Chapter 4 Randomized Blocks, Latin Squares, and Related Designs Solutions

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Chater 4 Randomized Blocks, Latin Squares, and Related Designs Solutions 4.. The ANOVA from a randomized comlete block exeriment outut is shown below. Source DF SS MS F P Treatment 4 00.56? 9.84? Block?? 64.765?? Error 0 69.33? Total 9 503.7 (a) Fill in the blanks. You may give bounds on the P-value. Comleted table is: Source DF SS MS F P Treatment 4 00.56 5.640 9.84 < 0.0000 Block 5 33.8 64.765 Error 0 69.33 8.467 Total 9 503.7 (b) How many blocks were used in this exeriment? Six blocks were used. (c) What conclusions can you draw? The treatment effect is significant; the means of the five treatments are not all equal. 4.. Consider the single-factor comletely randomized exeriment shown in Problem 3.4. Suose that this exeriment had been conducted in a randomized comlete block design, and that the sum of squares for blocks was 80.00. Modify the ANOVA for this exeriment to show the correct analysis for the randomized comlete block exeriment. The modified ANOVA is shown below: Source DF SS MS F P Treatment 4 987.7 46.93 46.3583 < 0.0000 Block 5 80.00 6.00 Error 0 06.53 5.33 Total 9 74.4 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.3. A chemist wishes to test the effect of four chemical agents on the strength of a articular tye of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and alies all four chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze the data from this exeriment (use α = 0.05) and draw aroriate conclusions. Bolt Chemical 3 4 5 73 68 74 7 67 73 67 75 7 70 3 75 68 78 73 68 4 73 7 75 75 69 Design Exert Outut Resonse: Strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 57.00 4 39.5 Model.95 3 4.3.38 0. not significant A.95 3 4.3.38 0. Residual.80.8 Cor Total 9.75 9 The "Model F-value" of.38 imlies the model is not significant relative to the noise. There is a. % chance that a "Model F-value" this large could occur due to noise. Std. Dev..35 R-Squared 0.377 Mean 7.75 Adj R-Squared 0.58 C.V..88 Pred R-Squared -0.746 PRESS 60.56 Adeq Precision 0.558 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 70.60 0.60-7.40 0.60 3-3 7.40 0.60 4-4 7.60 0.60 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -0.80 0.85-0.94 0.3665 vs 3 -.80 0.85 -. 0.0564 vs 4 -.00 0.85 -.35 0.0370 vs 3 -.00 0.85 -.7 0.635 vs 4 -.0 0.85 -.4 0.846 3 vs 4-0.0 0.85-0.3 0.885 There is no difference among the chemical tyes at α = 0.05 level. 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.4. Three different washing solutions are being comared to study their effectiveness in retarding bacteria growth in five-gallon milk containers. The analysis is done in a laboratory, and only three trials can be run on any day. Because days could reresent a otential source of variability, the exerimenter decides to use a randomized block design. Observations are taken for four days, and the data are shown here. Analyze the data from this exeriment (use α = 0.05) and draw conclusions. Days Solution 3 4 3 8 39 6 4 7 44 3 5 4 Design Exert Outut Resonse: Growth ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 06.9 3 368.97 Model 703.50 35.75 40.7 0.0003 significant A 703.50 35.75 40.7 0.0003 Residual 5.83 6 8.64 Cor Total 86.5 The Model F-value of 40.7 imlies the model is significant. There is only a 0.03% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..94 R-Squared 0.934 Mean 8.75 Adj R-Squared 0.9085 C.V. 5.68 Pred R-Squared 0.755 PRESS 07.33 Adeq Precision 9.687 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 3.00.47-5.5.47 3-3 8.00.47 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -.5.08 -.08 0.306 vs 3 5.00.08 7. 0.0004 vs 3 7.5.08 8.30 0.000 There is a difference between the means of the three solutions. The Fisher LSD rocedure indicates that solution 3 is significantly different than the other two. 4-3

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.5. Plot the mean tensile strengths observed for each chemical tye in Problem 4.3 and comare them to a scaled t distribution. What conclusions would you draw from the dislay? Scaled t Distribution () () (3,4) 70.0 7.0 7.0 73.0 Mean Strength S MS b. 8 5 E y i. = = = 0. 603 There is no obvious difference between the means. This is the same conclusion given by the analysis of variance. 4.6. Plot the average bacteria counts for each solution in Problem 4.4 and comare them to an aroriately scaled t distribution. What conclusions can you draw? Scaled t Distribution (3) ( ) ( ) 5 0 5 0 5 Bacteria Growth S MS b 8. 64 4 E y i. = = =. 47 4-4

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY There is no difference in mean bacteria growth between solutions and. However, solution 3 roduces significantly lower mean bacteria growth. This is the same conclusion reached from the Fisher LSD rocedure in Problem 4.4. 4.7. Consider the hardness testing exeriment described in Section 4.. Suose that the exeriment was conducted as described and the following Rockwell C-scale data (coded by subtracting 40 units) obtained: (a) Analyize the data from this exeriment. Couon Ti 3 4 9.3 9.4 9.6 0.0 9.4 9.3 9.8 9.9 3 9. 9.4 9.5 9.7 4 9.7 9.6 0.0 0. There is a difference between the means of the four tis. Design Exert Outut Resonse: Hardness ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Bock 0.8 3 0.7 Model 0.38 3 0.3 4.44 0.0009 significant A 0.38 3 0.3 4.44 0.0009 Residual 0.080 9 8.889E-003 Cor Total.9 5 The Model F-value of 4.44 imlies the model is significant. There is only a 0.09% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 0.094 R-Squared 0.880 Mean 9.63 Adj R-Squared 0.7706 C.V. 0.98 Pred R-Squared 0.4563 PRESS 0.5 Adeq Precision 5.635 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 9.57 0.047-9.60 0.047 3-3 9.45 0.047 4-4 9.88 0.047 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -0.05 0.067-0.38 0.763 vs 3 0.3 0.067.87 0.0935 vs 4-0.30 0.067-4.50 0.005 vs 3 0.5 0.067.5 0.050 vs 4-0.7 0.067-4. 0.006 3 vs 4-0.43 0.067-6.37 0.000 (b) Use the Fisher LSD method to make comarisons among the four tis to determine secifically which tis differ in mean hardness readings. Based on the LSD bars in the Design Exert lot below, the mean of ti 4 differs from the means of tis,, and 3. The LSD method identifies a marginal difference between the means of tis and 3. 4-5

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 0. One Factor Plot 9.95 Hardness 9.7 9.45 9. 3 4 A: Ti (c) Analyze the residuals from this exeriment. The residual lots below do not identify any violations to the assumtions. Normal Plot of Residuals Residuals vs. Predicted 0.5 99 Normal % Probability 95 90 80 70 50 30 0 0 5 Residuals 0.0875 0.05-0.0375-0. -0. -0.0375 0.05 0.0875 0.5 9. 9.47 9.7 9.96 0.0 Residual Predicted 4-6

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Residuals vs. Ti 0.5 0.0875 Residuals 0.05-0.0375-0. 3 4 Ti 4.8. A consumer roducts comany relies on direct mail marketing ieces as a major comonent of its advertising camaigns. The comany has three different designs for a new brochure and want to evaluate their effectiveness, as there are substantial differences in costs between the three designs. The comany decides to test the three designs by mailing 5,000 samles of each to otential customers in four different regions of the country. Since there are known regional differences in the customer base, regions are considered as blocks. The number of resonses to each mailing is shown below. (a) Analyze the data from this exeriment. Region Design NE NW SE SW 50 350 9 375 400 55 390 580 3 75 340 00 30 The residuals of the analsysis below identify concerns with the normality and equality of variance assumtions. As a result, a square root transformation was alied as shown in the second ANOVA table. The residuals of both analysis are resented for comarison in art (c) of this roblem. The analysis concludes that there is a difference between the mean number of resonses for the three designs. Design Exert Outut Resonse: Number of resonses ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Block 49035.67 3 6345. Model 90755.7 45377.58 50.5 0.000 significant A 90755.7 45377.58 50.5 0.000 Residual 548.83 6 904.8 Cor Total.45E+005 The Model F-value of 50.5 imlies the model is significant. There is only a 0.0% chance that a "Model F-Value" this large could occur due to noise. 4-7

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Std. Dev. 30.08 R-Squared 0.9436 Mean 35.7 Adj R-Squared 0.947 C.V. 8.57 Pred R-Squared 0.774 PRESS 75.33 Adeq Precision 6.97 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 98.50 5.04-473.75 5.04 3-3 8.5 5.04 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -75.5.7-8.4 0.000 vs 3 7.5.7 0.8 0.4483 vs 3 9.50.7 9.05 0.000 Design Exert Outut for Transformed Data Resonse: Number of resonses Transform: Square root Constant: 0 ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Block 35.89 3.96 Model 60.73 30.37 60.47 0.000 significant A 60.73 30.37 60.47 0.000 Residual 3.0 6 0.50 Cor Total 99.64 The Model F-value of 60.47 imlies the model is significant. There is only a 0.0% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 0.7 R-Squared 0.957 Mean 8.5 Adj R-Squared 0.9370 C.V. 3.83 Pred R-Squared 0.809 PRESS.05 Adeq Precision 8.9 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 7.7 0.35 -.69 0.35 3-3 6.69 0.35 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -4.5 0.50-9.0 0.000 vs 3 0.48 0.50 0.95 0.3769 vs 3 4.99 0.50 9.96 < 0.000 (b) Use the Fisher LSD method to make comarisons among the three designs to determine secifically which designs differ in mean resonse rate. Based on the LSD bars in the Design Exert lot below, designs and 3 do not differ; however, design is different than designs and 3. 4-8

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.083 One Factor Plot Sqrt(Number of resonses).598 9.3 6.67 4.4 3 A: Design (c) Analyze the residuals from this exeriment. The first set of residual lots resented below reresent the untransformed data. Concerns with normality as well as inequality of variance are resented. The second set of residual lots reresent transformed data and do not identify significant violations of the assumtions. The residuals vs. design lot indicates a slight inequality of variance; however, not a strong violation and an imrovement over the non-transformed data. Normal Plot of Residuals Residuals vs. Predicted 36.5833 99 Normal % Probability 95 90 80 70 50 30 0 0 5 Residuals 7 -.58333 -.667-4.75-4.75 -.667 -.58333 7 36.5833 99.75 85.88 37.00 458.3 544.5 Residual Predicted 4-9

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 36.5833 Residuals vs. Design 7 Residuals -.58333 -.667-4.75 3 Design The following are the square root transformed data residual lots. Normal Plot of Residuals Residuals vs. Predicted 0.94069 99 Normal % Probability 95 90 80 70 50 30 0 0 5 Residuals 0.4769 0.0054-0.45563-0.904-0.904-0.45563 0.0054 0.4769 0.94069 4.4 6.68 8.96.4 3.5 Residual Predicted 4-0

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Residuals vs. Design 0.94069 0.4769 Residuals 0.0054-0.45563-0.904 3 Design 4.9. The effect of three different lubricating oils on fuel economy in diesel truck engines is being studied. Fuel economy is measured using brake-secific fuel consumtion after the engine has been running for 5 minutes. Five different truck engines are available for the study, and the exerimenters conduct the following randomized comlete block design. (a) Analyize the data from this exeriment. Truck Oil 3 4 5 0.500 0.634 0.487 0.39 0.5 0.535 0.675 0.50 0.435 0.540 3 0.53 0.595 0.488 0.400 0.50 From the analysis below, there is a significant difference between lubricating oils with regards to fuel economy. Design Exert Outut Resonse: Fuel consumtion ANOVA for Selected Factorial Model Analysis of variance table [Terms added sequentially (first to last)] Sum of Mean F Source Squares DF Square Value Prob > F Block 0.09 4 0.03 Model 6.706E-003 3.353E-003 6.35 0.03 significant A 6.706E-003 3.353E-003 6.35 0.03 Residual 4.E-003 8 5.78E-004 Cor Total 0.0 4 The Model F-value of 6.35 imlies the model is significant. There is only a.3% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 0.03 R-Squared 0.636 Mean 0.5 Adj R-Squared 0.570 C.V. 4.49 Pred R-Squared -0.3583 PRESS 0.05 Adeq Precision 8.84 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 0.49 0.00-0.54 0.00 3-3 0.50 0.00 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -0.049 0.05-3.34 0.00 vs 3-8.800E-003 0.05-0.6 0.565 vs 3 0.040 0.05.74 0.055 (b) Use the Fisher LSD method to make comarisons among the three lubricating oils to determine secifically which oils differ in break-secific fuel consumtion. Based on the LSD bars in the Design Exert lot below, the means for break-secific fuel consumtion for oils and 3 do not differ; however, oil is different than oils and 3. 0.675 One Factor Plot 0.5885 Fuel consumtion 0.50 0.455 0.39 3 A: Oil (c) Analyze the residuals from this exeriment. The residual lots below do not identify any violations to the assumtions. 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Normal Plot of Residuals Residuals vs. Predicted 0.03333 99 Normal % Probability 95 90 80 70 50 30 0 0 5 Residuals 0.00678333-0.00876667-0.04367-0.0398667-0.0398667-0.04367-0.00876667 0.00678333 0.03333 0.37 0.44 0.5 0.59 0.66 Residual Predicted Residuals vs. Oil 0.03333 0.00678333 Residuals -0.00876667-0.04367-0.0398667 3 Oil 4.0. An article in the Fire Safety Journal ( The Effect of Nozzle Design on the Stability and Performance of Turbulent Water Jets, Vol. 4, August 98) describes an exeriment in which a shae factor was determined for several different nozzle designs at six levels of jet efflux velocity. Interest focused on otential differences between nozzle designs, with velocity considered as a nuisance variable. The data are shown below: Jet Efflux Velocity (m/s) Nozzle Design.73 4.37 6.59 0.43 3.46 8.74 0.78 0.80 0.8 0.75 0.77 0.78 0.85 0.85 0.9 0.86 0.8 0.83 3 0.93 0.9 0.95 0.89 0.89 0.83 4.4 0.97 0.98 0.88 0.86 0.83 5 0.97 0.86 0.78 0.76 0.76 0.75 4-3

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY (a) Does nozzle design affect the shae factor? Comare nozzles with a scatter lot and with an analysis of variance, using α = 0.05. Design Exert Outut Resonse: Shae 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 0.063 5 0.03 Model 0.0 4 0.06 8.9 0.0003 significant A 0.0 4 0.06 8.9 0.0003 Residual 0.057 0.865E-003 Cor Total 0. 9 The Model F-value of 8.9 imlies the model is significant. There is only a 0.03% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 0.054 R-Squared 0.6407 Mean 0.86 Adj R-Squared 0.5688 C.V. 6.3 Pred R-Squared 0.96 PRESS 0.3 Adeq Precision 9.438 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 0.78 0.0-0.85 0.0 3-3 0.90 0.0 4-4 0.94 0.0 5-5 0.8 0.0 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -0.07 0.03 -.3 0.03 vs 3-0. 0.03-3.88 0.0009 vs 4-0.6 0.03-5.3 < 0.000 vs 5-0.03 0.03 -.0 0.377 vs 3-0.048 0.03 -.56 0.335 vs 4-0.090 0.03 -.9 0.0086 vs 5 0.040 0.03.9 0.03 3 vs 4-0.04 0.03 -.35 0.96 3 vs 5 0.088 0.03.86 0.0097 4 vs 5 0.3 0.03 4. 0.0004 Nozzle design has a significant effect on shae factor. 4-4

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY.4 One Factor Plot.0436 Shae 0.94478 0.847076 0.749435 3 4 5 Nozzle Design (b) Analyze the residual from this exeriment. The lots shown below do not give any indication of serious roblems. Thre is some indication of a mild outlier on the normal robability lot and on the lot of residuals versus the redicted velocity. Normal lot of residuals Residuals vs. Predicted 0.333 99 Normal % robability 95 90 80 70 50 30 0 0 5 0.073333 Residuals 0.03333-0.086667-0.0786667-0.0786667-0.086667 0.03333 0.073333 0.333 0.73 0.80 0.87 0.95.0 Residual Predicted 4-5

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 0.333 Residuals vs. Nozzle Design 0.073333 Residuals 0.03333-0.086667-0.0786667 3 4 5 Nozzle Design (c) Which nozzle designs are different with resect to shae factor? Draw a grah of average shae factor for each nozzle tye and comare this to a scaled t distribution. Comare the conclusions that you draw from this lot to those from Duncan s multile range test. S MS b 0. 00865 6 E y i. = = = 0. 085 R = r 0.05 (,0) S yi. = (.95)(0.085)= 0.06446 R 3 = r 0.05 (3,0) S yi. = (3.0)(0.085)= 0.06774 R 4 = r 0.05 (4,0) S yi. = (3.8)(0.085)= 0.06949 R 5 = r 0.05 (5,0) S yi. = (3.5)(0.085)= 0.070 Mean Difference R vs 4 0.667 > 0.070 different vs 3 0.000 > 0.06949 different vs 0.0767 > 0.06774 different vs 5 0.0367 < 0.06446 5 vs 4 0.3000 > 0.06949 different 5 vs 3 0.08833 > 0.06774 different 5 vs 0.04000 < 0.06446 vs 4 0.09000 > 0.06774 different vs 3 0.04833 < 0.06446 3 vs 4 0.0467 < 0.06446 4-6

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Scaled t () (5) ( (3 (4 0.7 0.8 0.8 0.9 0.9 Shae 4.. Consider the ratio control algorithm exeriment described in Section 3.9. The exeriment was actually conducted as a randomized block design, where six time eriods were selected as the blocks, and all four ratio control algorithms were tested in each time eriod. The average cell voltage and the standard deviation of voltage (shown in arentheses) for each cell are as follows: Ratio Control Time Period Algorithms 3 4 5 6 4.93 (0.05) 4.86 (0.04) 4.75 (0.05) 4.95 (0.06) 4.79 (0.03) 4.88 (0.05) 4.85 (0.04) 4.9 (0.0) 4.79 (0.03) 4.85 (0.05) 4.75 (0.03) 4.85 (0.0) 3 4.83 (0.09) 4.88 (0.3) 4.90 (0.) 4.75 (0.5) 4.8 (0.08) 4.90 (0.) 4 4.89 (0.03) 4.77 (0.04) 4.94 (0.05) 4.86 (0.05) 4.79 (0.03) 4.76 (0.0) (a) Analyze the average cell voltage data. (Use α = 0.05.) Does the choice of ratio control algorithm affect the cell voltage? Design Exert Outut Resonse: Average ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 0.07 5 3.487E-003 Model.746E-003 3 9.53E-004 0.9 0.904 not significant A.746E-003 3 9.53E-004 0.9 0.904 Residual 0.07 5 4.8E-003 Cor Total 0.09 3 The "Model F-value" of 0.9 imlies the model is not significant relative to the noise. There is a 90.4 % chance that a "Model F-value" this large could occur due to noise. Std. Dev. 0.069 R-Squared 0.0366 Mean 4.84 Adj R-Squared -0.560 C.V..43 Pred R-Squared -.466 PRESS 0.8 Adeq Precision.688 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - 4.86 0.08-4.83 0.08 3-3 4.85 0.08 4-4 4.84 0.08 4-7

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs 0.07 0.040 0.67 0.556 vs 3 0.03 0.040 0.33 0.7438 vs 4 0.05 0.040 0.6 0.549 vs 3-0.03 0.040-0.33 0.7438 vs 4 -.667E-003 0.040-0.04 0.9674 3 vs 4 0.0 0.040 0.9 0.7748 The ratio control algorithm does not affect the mean cell voltage. (b) Perform an aroriate analysis of the standard deviation of voltage. (Recall that this is called ot noise. ) Does the choice of ratio control algorithm affect the ot noise? Design Exert Outut Resonse: StDev Transform: Natural log Constant: 0.000 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 0.94 5 0.9 Model 6.7 3.06 33.6 < 0.000 significant A 6.7 3.06 33.6 < 0.000 Residual 0.93 5 0.06 Cor Total 8.04 3 The Model F-value of 33.6 imlies the model is significant. There is only a 0.0% chance that a "Model F-Value" this large could occur due to noise. Std. Dev. 0.5 R-Squared 0.8693 Mean -3.04 Adj R-Squared 0.843 C.V. -8.8 Pred R-Squared 0.6654 PRESS.37 Adeq Precision.446 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error - -3.09 0.0 - -3.5 0.0 3-3 -.0 0.0 4-4 -3.36 0.0 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs 0.4 0.4.93 0.003 vs 3-0.89 0.4-6.9 < 0.000 vs 4 0.7 0.4.87 0.083 vs 3 -.3 0.4-9. < 0.000 vs 4-0.5 0.4 -.06 0.304 3 vs 4.6 0.4 8.06 < 0.000 A natural log transformation was alied to the ot noise data. The ratio control algorithm does affect the ot noise. 4-8

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY (c) Conduct any residual analyses that seem aroriate. Normal lot of residuals Residuals vs. Predicted 0.88958 99 Normal % robability 95 90 80 70 50 30 0 0 5 0.6945 Residuals -0.0350673-0.9708-0.359093-0.359093-0.9708-0.0350673 0.6945 0.88958-3.73-3.6 -.78 -.3 -.84 Residual Predicted Residuals vs. Algorithm 0.88958 0.6945 Residuals -0.0350673-0.9708-0.359093 3 4 Algorithm The normal robability lot shows slight deviations from normality; however, still accetable. (d) Which ratio control algorithm would you select if your objective is to reduce both the average cell voltage and the ot noise? Since the ratio control algorithm has little effect on average cell voltage, select the algorithm that minimizes ot noise, that is algorithm #. 4-9

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.. An aluminum master alloy manufacturer roduces grain refiners in ingot form. The comany roduces the roduct in four furnaces. Each furnace is known to have its own unique oerating characteristics, so any exeriment run in the foundry that involves more than one furnace will consider furnaces as a nuisance variable. The rocess engineers susect that stirring rate imacts the grain size of the roduct. Each furnace can be run at four different stirring rates. A randomized block design is run for a articular refiner and the resulting grain size data is as follows. Furnace Stirring Rate 3 4 5 8 4 5 6 0 4 5 6 9 5 4 6 9 0 7 9 3 6 (a) Is there any evidence that stirring rate imacts grain size? Design Exert Outut Resonse: Grain Size ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 65.9 3 55.06 Model.9 3 7.40 0.85 0.4995 not significant A.9 3 7.40 0.85 0.4995 Residual 78.06 9 8.67 Cor Total 65.44 5 The "Model F-value" of 0.85 imlies the model is not significant relative to the noise. There is a 49.95 % chance that a "Model F-value" this large could occur due to noise. Std. Dev..95 R-Squared 0.3 Mean 7.69 Adj R-Squared -0.038 C.V. 38.3 Pred R-Squared -.460 PRESS 46.7 Adeq Precision 5.390 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -5 5.75.47-0 8.50.47 3-5 7.75.47 4-0 8.75.47 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs -.75.08 -.3 0.93 vs 3 -.00.08-0.96 0.360 vs 4-3.00.08 -.44 0.836 vs 3 0.75.08 0.36 0.770 vs 4-0.5.08-0. 0.907 3 vs 4 -.00.08-0.48 0.645 The analysis of variance shown above indicates that there is no difference in mean grain size due to the different stirring rates. 4-0

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY (b) Grah the residuals from this exeriment on a normal robability lot. Interret this lot. Normal lot of residuals 99 Normal % robability 95 90 80 70 50 30 0 0 5-3.85 -.065-0.35.4375 3.875 Residual The lot indicates that normality assumtion is valid. (c) Plot the residuals versus furnace and stirring rate. Does this lot convey any useful information? Residuals vs. Stirring Rate Residuals vs. Furnace 3.875 3.875.4375.4375 Residuals -0.35 Residuals -0.35 -.065 -.065-3.85-3.85 3 4 3 4 Stirring Rate Furnace The variance is consistent at different stirring rates. Not only does this validate the assumtion of uniform variance, it also identifies that the different stirring rates do not affect variance. (d) What should the rocess engineers recommend concerning the choice of stirring rate and furnace for this articular grain refiner if small grain size is desirable? There really is no effect due to the stirring rate. 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.3. Analyze the data in Problem 4.4 using the general regression significance test. μ: μ + 4 τ + 4 τ + 4 τ3 + 3 β + 3 β + 3 β3 + 3 β4 = 5 τ: 4 μ + 4 τ + β + β + β3 + β4 = 9 τ : 4μ + 4τ + β + β + β3 + β4 = 0 τ 3: 4μ + 4τ3 + β + β + β3 + β4 = 3 β : 3μ + τ + τ + τ3 + 3β = 34 β : 3μ + τ + τ + τ3 + 3β = 50 β: 3 μ + τ + τ + τ 3 + 3β 3 = 36 β : 3 μ + τ + τ + τ + 3 β = 05 Alying the constraints τ β = i 3 4 = 0, we obtain: j 5 μ =, τ 5 =, τ 78 =, 9 89 5 8 95 τ 3 =, β =, β =, β 3 =, β 4 = 5 5 78 9 89 5 R μ, τ, β = 5 + 9 + 0 + 3 + 34 + 50 ( ) ( ) ( ) ( ) ( ) ( ) ( )+ ij = 608 ij = 608 609. 7 = y, SS y R( μ, τ, β ) Model Restricted to τ i = 0 : E 8 = 5. 83 95 ( 36) + ( 05) = 609. 7 μ: μ + 3 β + 3 β + 3β3 + 3β4 = 5 β : 3μ + 3β β : 3μ + 3β β 3: 3μ + 3β3 = 34 = 50 = 36 β : 3 μ + 3 β = 05 4 4 Alying the constraint β j = 0, we obtain: 5 μ =, 5 8 95 β = 89 /, β =, β 3 =, β 4 =. Now: 5 89 5 8 95 R μ, β = 5 + 34 + 50 + 36 + 05 = 535. ( ) ( ) ( ) ( ) ( ) ( ) 67 ( τ μ, β ) = R( μ, τ, β ) R( μ, β ) = 609. 7 535. 67 = 703. 50 SSTreatments R = Model Restricted to β = 0 : j 4-

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Alying the constraint τ i = 0, we obtain: μ: μ + 4 τ + 4τ + 4τ3 τ : 4μ + 4τ τ : 4μ + 4τ = 5 = 9 = 0 τ : 4 μ + 4 τ = 3 3 3 5 μ =, τ 5 =, τ 78 =, 9 τ 3 = 5 5 78 9 R μ, τ = 5 + 9 + 0 + 3 = 49. ( ) ( ) ( ) ( ) ( ) 5 ( β μ, τ ) = R( μ, τ, β ) R( μ, τ ) = 609. 7 49. 5 = 06. 9 SSBlocks R = 4.4. Assuming that chemical tyes and bolts are fixed, estimate the model arameters τ i and β j in Problem 4.3. Using Equations 4.8, alying the constraints, we obtain: 35 3 7 μ =, τ =, τ =, 0 0 0 τ 3 3 = 0, τ 7 4 = 0, 35 β =, 0 65 β =, 0 75 β 3 =, 0 0 β 4 =, β 0 5 65 = 0 4.5. Draw an oerating characteristic curve for the design in Problem 4.4. Does this test seem to be sensitive to small differences in treatment effects? Assuming that solution tye is a fixed factor, we use the OC curve in aendix V. Calculate using MS E to estimate σ. We have: b τ i 4 τ i aσ 38.64 ( ) Φ = = υ = a υ = ( a )( b ) = ( )( 3) 6. = = If τ i = σ = MSE, then: 4 Φ = =. 5 and β 0. 70 3 () If τ i σ = MS = E, then: ( ) () This test is not very sensitive to small differences. 4 Φ= =.63 and β 0. 55, etc. 3 4-3

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.6. Suose that the observation for chemical tye and bolt 3 is missing in Problem 4.3. Analyze the roblem by estimating the missing value. Perform the exact analysis and comare the results. y 3 is missing. ( a )( b ) ( ) ( ) ( )( ) ' ' ' ay. + by 4 8 5 7 360.3 y +.. y 3 = = = 75.5 3 4 Therefore, y. =357.5, y.3 =30.5, and y.. =435.5 Source SS DF MS F 0 Chemicals.7844 3 4.65.54 Bolts 58.8875 4 Error.765.9784 Total 93.4344 8 F 0.05,3, =3.59, Chemicals are not significant. This is the same result as found in Problem 4.3. 4.7. Consider the hardness testing exeriment in Problem 4.7. Suose that the observation for ti in couon 3 is missing. Analyze the roblem by estimating the missing value. y 3 is missing. ( a )( b ) ( ) ( ) ( )( ) ' ' ' ay. + by 4 8.6 4 9. 44..3 y +.. y 3 = = = 9.6 3 3 Therefore, y. =38., y.3 =38.7, and y.. =53.8 Source SS DF MS F 0 Ti 0.40 3 0.33333 9.9 Couon 0.80 3 Error 0.06 9 0.00694 Total.6 5 F 0.05,3,9 =3.86, Tis are significant. This is the same result as found in Problem 4.7. 4.8. Two missing values in a randomized block. Suose that in Problem 4.3 the observations for chemical tye and bolt 3 and chemical tye 4 and bolt 4 are missing. (a) Analyze the design by iteratively estimating the missing values as described in Section 4..3. ŷ 3 '. 4y + 5y 3 y = and '. '.. ŷ 44 = ' ' 4y4.. 4 + 5y '.. y 0 y 3 Data is coded y-70. As an initial guess, set equal to the average of the observations available for 0 chemical. Thus, y 3 = = 0. 5. Then, 4 0 4( 8) + 5( 6) 5. 5 ŷ 44 = = 3. 04 4( ) + 5( 7) 8. 04 ŷ 3 = = 5. 4 4-4

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY ( 8) + 5( 6) 4 30. 4 ŷ 44 = =. 63 4( ) + 5( 7) 7. 63 ŷ 44 = = 5. 44 4( 8) + 5( 6) 30. 44 ŷ 44 = =. 63 ŷ = 5 44 ŷ = 63 3. 44. Design Exert Outut ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 56.83 4 39. Model 9.59 3 3.0.08 0.560 not significant A 9.59 3 3.0.08 0.560 Residual 8.4.53 Cor Total 84.83 9 (b) Differentiate SS E with resect to the two missing values, equate the results to zero, and solve for estimates of the missing values. Analyze the design using these two estimates of the missing values. SS SSE SSE From = = 0, we obtain: y y 3 44 yij y + 5 i. y 4. j E = y 0.. = 3 44 3 44 3 44 SS E 0. 6y + 0. 6y 6. 8y 3. 7 y + 0. y y + R. ŷ 0. ŷ 3 3 + 0. ŷ +. ŷ 44 44 = 6. 8 = 3. 7 ŷ = 5 45, ŷ = 63 3. 44. These quantities are almost identical to those found in art (a). The analysis of variance using these new data does not differ substantially from art (a). (c) Derive general formulas for estimating two missing values when the observations are in different blocks. SS y y ( y. + y ) + ( y. + y ) ( y. + y ) + ( y. + y ) ( y.. + y + y ) i iu k kv u iu v kv iu kv E = iu + kv + SSE SSE From = = 0, we obtain: y y 3 44 ŷ ŷ iu kv b ( a )( b ) ab ay' i. + by'. j = ab ( a )( b ) ay' k. + by' = ab ab y'.v a.. y' ŷkv ab.. ŷiu ab ab 4-5

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY whose simultaneous solution is: y iu ( ) ( ) ( ) ( ) ( ) ( ) u ( a )( b ) ( a ) ( b ) ab[ ay ' k. + by '. v y '..] ( a ) ( b ) + ( )( )[ + ] ( a ) ( b ) y' i. a a b ab + y'. b a b ab y'.. ab a b = + y kv ay' by' y' b a ay ' by' y ' = i.. u.. k.. v.. (d) Derive general formulas for estimating two missing values when the observations are in the same block. Suose that two observations y ij and y kj are missing, i k (same block j). SS E = y SSE SSE From = = 0, we obtain y y ij whose simultaneous solution is: y ij kj ( y + y ) + ( y + y ) ( y + y + y ) ( y + y y ) i. ij k. kj. j ij kj.. ij + ij + ykj + ( a )( b ) ŷ ŷ ij kj b ayi. + by. j y.. = + ŷ b ( a )( b ) ( a )( b ) kj a ( a )( ) ayk. + by. j y.. = + ŷ b ij ( a )( ) ( b ) ay k. + by. j y.. + ( a )( b ) ( ay i. + by. j y..) ay + by y = + i.. j.. 4 y kj ( a ) ( b ) ( ) ( ) 4 ( a )( b ) ( a ) ( b ) ay k. + by. j y.. b a ay i. + by. j y.. = ab kj 4.9. An industrial engineer is conducting an exeriment on eye focus time. He is interested in the effect of the distance of the object from the eye on the focus time. Four different distances are of interest. He has five subjects available for the exeriment. Because there may be differences among individuals, he decides to conduct the exeriment in a randomized block design. The data obtained follow. Analyze the data from this exeriment (use α = 0.05) and draw aroriate conclusions. Subject Distance (ft) 3 4 5 4 0 6 6 6 6 6 7 6 6 6 8 5 3 3 5 0 6 4 4 3 4-6

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Design Exert Outut Resonse: Focus Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Block 36.30 4 9.07 Model 3.95 3 0.98 8.6 0.005 significant A 3.95 3 0.98 8.6 0.005 Residual 5.30.7 Cor Total 84.55 9 The Model F-value of 8.6 imlies the model is significant. There is only a 0.5% chance that a "Model F-Value" this large could occur due to noise. Std. Dev..3 R-Squared 0.689 Mean 4.85 Adj R-Squared 0.6036 C.V. 3.8 Pred R-Squared 0.9 PRESS 4.50 Adeq Precision 0.43 Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error -4 6.80 0.50-6 5.0 0.50 3-8 3.60 0.50 4-0 3.80 0.50 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > t vs.60 0.7.4 0.0448 vs 3 3.0 0.7 4.48 0.0008 vs 4 3.00 0.7 4.0 0.00 vs 3.60 0.7.4 0.0448 vs 4.40 0.7.96 0.0736 3 vs 4-0.0 0.7-0.8 0.784 Distance has a statistically significant effect on mean focus time. 4.0. The effect of five different ingredients (A, B, C, D, E) on reaction time of a chemical rocess is being studied. Each batch of new material is only large enough to ermit five runs to be made. Furthermore, each run requires aroximately / hours, so only five runs can be made in one day. The exerimenter decides to run the exeriment as a Latin square so that day and batch effects can be systematically controlled. She obtains the data that follow. Analyze the data from this exeriment (use α = 0.05) and draw conclusions. Day Batch 3 4 5 A=8 B=7 D= C=7 E=3 C= E= A=7 D=3 B=8 3 B=4 A=9 C=0 E= D=5 4 D=6 C=8 E=6 B=6 A=0 5 E=4 D= B=3 A=8 C=8 The Minitab outut below identifies the ingredients as having a significant effect on reaction time. 4-7

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Minitab Outut General Linear Model Factor Tye Levels Values Batch random 5 3 4 5 Day random 5 3 4 5 Catalyst fixed 5 A B C D E Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Catalyst 4 4.440 4.440 35.360.3 0.000 Batch 4 5.440 5.440 3.860.3 0.348 Day 4.40.40 3.060 0.98 0.455 Error 37.50 37.50 3.7 Total 4 06.640 4.. An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the assembly time for a color television comonent. Four oerators are selected for the study. Furthermore, the engineer knows that each assembly method roduces such fatigue that the time required for the last assembly may be greater than the time required for the first, regardless of the method. That is, a trend develos in the required assembly time. To account for this source of variability, the engineer uses the Latin square design shown below. Analyze the data from this exeriment (α = 0.05) draw aroriate conclusions. Order of Oerator Assembly 3 4 C=0 D=4 A=7 B=8 B=7 C=8 D= A=8 3 A=5 B=0 C= D=9 4 D=0 A=0 B= C=4 The Minitab outut below identifies assembly method as having a significant effect on assembly time. Minitab Outut General Linear Model Factor Tye Levels Values Order random 4 3 4 Oerator random 4 3 4 Method fixed 4 A B C D Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Method 3 7.500 7.500 4.67 3.8 0.004 Order 3 8.500 8.500 6.67 3.5 0.089 Oerator 3 5.500 5.500 7.67 9.8 0.00 Error 6 0.500 0.500.750 Total 5 53.000 4.. Suose that in Problem 4.0 the observation from batch 3 on day 4 is missing. Estimate the missing value from Equation 4.4, and erform the analysis using this value. y 354 is missing. ŷ [ y + y + y ] ( )( ) y [ + 5 + 4] ( 46) ()() 3 4 5 8 = i... j...k... 354 =. = 3 58 4-8

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Let y ijk be missing. Then ( y i.. + yijk ) ( y. j. + yijk ) ( y.. k + yijk ) ( y... + yijk ) SSE = yijk + + R SS E where R is all terms without y ijk.. From = 0, we obtain: y ijk y ijk ( )( ) ( y' i.. + y'. j. + y'..k ) = y'..., or y ijk = ( y' + y' + y' ) i... j...k ( )( ) y'... 4.5. Designs involving several Latin squares. [See Cochran and Cox (957), John (97).] The x Latin square contains only observations for each treatment. To obtain more relications the exerimenter may use several squares, say n. It is immaterial whether the squares used are the same are different. The aroriate model is y ijkh = + ρ h + α i( h ) + τ j + β k( h ) + ( τρ ) jh μ + ε ijkh i =,,..., j =,,..., k =,,..., h =,,...,n where y ijkh is the observation on treatment j in row i and column k of the hth square. Note that α ih ( ) and β k( h) are row and column effects in the hth square, and ρ h is the effect of the hth square, and ( τρ) jh is the interaction between treatments and squares. (a) Set u the normal equations for this model, and solve for estimates of the model arameters. Assume that aroriate side conditions on the arameters are ρ = 0, α ( ) = 0, and β ( ) = 0 j for each h, τ = 0, ( ρ ) = 0 for each h, and j j jh h h τ ( ) = 0 μ = y... ρ = y y h... h... τ = y y j α β. j..... = y y i( h) i.. h... h = y y k( h).. kh... h ^ τρ jh = y y y + y h jh. jh.. j..... h... i i h τ ρ for each j. k k h 4-30

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY (b) Write down the analysis of variance table for this design. Source SS DF y.... j.. Treatments y n n y h Squares y... n -... n- y. j.h y... Treatment x Squares n SSTreatments SSSquares (-)(n-) Rows yi.. h y... h n n(-) Columns y.. kh y... h n n(-) Error subtraction n(-)(-) y... Total y ijkh n n - 4.6. Discuss how the oerating characteristics curves in the Aendix may be used with the Latin square design. For the fixed effects model use: Φ τ j = τ = j σ σ, = υ υ = ( )( ) For the random effects model use: λ = + σ τ σ, υ = υ = ( )( ) 4.7. Suose that in Problem 4.0 the data taken on day 5 were incorrectly analyzed and had to be discarded. Develo an aroriate analysis for the remaining data. Two methods of analysis exist: () Use the general regression significance test, or () recognize that the design is a Youden square. The data can be analyzed as a balanced incomlete block design with a = b = 5, r = k = 4 and λ = 3. Using either aroach will yield the same analysis of variance. Minitab Outut General Linear Model Factor Tye Levels Values Catalyst fixed 5 A B C D E Batch random 5 3 4 5 Day random 4 3 4 4-3

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Catalyst 4 9.800 0.67 30.04 7.48 0.008 Batch 4.667.667.97 0.73 0.598 Day 3 6.950 6.950.37 0.58 0.646 Error 8 3.33 3.33 4.07 Total 9 70.550 4.8. The yield of a chemical rocess was measured using five batches of raw material, five acid concentrations, five standing times, (A, B, C, D, E) and five catalyst concentrations (α, β, γ, δ, ε). The Graeco-Latin square that follows was used. Analyze the data from this exeriment (use α = 0.05) and draw conclusions. Acid Concentration Batch 3 4 5 Aα=6 Bβ=6 Cγ=9 Dδ=6 Eε=3 Bγ=8 Cδ= Dε=8 Eα= Aβ= 3 Cε=0 Dα= Eβ=6 Aγ=5 Bδ=3 4 Dβ=5 Eγ=5 Aδ= Bε=4 Cα=7 5 Eδ=0 Aε=4 Bα=7 Cβ=7 Dγ=4 The Minitab outut below identifies standing time as having a significant effect on yield. Minitab Outut General Linear Model Factor Tye Levels Values Time fixed 5 A B C D E Catalyst random 5 a b c d e Batch random 5 3 4 5 Acid random 5 3 4 5 Analysis of Variance for Yield, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Time 4 34.800 34.800 85.700 4.65 0.00 Catalyst 4.000.000 3.000 0.5 0.79 Batch 4 0.000 0.000.500 0.43 0.785 Acid 4 4.400 4.400 6.00.04 0.443 Error 8 46.800 46.800 5.850 Total 4 436.000 4.9. Suose that in Problem 4. the engineer susects that the worklaces used by the four oerators may reresent an additional source of variation. A fourth factor, worklace (α, β, γ, δ) may be introduced and another exeriment conducted, yielding the Graeco-Latin square that follows. Analyze the data from this exeriment (use α = 0.05) and draw conclusions. Order of Oerator Assembly 3 4 Cβ= Bγ=0 Dδ=4 Aα=8 Bα=8 Cδ= Aγ=0 Dβ= 3 Aδ=9 Dα= Bβ=7 Cγ=5 4 Dγ=9 Aβ=8 Cα=8 Bδ=6 4-3

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY Minitab Outut General Linear Model Factor Tye Levels Values Method fixed 4 A B C D Order random 4 3 4 Oerator random 4 3 4 Worklac random 4 a b c d Analysis of Variance for Time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Method 3 95.500 95.500 3.833 3.47 0.67 Order 3 0.500 0.500 0.67 0.0 0.996 Oerator 3 9.000 9.000 6.333 0.69 0.66 Worklac 3 7.500 7.500.500 0.7 0.843 Error 3 7.500 7.500 9.67 Total 5 50.000 Method and worklace do not have a significant effect on assembly time. However, there are only three degrees of freedom for error, so the test is not very sensitive. 4.30. Construct a 5 x 5 hyersquare for studying the effects of five factors. Exhibit the analysis of variance table for this design. Three 5 x 5 orthogonal Latin Squares are: ABCDE BCDEA CDEAB DEABC EABCD αβγδε γδεαβ εαβγδ βγδεα δεαβγ 345 453 345 534 345 Let rows = factor, columns = factor, Latin letters = factor 3, Greek letters = factor 4 and numbers = factor 5. The analysis of variance table is: Source SS DF Rows 5 y... yi... 5 i= 5 4 Columns 5 y... y... m 5 m= 5 4 Latin Letters 5 y... y. j... 5 j= 5 4 Greek Letters 5 y... y.. k.. 5 k = 5 4 Numbers 5 y... y... l. 5 4 l= 5 Error SS E by subtraction 4 5 5 5 5 5 y... Total y 4 ijklm i= j= k= l= m= 5 4-33

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.3. Consider the data in Problems 4. and 4.9. Suressing the Greek letters in 4.9, analyze the data using the method develoed in Problem 4.5. Square - Oerator Batch 3 4 Row Total C=0 D=4 A=7 B=8 (39) B=7 C=8 D= A=8 (44) 3 A=5 B=0 C= D=9 (35) 4 D=0 A=0 B= C=4 (46) (3) (5) (4) (36) 64=y Square - Oerator Batch 3 4 Row Total C= B=0 D=4 A=8 (43) B=8 C= A=0 D= (4) 3 A=9 D= B=7 C=5 (4) 4 D=9 A=8 C=8 B=6 (4) (37) (4) (49) (4) 68=y Assembly Methods Totals A y... =65 B y... =68 C y.3.. =09 D y.4.. =90 Source SS DF MS F 0 Assembly Methods 59.5 3 53.08 Squares 0.50 0.50 4.00* A x S 8.75 3.9 0.77 Assembly Order (Rows) 9.00 6 3.7 Oerators (columns) 70.50 6.75 Error 45.50 3.79 Total 303.50 3 Significant at %. 4.3. Consider the randomized block design with one missing value in Problem 4.7. Analyze this data by using the exact analysis of the missing value roblem discussed in Section 4..4. Comare your results to the aroximate analysis of these data given in Table 4.7. To simlify the calculations, the data in Problems 4.7 and 4.7, the data was transformed by multilying by 0 and substracting 95. 4-34

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY μ: 5 μ + 4 τ + 3 τ + 4 τ3 + 4 τ4 + 4 β + 4 β + 3 β 3 + 4β4 = 7 τ: 4 μ + 4 τ + β + β + β3 + β4 = 3 τ : 3μ + 3τ + β + β + β 4 = τ 3: 4μ + 4τ3 + β + β + β3 + β4 = τ 4 : 4μ + 4τ4 + β + β + β3 + β4 = 5 β : 4μ + τ + τ + τ3 + τ4 + 4β = 4 β : 4μ + τ + τ + τ 3 + τ 4 + 4β β 3: 3μ + τ + τ3 + τ4 + 3β3 = 3 = 6 β : 4 μ + τ + τ + τ + τ + 4 β = 8 4 3 4 4 Alying the constraints τ β = i = 0, we obtain: j 4 4 4 59 94 μ =, τ =, τ =, τ 3 =, τ 4 =, 36 36 36 36 36 77 β =, 36 68 β =, 36 4 β 3 =, 36 β 4 = 36 R With 7 degrees of freedom. 4 4.. β j y. j =. i= j= ( μ, τ, β ) = μy + τi yi. + = 45. 00 38 78 ij = 45. 00 38. 78 = y ij, SS y R( μ, τ, β ) E = 6. which is identical to SS E obtained in the aroximate analysis. In general, the SS E in the exact and aroximate analyses will be the same. To test H o : τ = 0 the reduced model is y = μ + β + ε. The normal equations used are: i ij j ij μ: 5 μ + 4 β + 4 β + 3β3 + 4β4 = 7 β : 4μ + 4β β : 4μ + 4β β 3: 3μ + 3β3 = 4 = 3 = 6 β : 4 μ + 4 β = 8 4 4 Alying the constraint β j = 0, we obtain: 9 μ =, 6 35 β =, 6 3 β =, 6 3 β 3 =, 6 4 53 β 4 =. Now R( μ, β ) = μy +.. β j y. j =. 6 j= 99 5 with 4 degrees of freedom. ( τ μ, β ) = R( μ, τ, β ) R( μ, β ) = 38. 78 99. 5 = 39. 53 SSTreatments R = 4-35

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY with 7-4=3 degrees of freedom. ( τ μ, β ) R is used to test H o :τ i = 0. The sum of squares for blocks is found from the reduced model used are: yij = μ + τ + ε. The normal equations i ij Model Restricted to β j = 0 : μ: 5 μ + 4 τ + 3 τ + 4τ3 + 4τ4 τ : 4μ + 4τ τ : 3μ + 3τ τ 3: 4μ + 4τ3 = 7 = 3 = = τ : 4 μ + 4 τ = 5 Alying the constraint with 4 degrees of freedom. τ i = 0 with 7-4=3 degrees of freedom. 4 4, we obtain: 3 4 9 9 3 μ =, τ =, τ =, τ3 =, τ 4 = R ( μ, τ ) = μy + 4.. τi yi. = 59. i= ( β μ, τ ) = R( μ, τ, β ) R( μ, τ ) = 38. 78 59. 83 = 78. 95 SS Blocks R = 83 Source DF SS(exact) SS(aroximate) Tis 3 39.53 39.98 Blocks 3 78.95 79.53 Error 8 6. 6. Total 4 5.74 5.73 Note that for the exact analysis, SS SS + SS + SS. T Tis Blocks E 4.33. An engineer is studying the mileage erformance characteristics of five tyes of gasoline additives. In the road test he wishes to use cars as blocks; however, because of a time constraint, he must use an incomlete block design. He runs the balanced design with the five blocks that follow. Analyze the data from this exeriment (use α = 0.05) and draw conclusions. Car Additive 3 4 5 7 4 3 4 4 3 0 3 3 9 4 3 5 0 8 4-36

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY There are several comuter software ackages that can analyze the incomlete block designs discussed in this chater. The Minitab General Linear Model rocedure is a widely available ackage with this caability. The outut from this routine for Problem 4.33 follows. The adjusted sums of squares are the aroriate sums of squares to use for testing the difference between the means of the gasoline additives. The gasoline additives have a significant effect on the mileage. Minitab Outut General Linear Model Factor Tye Levels Values Additive fixed 5 3 4 5 Car random 5 3 4 5 Analysis of Variance for Mileage, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Additive 4 3.7000 35.7333 8.9333 9.8 0.00 Car 4 35.333 35.333 8.8083 9.67 0.00 Error 0.067 0.067 0.906 Total 9 76.9500 4.34. Construct a set of orthogonal contrasts for the data in Problem 4.33. Comute the sum of squares for each contrast. One ossible set of orthogonal contrasts is: H 0 : μ 4 + μ5 = μ + μ () H 0 : μ = μ () H 0 : μ 4 = μ5 (3) H 0 : 4μ 3 = μ4 + μ5 + μ + μ (4) The sums of squares and F-tests are: Brand -> 3 4 5 Q i 33/4 /4-3/4-4/4-7/4 cq i i SS F 0 () - - 0-85/4 30.0 33.06 () - 0 0 0 /4 4.03 4.46 (3) 0 0 0 - -3/4.4.55 (4) - - 4 - - -5/4 0.9 0. Contrasts () and () are significant at the % and 5% levels, resectively. 4-37

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY 4.35. Seven different hardwood concentrations are being studied to determine their effect on the strength of the aer roduced. However the ilot lant can only roduce three runs each day. As days may differ, the analyst uses the balanced incomlete block design that follows. Analyze this exeriment (use α = 0.05) and draw conclusions. Hardwood Days Concentration (%) 3 4 5 6 7 4 0 7 4 6 0 9 6 37 7 34 8 4 9 49 0 45 50 43 0 8 3 4 36 30 7 There are several comuter software ackages that can analyze the incomlete block designs discussed in this chater. The Minitab General Linear Model rocedure is a widely available ackage with this caability. The outut from this routine for Problem 4.35 follows. The adjusted sums of squares are the aroriate sums of squares to use for testing the difference between the means of the hardwood concentrations. Minitab Outut General Linear Model Factor Tye Levels Values Concentr fixed 7 4 6 8 0 4 Days random 7 3 4 5 6 7 Analysis of Variance for Strength, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Concentr 6 037.6 37.43 9.57 0.4 0.00 Days 6 394.0 394.0 65.68 3. 0.070 Error 8 68.57 68.57.07 Total 0 600.9 4.36. Analyze the data in Examle 4.5 using the general regression significance test. μ: μ + 3 τ + 3 τ + 3 τ3 + 3 τ4 + 3 β + 3 β + 3 β 3 + 3β4 = 870 τ: 3 μ + 3 τ + β + β + β4 = 8 τ : 3μ + 3τ + β + β3 + β4 = 4 τ 3: 3μ + 3τ3 + β + β + β3 = 6 τ 4 : 3μ + 3τ4 + β + β3 + β4 = β : 3μ + τ + τ3 + τ4 + 3β = β : 3 μ + τ + τ + τ 3 + 3 β = 4 β 3: 3μ + τ + τ3 + τ4 + 3β3 = 07 β 4 : 3μ + τ + τ + τ4 + 3β4 = 8 Alying the constraints $ $ τ = i β j 0, we obtain: $ μ = 870 /, $ τ = 9/ 8, $ τ = 7/ 8, $ τ 3 = 4/ 8, $ τ 4 = 0 / 8, 4-38

Solutions from Montgomery, D. C. (008) Design and Analysis of Exeriments, Wiley, NY $ β = 7/ 8, β = 4/8, β 3 = 3/8, $ β 4 = 0/ 8 ( ) with 7 degrees of freedom. 4 4 R μτ,, β = μy + τy + β y = 63,5.75 y ij,. E ij = 6356 00.. i i. j. j i= j= SS = y R( μτ,, β ) = 6356. 00 635. 75 = 3. 5. To test H o :τ i = 0 the reduced model is y ij = μ + β j + ε ij. The normal equations used are: μ: μ + 3 β + 3 β + 3β3 + 3β4 = 870 β : 3μ + 3β β : 3μ + 3β β 3: 3μ + 3β3 = = 4 = 07 β : 3 μ + 3 β = 8 4 4 Alying the constraint β j = 0, we obtain: 870 μ =, R with 4 degrees of freedom. 7 β =, 3 β =, β3 =, 6 6 6 ( μ, β ) = μy + 4.. β j y. j = 6330,. j= 00 β 4 = 6 ( τ μ, β ) = R( μ, τ, β ) R( μ, β ) = 635. 75 6330. 00 =. 75 SSTreatments R = with 7 4 = 3 degrees of freedom. ( τ μ, β ) R is used to test H o : τ = 0. i The sum of squares for blocks is found from the reduced model used are: yij = μ + τ + ε. The normal equations i ij Model Restricted to β j = 0 : μ: μ + 3 τ + 3 τ + 3τ3 + 3τ4 τ : 3μ + 3τ τ : 3μ + 3τ τ 3: 3μ + 3τ3 = 870 = 8 = 4 = 6 τ : 3 μ + 3 τ = 4 4 The sum of squares for blocks is found as in Examle 4.5. We may use the method shown above to find an adjusted sum of squares for blocks from the reduced model, y = μ + τ + ε. ij i ij 4-39