5 3.5 7. 8.5 36. Mallet Breaking Downhill 8. 4.5 6.5. 3. 4. 4. 3 Mallet Breaking Downhill.5.5 6.5. 5.5 4. 6. Cavity-back Breaking Downhill... 4.5. 4. 6.5 3 Cavity-back Breaking Downhill 8. 5. 7.. 3.5 8.5 8. (a) Analyze the data from this experiment. Which factors significantly affect putting performance? The half normal probability plot of effects identifies only factors A and B, length of putt and type of putter, as having a potentially significant affect on putting performance. The analysis of variance with only these significant factors is presented as well and confirms significance. 6-35

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY DESIGN-EXPERT Plot Distance from cup Half Normal plot A: Length of putt B: Ty pe of putter C: Break of putt D: Slope of putt A Half Normal % probability 97 9 85 8 7 6 B 4..43.86 4.9 5.7 Effect with Only Factors A and B Response: Distance from cup 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 Model 35.9 65.65 7.69.8 significant A 97.5 97.5.8.4 B 388.5 388.5 4.57.347 Residual 948.94 9 84.85 Lack of Fit 933.5 3 7.78.83.69 not significant Pure Error 835.79 96 86.6 Cor Total 554.3 The Model F-value of 7.69 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. Std. Dev. 9. R-Squared.37 Mean.3 Adj R-Squared.76 C.V. 74.9 Pred R-Squared.748 PRESS 9765.6 Adeq Precision 6.66 (b) Analyze the residuals from this experiment. Are there any indications of model inadequacy? The residual plots for the model containing only the significant factors A and B are shown below. The normality assumption appears to be violated. Also, as a golfer might expect, there is a slight inequality of variance with regards to the length of putt. A square root transformation is applied which corrects the violations. The analysis of variance and corrected residual plots are also presented. Finally, an effects plot identifies a foot putt and the cavity-back putter reduce the mean distance from the cup. 6-36

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY Normal Plot of vs. Predicted 9.79 Normal % Probability 9 8 7 5 3 5 8. 6.33-5.345 6 3 4 3 3-7.3-7.3-5.345 6.33 8. 9.79 7.58 9.94.3 4.66 7. Residual Predicted vs. Length of putt vs. Type of putter 9.79 9.79 8. 8. 6.33-5.345 3 6 4 6.33-5.345 3 4 6 3 3 3 3-7.3-7.3 3 7 3 7 3 Length of putt Type of putter vs. Break of putt vs. Slope of putt 9.79 9.79 8. 8. 6.33-5.345 4 5 6.33-5.345 3 3 4 3-7.3-7.3 Break of putt Slope of putt 6-37

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY with Only Factors A and B and a Square Rot Transformation Response: Distance from cuptransform:square root Constant: 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 Model 37.6 8.63 7.85.7 significant A.6.6 9..3 B 5.64 5.64 6.59.6 Residual 58.63 9.37 Lack of Fit 3.9 3.3.98.487 not significant Pure Error 8.45 96.38 Cor Total.89 The Model F-value of 7.85 implies the model is significant. There is only a.7% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than.5 indicate model terms are significant. In this case A, B, are significant model terms. Std. Dev..54 R-Squared.59 Mean 3. Adj R-Squared. C.V. 49.57 Pred R-Squared.77 PRESS 73.6 Adeq Precision 6.45 Normal Plot of vs. Predicted 3.366 Normal % Probability 9 8 7 5 3 5.5464 -.79783 -. 6 3 4 3 3-3.963-3.963 -. -.79783.5464 3.366.9.7 3. 3.5 3.9 Residual Predicted 6-38

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY vs. Length of putt vs. Type of putter 3.366 3.366.5464 -.79783 3 4.5464 -.79783 3 4 -. 6 -. 6 3 3 3 3-3.963-3.963 3 7 3 7 3 Length of putt Type of putter vs. Break of putt vs. Slope of putt 3.366 3.366.5464 -.79783 4.5464 -.79783 3 -. 5 -. 4 3-3.963-3.963 Break of putt Slope of putt DESIGN-EXPERT Plot Sqrt(Distance f rom cup) X = A: Length of putt Y = B: Ty pe of putter 6 B Mallet B Cav ity -back Actual Factors C: Break of putt = Straight D: Slope of putt = Lev el Distance from cup 8 Interaction Graph B: Type of putter 4. 5.. 5. 3. A: Length of putt 6-39

Solutions from Montgomery, D. C. () Design and Analysis of Experiments, Wiley, NY 6.. Semiconductor manufacturing processes have long and complex assembly flows, so matrix marks and automated d-matrix readers are used at several process steps throughout factories. Unreadable matrix marks negatively effect factory run rates, because manual entry of part data is required before manufacturing can resume. A 4 factorial experiment was conducted to develop a d-matrix laser mark on a metal cover that protects a substrate mounted die. The design factors are A = laser power (9W, 3W), B = laser pulse frequency (4 Hz, Hz), C = matrix cell size (.7 in,. in), and D = writing speed ( in/sec, in/sec), and the response variable is the unused error correction (UEC). This is a measure of the unused portion of the redundant information embedded in the d matrix. A UEC of represents the lowest reading that still results in a decodable matrix while a value of is the highest reading. A DMX Verifier was used to measure UEC. The data from this experiment are shown below. Standard Order Run Order Laser Power Pulse Frequency Cell Size Writing Speed UEC 8 -.8 - -.8 3 -.79 9 4 - - -.6 7 5 - -.65 5 6 -.55 7 - - -.98 6 8 - -.67 6 9.69 3 - -.56 5 - - -.63 4 -.65 3 - - - -.75 3 4 - - -.7 4 5 - -.98 6 - -.63 (a) Analyze the data from this experiment. Which factors significantly affect UEC? The normal probability plot of effects identifies A, C, D, and the AC interaction as significant. The Design Expert output including the analysis of variance confirms the significance and identifies the corresponding model. Contour plots identify factors A and C with B held constant at zero and D toggled from - to +. 6-4