Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Chapter 13 Experiments with Random Factors Solutions 13.. An article by Hoof and Berman ( Statistical Analysis of Power Module Thermal Test Equipment Performance, IEEE Transactions on Components, Hybrids, and Manufacturing Technology Vol. 11, pp. 516-50, 1988) describes an experiment conducted to investigate the capability of measurements on thermal impedance (Cº/W x 100) on a power module for an induction motor starter. There are 10 parts, three operators, and three replicates. The data are shown in Table P13.. Table P13. Part Inspector 1 Inspector Inspector 3 Number Test 1 Test Test 3 Test 1 Test Test 3 Test 1 Test Test 3 1 37 38 37 41 41 40 41 4 41 4 41 43 4 4 4 43 4 43 3 30 31 31 31 31 31 9 30 8 4 4 43 4 43 43 43 4 4 4 5 8 30 9 9 30 9 31 9 9 6 4 4 43 45 45 45 44 46 45 7 5 6 7 8 8 30 9 7 7 8 40 40 40 43 4 4 43 43 41 9 5 5 5 7 9 8 6 6 6 10 35 34 34 35 35 34 35 34 35 (a) Analyze the data from this experiment, assuming both parts and operators are random effects. ANOVA: Impedance versus Inspector, Part Inspecto random 3 1 3 Part random 10 1 3 4 5 6 7 8 9 10 Analysis of Variance for Impedanc Source DF SS F P Inspecto 39.7 19.63 7.8 0.005 Part 9 3935.96 437.33 16.7 0.000 Inspecto*Part 18 48.51.70 5.7 0.000 Error 60 30.67 0.51 Total 89 4054.40 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Inspecto 0.5646 3 (4) + 3(3) + 30(1) Part 48.96 3 (4) + 3(3) + 9() 3 Inspecto*Part 0.780 4 (4) + 3(3) 4 Error 0.5111 (4) (b) Estimate the variance components using the analysis of variance method. 13-1
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY ˆ E n an bn E ˆ B ˆ A ˆ All estimates agree with the Minitab output. ˆ 0.51.70 0.51 ˆ 0.73 3 437.33.70 ˆ 48.9 33 19.63.70 ˆ 0.56 10 3 13.3. Reconsider the data in Problem 5.8. Suppose that both factors, machines and operators, are chosen at random. (a) Analyze the data from this experiment. Machine Operator 1 3 4 1 109 110 108 110 110 115 109 108 110 110 111 114 11 111 109 11 3 116 11 114 10 114 115 119 117 The following Minitab output contains the analysis of variance and the variance component estimates: ANOVA: Strength versus Operator, Machine Operator random 3 1 3 Machine random 4 1 3 4 Analysis of Variance for Strength Source DF SS F P Operator 160.333 80.167 10.77 0.010 Machine 3 1.458 4.153 0.56 0.66 Operator*Machine 6 44.667 7.444 1.96 0.151 Error 1 45.500 3.79 Total 3 6.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator 9.0903 3 (4) + (3) + 8(1) Machine -0.5486 3 (4) + (3) + 6() 3 Operator*Machine 1.864 4 (4) + (3) 4 Error 3.7917 (4) (b) Find point estimates of the variance components using the analysis of variance method. ˆ E ˆ 3.79167 13-
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY n an A ˆ bn E ˆ B ˆ 7.44444 3.79167 ˆ 1.8639 4.1578 7.44444 ˆ 0, assume 3() 80.16667 7.44444 ˆ 9.0908 4() These results agree with the Minitab variance component analysis. ˆ 0 13.4. Reconsider the data in Problem 5.15. Suppose that both factors are random. (a) Analyze the data from this experiment. General Linear Model: Response versus Row, Column Row random 3 1 3 Column random 4 1 3 4 Column Factor Row Factor 1 3 4 1 36 39 36 3 18 0 0 3 30 37 33 34 Analysis of Variance for Response, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj F P Row 580.500 580.500 90.50 60.40 ** Column 3 8.917 8.917 9.639.01 ** Row*Column 6 8.833 8.833 4.806 ** Error 0 0.000 0.000 0.000 Total 11 638.50 ** Denominator of F-test is zero. Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term 1 Row (4) + (3) + 4.0000(1) Column (4) + (3) + 3.0000() 3 Row*Column (4) + (3) 4 Error (4) Error Terms for Tests, using Adjusted SS Source Error DF Error Synthesis of Error 1 Row * 4.806 (3) Column * 4.806 (3) 3 Row*Column * * (4) Variance Components, using Adjusted SS Source Estimated Value Row 71.3611 Column 1.6111 Row*Column 4.8056 Error 0.0000 13-3
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY (b) Estimate the variance components. Because the experiment is unreplicated and the interaction term was included in the model, there is no estimate of E, and therefore, no estimate of. n an E ˆ B ˆ bn A ˆ These estimates agree with the Minitab output. 4.8056 0 ˆ 4.8056 1 9.6389 4.8056 ˆ 1.6111 31 90.500 4.8056 ˆ 71.3611 41 13.5. Suppose that in Problem 5.13 the furnace positions were randomly selected, resulting in a mixed model experiment. Reanalyze the data from this experiment under this new assumption. Estimate the appropriate model components. The following analysis assumes a restricted model: ANOVA: Density versus Position, Temperature Position random 1 Temperat fixed 3 800 85 850 Analysis of Variance for Density Temperature ( C) Position 800 85 850 570 1063 565 1 565 1080 510 583 1043 590 58 988 56 547 106 538 51 1004 53 Source DF SS F P Position 1 7160 7160 16.00 0.00 Temperat 94534 47671 1155.5 0.001 Position*Temperat 818 409 0.91 0.47 Error 1 5371 448 Total 17 958691 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Position 745.83 4 (4) + 9(1) Temperat 3 (4) + 3(3) + 6Q[] 3 Position*Temperat -1.83 4 (4) + 3(3) 4 Error 447.56 (4) ˆ E ˆ 447.56 13-4
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY n bn E ˆ A E ˆ These results agree with the Minitab output. 409 448 ˆ 0 assume ˆ 0 3 7160 448 ˆ 745.83 33 13.6. Reanalyze the measurement systems experiment in Problem 13.1, assuming that operators are a fixed factor. Estimate the appropriate model components. Table P13.1 Operator 1 Operator Part Measurements Measurements Number 1 3 1 3 1 50 49 50 50 48 51 5 5 51 51 51 51 3 53 50 50 54 5 51 4 49 51 50 48 50 51 5 48 49 48 48 49 48 6 5 50 50 5 50 50 7 51 51 51 51 50 50 8 5 50 49 53 48 50 9 50 51 50 51 48 49 10 47 46 49 46 47 48 The following analysis assumes a restricted model: ANOVA: Measurement versus Part, Operator Part random 10 1 3 4 5 6 7 8 9 10 Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part 9 99.017 11.00 7.33 0.000 Operator 1 0.417 0.417 0.69 0.47 Part*Operator 9 5.417 0.60 0.40 0.97 Error 40 60.000 1.500 Total 59 164.850 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Part 1.5836 4 (4) + 6(1) Operator 3 (4) + 3(3) + 30Q[] 3 Part*Operator -0.994 4 (4) + 3(3) 4 Error 1.5000 (4) n E ˆ ˆ E ˆ 1.5000 0.60185 1.5000 ˆ 0 assume ˆ 0 3 13-5
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY bn A E ˆ 11.00185 1.50000 ˆ 1.58364 3 These results agree with the Minitab output. 13.7. Reanalyze the measurement system experiment in Problem 13., assuming that operators are a fixed factor. Estimate the appropriate model components. Table P13. Part Inspector 1 Inspector Inspector 3 Number Test 1 Test Test 3 Test 1 Test Test 3 Test 1 Test Test 3 1 37 38 37 41 41 40 41 4 41 4 41 43 4 4 4 43 4 43 3 30 31 31 31 31 31 9 30 8 4 4 43 4 43 43 43 4 4 4 5 8 30 9 9 30 9 31 9 9 6 4 4 43 45 45 45 44 46 45 7 5 6 7 8 8 30 9 7 7 8 40 40 40 43 4 4 43 43 41 9 5 5 5 7 9 8 6 6 6 10 35 34 34 35 35 34 35 34 35 ANOVA: Impedance versus Inspector, Part Inspecto fixed 3 1 3 Part random 10 1 3 4 5 6 7 8 9 10 Analysis of Variance for Impedanc Source DF SS F P Inspecto 39.7 19.63 7.8 0.005 Part 9 3935.96 437.33 855.64 0.000 Inspecto*Part 18 48.51.70 5.7 0.000 Error 60 30.67 0.51 Total 89 4054.40 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Inspecto 3 (4) + 3(3) + 30Q[1] Part 48.5353 4 (4) + 9() 3 Inspecto*Part 0.780 4 (4) + 3(3) 4 Error 0.5111 (4) ˆ ˆ E 0.51.70 0.51 ˆ 0.73 n 3 437.33 0.51 ˆ 48.54 an 33 E ˆ B E ˆ These results agree with the Minitab output. 13-6
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 13.8. In problem 5.8, suppose that there are only four machines of interest, but the operators were selected at random. (a) What type of model is appropriate? A mixed model is appropriate. Operator Machine 1 3 4 1 109 110 108 110 110 115 109 108 110 110 111 114 11 111 109 11 3 116 11 114 10 114 115 119 117 (b) Perform the analysis and estimate the model components. The following analysis assumes a restricted model: ANOVA: Strength versus Operator, Machine Operator random 3 1 3 Machine fixed 4 1 3 4 Analysis of Variance for Strength Source DF SS F P Operator 160.333 80.167 1.14 0.000 Machine 3 1.458 4.153 0.56 0.66 Operator*Machine 6 44.667 7.444 1.96 0.151 Error 1 45.500 3.79 Total 3 6.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator 9.547 4 (4) + 8(1) Machine 3 (4) + (3) + 6Q[] 3 Operator*Machine 1.86 4 (4) + (3) 4 Error 3.79 (4) ˆ n bn E ˆ A E ˆ These results agree with the Minitab output. E ˆ 3.79 7.444 3.79 ˆ 1.86 80.167 3.79 ˆ 9.547 4 13.9. Rework Problem 13.5 using the REML method. 13-7
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Temperature ( C) Position 800 85 850 570 1063 565 1 565 1080 510 583 1043 590 58 988 56 547 106 538 51 1004 53 The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.5. JMP Output RSquare 0.993347 RSquare Adj 0.9946 Root Mean Square Error 1.15551 Mean of Response 709.9444 Observations (or Sum Wgts) 18 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 709.94444 19.94444 1 35.60 0.0179* Temperature[800] -157.6111 6.741707-3.38 0.0018* Temperature[85] 34.05556 6.741707 48.07 0.0004* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Position 1.6760179 750.11111 116.0118-1456.83 957.0538 63.309 Position*Temperature -0.08674-1.83333 149.33585-305.56 79.85956-1.083 Residual 447.55556 18.71379 30.13858 119.556 37.774 Total 1184.8333 100.000 Covariance Matrix of Variance Component Estimates Random Effect Position Position*Temperature Residual Position 16790.7-6197.76-1.41e-9 Position*Temperature -6197.76 301.197-1118.11 Residual -1.41e-9-1118.11 33384.39 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Temperature 1155.518 0.0009* 13.10. Rework Problem 13.6 using the REML method. 13-8
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Table P13.1 Operator 1 Operator Part Measurements Measurements Number 1 3 1 3 1 50 49 50 50 48 51 5 5 51 51 51 51 3 53 50 50 54 5 51 4 49 51 50 48 50 51 5 48 49 48 48 49 48 6 5 50 50 5 50 50 7 51 51 51 51 50 50 8 5 50 49 53 48 50 9 50 51 50 51 48 49 10 47 46 49 46 47 48 The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.6. JMP Output RSquare 0.40766 RSquare Adj 0.410779 Root Mean Square Error 1.4745 Mean of Response 49.95 Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 49.95 0.481 9 116.65 <.0001* Operator[Operator 1] 0.0833333 0.100154 9 0.83 0.469 REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number 1.1555556 1.7333333 0.8656795 0.036636 3.430034 59.078 Part Number*Operator -0.199588-0.99383 0.146437-0.586394-0.01371-10.04 Residual 1.5 0.335410 1.0110933.455691 51.16 Total.9339506 100.000 Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Operator Residual Part Number 0.749401-0.00447-1.43e-14 Part Number*Operator -0.00447 0.014438-0.0375 Residual -1.43e-14-0.0375 0.115 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Operator 1 1 9 0.693 0.469 13.11. Rework Problem 13.7 using the REML method. Table P13. Part Inspector 1 Inspector Inspector 3 Number Test 1 Test Test 3 Test 1 Test Test 3 Test 1 Test Test 3 1 37 38 37 41 41 40 41 4 41 4 41 43 4 4 4 43 4 43 3 30 31 31 31 31 31 9 30 8 4 4 43 4 43 43 43 4 4 4 13-9
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 5 8 30 9 9 30 9 31 9 9 6 4 4 43 45 45 45 44 46 45 7 5 6 7 8 8 30 9 7 7 8 40 40 40 43 4 4 43 43 41 9 5 5 5 7 9 8 6 6 6 10 35 34 34 35 35 34 35 34 35 The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.7. JMP Output RSquare 0.99005 RSquare Adj 0.99181 Root Mean Square Error 0.7149 Mean of Response 35.8 Observations (or Sum Wgts) 90 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 35.8.0436 9 16.4 <.0001* Inspector[Inspector 1] -0.9 0.4475 18-3.68 0.0017* Inspector[Inspector ] 0.6666667 0.4475 18.7 0.0139* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number 94.485507 48.9593.90677 3.396334 93.18895 97.498 Part Number*Inspector 1.443156 0.779835 0.301065 0.1379119 1.318055 1.470 Residual 0.5111111 0.0933157 0.3681575 0.757543 1.03 Total 49.531687 100.000 Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Inspector Residual Part Number 54.71813-0.0989 -.76e-13 Part Number*Inspector -0.0989 0.0906386-0.00903 Residual -.76e-13-0.00903 0.0087078 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Inspector 18 7.849 0.0048* 13.1. Rework Problem 13.8 using the REML method. Operator Machine 1 3 4 1 109 110 108 110 110 115 109 108 110 110 111 114 11 111 109 11 3 116 11 114 10 114 115 119 117 The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.8. JMP Output 13-10
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY RSquare 0.78154 RSquare Adj 0.748771 Root Mean Square Error 1.947 Mean of Response 11.917 Observations (or Sum Wgts) 4 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 11.9167 1.87643 61.44 0.0003* Machine[1] -0.458333 0.964653 6-0.48 0.6515 Machine[] -0.15 0.964653 6-0.13 0.9011 Machine[3] -0.65 0.964653 6-0.65 0.5410 REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Operator.3974359 9.090778 10.0355-10.5784 8.758958 61.804 Operator*Machine 0.481685 1.863889.841505 -.650464 6.303416 1.417 Residual 3.7916667 1.5479414 1.949717 10.33013 5.779 Total 14.708333 100.000 Covariance Matrix of Variance Component Estimates Random Effect Operator Operator*Machine Residual Operator 100.70575-1.154578 1.686e-1 Operator*Machine -1.154578 5.173434-1.198061 Residual 1.686e-1-1.198061.39617 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Machine 3 3 6 0.5578 0.6619 13.13. By application of the expectation operator, develop the expected mean squares for the two-factor factorial, mixed model. Use the restricted model assumptions. Check your results with the expected mean squares given in Equation 13.9 to see that they agree. The sums of squares may be written as SS n SS a A bn b a yi.. y... i1, SSB any. j. y... b j1 y ij. yi.. y. j. y..., SSE y ijk y... i1 j1 a b n i1 j1 k 1 Using the model yijk i j ijk y y y y ij, we may find that i.. i i. i... j. j. j. ij. i j ij ij........ Using the assumptions for the restricted form of the mixed model,. 0 0... Substituting these expressions into the sums of squares yields, 0. j, which imply that 13-11
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY SS SS SS SS a b a b a b n bn A i i. i..... i1 B j. j.... j1 an ) n ij i. ij. i... j.... i1 j1 E ijk ij. i1 j1 k 1 E ijk 0 Using the assumption that, V ijk E ijk i' j' k' squares by its degrees of freedom and take the expectation to produce E E E E ( ) 0 a 1b 1, and 0 A i i. a 1 i1 B b j b 1 j1 a b E ij i. i1 j1 E a bn E an n, we may divide each sum of Note that E B and E E are the results given in Table 8-3. We need to simplify A E. Consider E since ij E E A E E crossprodu cts 0 E A A a 1 is 0, a bn a 1 1 bn a i1 A n a i1 i i1 i1 a bn a 1 a i a i. a 1 a i a b NID. Consider E E E E a b E ij i. a 1 b 1 i1 j1 n b 1 a 1 a b b a 1 1 i1 j1 a b n n E and 13-1
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Thus E A and E agree with Equation 13.9. 13.14. Consider the three-factor factorial design in Example 13.5. Propose appropriate test statistics for all main effects and interactions. Repeat for the case where A and B are fixed and C is random. If all three factors are random there are no exact tests on main effects. We could use the following: A : F B : F C : F A B C AC C AC C If A and B are fixed and C is random, the expected mean squares are (assuming the restricted form of the model): F F R R a b c n Factor i j k l E() i 0 b c n j a 0 c n BC C k a b 1 n abn ij 0 0 c n ik jk ijk BC i bn bcn a 1 an acn b 1 n 0 b 1 n bn a 0 1 n an 0 0 1 n n ijkl 1 1 1 1 cn These are exact tests for all effects. j ij a 1b 1 13.15. Consider the experiment in Example 13.6. Analyze the data for the case where A, B, and C are random. ANOVA: Drop versus Temp, Operator, Gauge Temp random 3 60 75 90 Operator random 4 1 3 4 Gauge random 3 1 3 Analysis of Variance for Drop Source DF SS F P Temp 103.36 511.68.30 0.171 x 13-13
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Operator 3 43.8 141.7 0.63 0.616 x Gauge 7.19 3.60 0.06 0.938 x Temp*Operator 6 111.97 0.00 14.59 0.000 Temp*Gauge 4 137.89 34.47.49 0.099 Operator*Gauge 6 09.47 34.91.5 0.081 Temp*Operator*Gauge 1 166.11 13.84 0.65 0.788 Error 36 770.50 1.40 Total 71 3950.3 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Temp 1.044 * (8) + (7) + 8(5) + 6(4) + 4(1) Operator -4.544 * (8) + (7) + 6(6) + 6(4) + 18() 3 Gauge -.164 * (8) + (7) + 6(6) + 8(5) + 4(3) 4 Temp*Operator 31.359 7 (8) + (7) + 6(4) 5 Temp*Gauge.579 7 (8) + (7) + 8(5) 6 Operator*Gauge 3.51 7 (8) + (7) + 6(6) 7 Temp*Operator*Gauge -3.780 8 (8) + (7) 8 Error 1.403 (8) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 Temp 6.97.63 (4) + (5) - (7) Operator 7.09 3.06 (4) + (6) - (7) 3 Gauge 5.98 55.54 (5) + (6) - (7) Since all three factors are random there are no exact tests on main effects. Minitab uses an approximate F test for the these factors. 13.16. Derive the expected mean squares shown in Table 13.11. F R R R a b c n Factor i j k l E() i j 0 b c n n bn a 1 c n an acn k a b 1 n an abn ij ik jk ijk 0 1 c n n cn 0 b 1 n n bn a 1 1 n an 0 1 1 n n ijkl 1 1 1 1 cn i bcn a 1 13.17. Consider a four-factor factorial experiment where factor A is at a levels, factor B is at b levels, factor C is at c levels, factor D is at d levels, and there are n replicates. Write down the sums of squares, the degrees of freedom, and the expected mean squares for the following cases. Assume the restricted model for all mixed models. You may use a computer package such as Minitab. Do exact tests exist for all effects? If not, propose test statistics for those effects that cannot be directly tested. The four factor model is: 13-14
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY y ijklh i j k l ij ik il jk jl kl ijklh ijk ijl jkl ikl ijkl To simplify the expected mean square derivations, let capital Latin letters represent the factor effects or variance components. For example, A bcdn i, or B acdn. a 1 (a) A, B, C, and D are fixed factors. F F F F R a b c d n Factor i j k l h E() 0 b c d n A a 0 c d n a b 0 d n a b c 0 n i j k l ( ) ij B C D 0 0 c d n 0 b 0 d n 0 b c 0 n AD ( ) jk a 0 0 d n BC ( ) ik ( ) il AC ( ) jl a 0 c 0 n BD ( ) kl a b 0 0 n CD ( ) ijk 0 0 0 d n C ( ) ijl 0 0 c 0 n D ( ) jkl a 0 0 0 n BCD ( ) ikl 0 b 0 0 n ACD ( ) ijkl 0 0 0 0 n CD ( ijkl) h 1 1 1 1 1 There are exact tests for all effects. The results can also be generated in Minitab as follows: ANOVA: y versus A, B, C, D A fixed H L B fixed H L C fixed H L D fixed H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 0.49 0.49 B 1 0.13 0.13 0.01 0.91 C 1 1.13 1.13 0.09 0.767 D 1 0.13 0.13 0.01 0.91 A*B 1 3.13 3.13 0.5 0.6 A*C 1 3.13 3.13 0.5 0.6 A*D 1 3.13 3.13 0.5 0.6 B*C 1 3.13 3.13 0.5 0.6 B*D 1 3.13 3.13 0.5 0.6 C*D 1 3.13 3.13 0.5 0.6 A*B*C 1 3.13 3.13 0.5 0.6 13-15
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY A*B*D 1 8.13 8.13.7 0.151 A*C*D 1 3.13 3.13 0.5 0.6 B*C*D 1 3.13 3.13 0.5 0.6 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A 16 (16) + 16Q[1] B 16 (16) + 16Q[] 3 C 16 (16) + 16Q[3] 4 D 16 (16) + 16Q[4] 5 A*B 16 (16) + 8Q[5] 6 A*C 16 (16) + 8Q[6] 7 A*D 16 (16) + 8Q[7] 8 B*C 16 (16) + 8Q[8] 9 B*D 16 (16) + 8Q[9] 10 C*D 16 (16) + 8Q[10] 11 A*B*C 16 (16) + 4Q[11] 1 A*B*D 16 (16) + 4Q[1] 13 A*C*D 16 (16) + 4Q[13] 14 B*C*D 16 (16) + 4Q[14] 15 A*B*C*D 16 (16) + Q[15] 16 Error 1.38 (16) (b) A, B, C, and D are random factors. R R R R R a b c d n Factor i j k l h E() 1 b c d n CD ACD D C AD AC A a 1 c d n CD BCD D C BD BC B a b 1 d n CD ACD BCD C AC BC CD C a b c 1 n CD ACD BCD D BD AD CD D i j k l ( ) ij 1 1 c d n CD C D ( ) ik 1 b 1 d n CD C ACD AC 1 b c 1 n CD D ACD AD ( ) jk a 1 1 d n ( ) il CD C BCD BC ( ) jl a 1 c 1 n CD D BCD BD ( ) kl a b 1 1 n CD ACD BCD CD ( ) ijk 1 1 1 d n CD C ( ) ijl 1 1 c 1 n CD D ( ) jkl a 1 1 1 n CD BCD ( ) ikl 1 b 1 1 n CD ACD ( ) ijkl 1 1 1 1 n CD ( ijkl) h 1 1 1 1 1 No exact tests exist on main effects or two-factor interactions. For main effects use statistics such as: A: F A C D ACD AC AD CD For testing two-factor interactions use statistics such as: : F C CD D 13-16
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY The results can also be generated in Minitab as follows: ANOVA: y versus A, B, C, D A random H L B random H L C random H L D random H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 ** B 1 0.13 0.13 ** C 1 1.13 1.13 0.36 0.843 x D 1 0.13 0.13 ** A*B 1 3.13 3.13 0.11 0.796 x A*C 1 3.13 3.13 1.00 0.667 x A*D 1 3.13 3.13 0.11 0.796 x B*C 1 3.13 3.13 1.00 0.667 x B*D 1 3.13 3.13 0.11 0.796 x C*D 1 3.13 3.13 1.00 0.667 x A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 1.00 0.500 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A 1.7500 * (16) + (15) + 4(13) + 4(1) + 4(11) + 8(7) + 8(6) + 8(5) + 16(1) B 1.3750 * (16) + (15) + 4(14) + 4(1) + 4(11) + 8(9) + 8(8) + 8(5) + 16() 3 C -0.150 * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + (15) + 4(14) + 4(13) + 4(1) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B -3.150 * (16) + (15) + 4(1) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D -3.150 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 0.0000 * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D -3.150 * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D 0.0000 * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + (15) + 4(11) 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D 0.0000 15 (16) + (15) + 4(13) 14 B*C*D 0.0000 15 (16) + (15) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A 0.56 * (5) + (6) + (7) - (11) - (1) - (13) + (15) B 0.56 * (5) + (8) + (9) - (11) - (1) - (14) + (15) 3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B 0.98 8.13 (11) + (1) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 13-17
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 7 A*D 0.98 8.13 (1) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15) 9 B*D 0.98 8.13 (1) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15) (c) A is fixed and B, C, and D are random. F R R R R a b c d n Factor i j k l h E() 0 b c d n CD ACD D C AD AC A a 1 c d n BCD BD BC B a b 1 d n BCD BC CD C a b c 1 n BCD BD CD D i j k l ( ) ij ( ) ik ( ) il ( ) jk ( ) jl ( ) kl ( ) ijk 0 1 c d n CD C D 0 b 1 d n CD C ACD AC 0 b c 1 n CD D ACD AD a 1 1 d n BCD BC a 1 c 1 n BCD BD a b 1 1 n BCD CD 0 1 1 d n CD C ( ) ijl 0 1 c 1 n CD D ( ) jkl a 1 1 1 n BCD ( ) ikl 0 b 1 1 n CD ACD ( ) ijkl 0 1 1 1 n CD ( ijkl) h 1 1 1 1 1 No exact tests exist on main effects or two-factor interactions involving the fixed factor A. To test the fixed factor A use A: F A C D ACD AC AD CD D BCD Random main effects could be tested by, for example: D : F BD CD For testing two-factor interactions involving A use: : F C CD D The results can also be generated in Minitab as follows: 13-18
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY ANOVA: y versus A, B, C, D A fixed H L B random H L C random H L D random H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 ** B 1 0.13 0.13 0.04 0.907 x C 1 1.13 1.13 0.36 0.761 x D 1 0.13 0.13 0.04 0.907 x A*B 1 3.13 3.13 0.11 0.796 x A*C 1 3.13 3.13 1.00 0.667 x A*D 1 3.13 3.13 0.11 0.796 x B*C 1 3.13 3.13 1.00 0.500 B*D 1 3.13 3.13 1.00 0.500 C*D 1 3.13 3.13 1.00 0.500 A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 0.5 0.6 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A * (16) + (15) + 4(13) + 4(1) + 4(11) + 8(7) + 8(6) + 8(5) + 16Q[1] B -0.1875 * (16) + 4(14) + 8(9) + 8(8) + 16() 3 C -0.150 * (16) + 4(14) + 8(10) + 8(8) + 16(3) 4 D -0.1875 * (16) + 4(14) + 8(10) + 8(9) + 16(4) 5 A*B -3.150 * (16) + (15) + 4(1) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D -3.150 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 0.0000 14 (16) + 4(14) + 8(8) 9 B*D 0.0000 14 (16) + 4(14) + 8(9) 10 C*D 0.0000 14 (16) + 4(14) + 8(10) 11 A*B*C 0.0000 15 (16) + (15) + 4(11) 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D 0.0000 15 (16) + (15) + 4(13) 14 B*C*D -.315 16 (16) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A 0.56 * (5) + (6) + (7) - (11) - (1) - (13) + (15) B 0.33 3.13 (8) + (9) - (14) 3 C 0.33 3.13 (8) + (10) - (14) 4 D 0.33 3.13 (9) + (10) - (14) 5 A*B 0.98 8.13 (11) + (1) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 8.13 (1) + (13) - (15) 13-19
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY (d) A and B are fixed and C and D are random. F F R R R a b c d n Factor i j k l h E() 0 b c d n ACD AD AC A a 0 c d n BCD BC BD B l a b 1 d n CD C a b c 1 n CD D i j k ( ) ij ( ) ik ( ) il ( ) jk ( ) jl ( ) kl ( ) ijk ( ) ijl ( ) jkl 0 0 c d n CD C D 0 b 1 d n ACD AC 0 b c 1 n ACD AD a 0 1 d n BCD BC a 0 c 1 n BCD BD a b 1 1 n 0 0 1 d n CD C CD 0 0 c 1 n CD D a 0 1 1 n 0 b 1 1 n ( ) ijkl 0 0 1 1 n ( ) ikl ( ijkl) h 1 1 1 1 1 BCD ACD CD There are no exact tests on the fixed factors A and B, or their two-factor interaction. The appropriate test statistics are: A: F B: F : F The results can also be generated in Minitab as follows: ANOVA: y versus A, B, C, D A fixed H L B fixed H L C random H L D random H L Analysis of Variance for y A AC B BC C ACD AD BCD BD Source DF SS F P A 1 6.13 6.13 1.96 0.604 x B 1 0.13 0.13 0.04 0.907 x C 1 1.13 1.13 0.36 0.656 D 1 0.13 0.13 0.04 0.874 A*B 1 3.13 3.13 0.11 0.796 x A*C 1 3.13 3.13 1.00 0.500 A*D 1 3.13 3.13 1.00 0.500 B*C 1 3.13 3.13 1.00 0.500 B*D 1 3.13 3.13 1.00 0.500 CD D 13-0
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY C*D 1 3.13 3.13 0.5 0.6 A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 0.5 0.6 B*C*D 1 3.13 3.13 0.5 0.6 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A * (16) + 4(13) + 8(7) + 8(6) + 16Q[1] B * (16) + 4(14) + 8(9) + 8(8) + 16Q[] 3 C -0.150 10 (16) + 8(10) + 16(3) 4 D -0.1875 10 (16) + 8(10) + 16(4) 5 A*B * (16) + (15) + 4(1) + 4(11) + 8Q[5] 6 A*C 0.0000 13 (16) + 4(13) + 8(6) 7 A*D 0.0000 13 (16) + 4(13) + 8(7) 8 B*C 0.0000 14 (16) + 4(14) + 8(8) 9 B*D 0.0000 14 (16) + 4(14) + 8(9) 10 C*D -1.1563 16 (16) + 8(10) 11 A*B*C 0.0000 15 (16) + (15) + 4(11) 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D -.315 16 (16) + 4(13) 14 B*C*D -.315 16 (16) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A 0.33 3.13 (6) + (7) - (13) B 0.33 3.13 (8) + (9) - (14) 5 A*B 0.98 8.13 (11) + (1) - (15) (e) A, B and C are fixed and D is random. F F F R R a b c d n Factor i j k l h E() 0 b c d n AD A a 0 c d n BD B a b 0 d n CD C a b c 1 n D i j k l ( ) ij 0 0 c d n D ( ) ik 0 b 0 d n ACD AC ( ) il 0 b c 1 n AD ( ) jk a 0 0 d n BCD BC ( ) jl a 0 c 1 n BD ( ) kl a b 0 1 n CD ( ) ijk 0 0 0 d n CD C ( ) ijl 0 0 c 1 n D ( ) jkl a 0 0 1 n BCD ( ) ikl 0 b 0 1 n ACD ( ) ijkl 0 0 0 1 n CD ( ijkl) h 1 1 1 1 1 13-1
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY There are exact tests for all effects. The results can also be generated in Minitab as follows: ANOVA: y versus A, B, C, D A fixed H L B fixed H L C fixed H L D random H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 1.96 0.395 B 1 0.13 0.13 0.04 0.874 C 1 1.13 1.13 0.36 0.656 D 1 0.13 0.13 0.01 0.91 A*B 1 3.13 3.13 0.11 0.795 A*C 1 3.13 3.13 1.00 0.500 A*D 1 3.13 3.13 0.5 0.6 B*C 1 3.13 3.13 1.00 0.500 B*D 1 3.13 3.13 0.5 0.6 C*D 1 3.13 3.13 0.5 0.6 A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13.7 0.151 A*C*D 1 3.13 3.13 0.5 0.6 B*C*D 1 3.13 3.13 0.5 0.6 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A 7 (16) + 8(7) + 16Q[1] B 9 (16) + 8(9) + 16Q[] 3 C 10 (16) + 8(10) + 16Q[3] 4 D -0.7656 16 (16) + 16(4) 5 A*B 1 (16) + 4(1) + 8Q[5] 6 A*C 13 (16) + 4(13) + 8Q[6] 7 A*D -1.1563 16 (16) + 8(7) 8 B*C 14 (16) + 4(14) + 8Q[8] 9 B*D -1.1563 16 (16) + 8(9) 10 C*D -1.1563 16 (16) + 8(10) 11 A*B*C 15 (16) + (15) + 4Q[11] 1 A*B*D 3.9375 16 (16) + 4(1) 13 A*C*D -.315 16 (16) + 4(13) 14 B*C*D -.315 16 (16) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) 13.18. Reconsider cases (c), (d) and (e) of Problem 13.17. Obtain the expected mean squares assuming the unrestricted model. You may use a computer package such as Minitab. Compare your results with those for the restricted model. A is fixed and B, C, and D are random. ANOVA: y versus A, B, C, D A fixed H L B random H L C random H L D random H L 13-
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 ** B 1 0.13 0.13 ** C 1 1.13 1.13 0.36 0.843 x D 1 0.13 0.13 ** A*B 1 3.13 3.13 0.11 0.796 x A*C 1 3.13 3.13 1.00 0.667 x A*D 1 3.13 3.13 0.11 0.796 x B*C 1 3.13 3.13 1.00 0.667 x B*D 1 3.13 3.13 0.11 0.796 x C*D 1 3.13 3.13 1.00 0.667 x A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 1.00 0.500 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) 1 A * (16) + (15) + 4(13) + 4(1) + 4(11) + 8(7) + 8(6) + 8(5) + Q[1] B 1.3750 * (16) + (15) + 4(14) + 4(1) + 4(11) + 8(9) + 8(8) + 8(5) + 16() 3 C -0.150 * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + (15) + 4(14) + 4(13) + 4(1) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B -3.150 * (16) + (15) + 4(1) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D -3.150 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 0.0000 * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D -3.150 * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D 0.0000 * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + (15) + 4(11) 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D 0.0000 15 (16) + (15) + 4(13) 14 B*C*D 0.0000 15 (16) + (15) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A 0.56 * (5) + (6) + (7) - (11) - (1) - (13) + (15) B 0.56 * (5) + (8) + (9) - (11) - (1) - (14) + (15) 3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B 0.98 8.13 (11) + (1) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 8.13 (1) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15) 9 B*D 0.98 8.13 (1) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15) A and B are fixed and C and D are random. 13-3
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY ANOVA: y versus A, B, C, D A fixed H L B fixed H L C random H L D random H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 1.96 0.604 x B 1 0.13 0.13 0.04 0.907 x C 1 1.13 1.13 0.36 0.843 x D 1 0.13 0.13 ** A*B 1 3.13 3.13 0.11 0.796 x A*C 1 3.13 3.13 1.00 0.667 x A*D 1 3.13 3.13 0.11 0.796 x B*C 1 3.13 3.13 1.00 0.667 x B*D 1 3.13 3.13 0.11 0.796 x C*D 1 3.13 3.13 1.00 0.667 x A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 1.00 0.500 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) 1 A * (16) + (15) + 4(13) + 4(1) + 4(11) + 8(7) + 8(6) + Q[1,5] B * (16) + (15) + 4(14) + 4(1) + 4(11) + 8(9) + 8(8) + Q[,5] 3 C -0.150 * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + (15) + 4(14) + 4(13) + 4(1) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B * (16) + (15) + 4(1) + 4(11) + Q[5] 6 A*C 0.0000 * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D -3.150 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 0.0000 * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D -3.150 * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D 0.0000 * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + (15) + 4(11) 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D 0.0000 15 (16) + (15) + 4(13) 14 B*C*D 0.0000 15 (16) + (15) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A 0.33 3.13 (6) + (7) - (13) B 0.33 3.13 (8) + (9) - (14) 3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B 0.98 8.13 (11) + (1) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 8.13 (1) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15) 9 B*D 0.98 8.13 (1) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15) 13-4
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY (e) A, B and C are fixed and D is random. ANOVA: y versus A, B, C, D A fixed H L B fixed H L C fixed H L D random H L Analysis of Variance for y Source DF SS F P A 1 6.13 6.13 1.96 0.395 B 1 0.13 0.13 0.04 0.874 C 1 1.13 1.13 0.36 0.656 D 1 0.13 0.13 ** A*B 1 3.13 3.13 0.11 0.795 A*C 1 3.13 3.13 1.00 0.500 A*D 1 3.13 3.13 0.11 0.796 x B*C 1 3.13 3.13 1.00 0.500 B*D 1 3.13 3.13 0.11 0.796 x C*D 1 3.13 3.13 1.00 0.667 x A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 8.13 8.13 9.00 0.05 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 1.00 0.500 A*B*C*D 1 3.13 3.13 0.5 0.6 Error 16 198.00 1.38 Total 31 64.88 x Not an exact F-test. ** Denominator of F-test is zero. Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) 1 A 7 (16) + (15) + 4(13) + 4(1) + 8(7) + Q[1,5,6,11] B 9 (16) + (15) + 4(14) + 4(1) + 8(9) + Q[,5,8,11] 3 C 10 (16) + (15) + 4(14) + 4(13) + 8(10) + Q[3,6,8,11] 4 D 1.3750 * (16) + (15) + 4(14) + 4(13) + 4(1) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B 1 (16) + (15) + 4(1) + Q[5,11] 6 A*C 13 (16) + (15) + 4(13) + Q[6,11] 7 A*D -3.150 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 14 (16) + (15) + 4(14) + Q[8,11] 9 B*D -3.150 * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D 0.0000 * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C 15 (16) + (15) + Q[11] 1 A*B*D 6.500 15 (16) + (15) + 4(1) 13 A*C*D 0.0000 15 (16) + (15) + 4(13) 14 B*C*D 0.0000 15 (16) + (15) + 4(14) 15 A*B*C*D -4.650 16 (16) + (15) 16 Error 1.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 7 A*D 0.98 8.13 (1) + (13) - (15) 9 B*D 0.98 8.13 (1) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15) 13-5
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 13.19. In Problem 5.19, assume that the three operators were selected at random. Analyze the data under these conditions and draw conclusions. Estimate the variance components. ANOVA: Score versus Cycle Time, Operator, Temperature Cycle Ti fixed 3 40 50 60 Operator random 3 1 3 Temperat fixed 300 350 Analysis of Variance for Score Source DF SS F P Cycle Ti 436.000 18.000.45 0.0 Operator 61.333 130.667 39.86 0.000 Temperat 1 50.074 50.074 8.89 0.096 Cycle Ti*Operator 4 355.667 88.917 7.13 0.000 Cycle Ti*Temperat 78.815 39.407 3.41 0.137 Operator*Temperat 11.59 5.630 1.7 0.194 Cycle Ti*Operator*Temperat 4 46.185 11.546 3.5 0.016 Error 36 118.000 3.78 Total 53 1357.333 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Cycle Ti 4 (8) + 6(4) + 18Q[1] Operator 7.077 8 (8) + 18() 3 Temperat 6 (8) + 9(6) + 7Q[3] 4 Cycle Ti*Operator 14.731 8 (8) + 6(4) 5 Cycle Ti*Temperat 7 (8) + 3(7) + 9Q[5] 6 Operator*Temperat 0.613 8 (8) + 9(6) 7 Cycle Ti*Operator*Temperat.756 8 (8) + 3(7) 8 Error 3.778 (8) The following calculations agree with the Minitab results: ˆ n cn C E ˆ E ˆ an BC E ˆ acn B E ˆ E ˆ 3.7778 11.54696 3.77778 ˆ.756 3 88.91667 3.77778 ˆ 14.7315 3 5.69630 3.77778 ˆ 0.613 33 130.66667 3.77778 ˆ 7.07716 3 3 13.0. Consider the three-factor factorial model y ijk i j k ij jk ijk Assuming that all the factors are random, develop the analysis of variance table, including the expected mean squares. Propose appropriate test statistics for all effects. 13-6
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Source DF E() A a-1 c bc B b-1 c a ac C c-1 a ab (a-1)(b-1) BC (b-1)(c-1) Error (AC + C) b(a-1)(c-1) Total abc-1 c a There are exact tests for all effects except B. To test B, use the statistic F B E BC 13.1. The three-factor factorial model for a single replicate is y ( ) ( ) ( ) ( ) ijk i j k ij jk ik ijk ijk If all the factors are random, can any effects be tested? If the three-factor interaction and the interaction do not exist, can all the remaining effects be tested? ( ) ij The expected mean squares are found by referring to Table 13.9, deleting the line for the error term ( ijk ) l and setting n=1. The three-factor interaction now cannot be tested; however, exact tests exist for the twofactor interactions and approximate F tests can be conducted for the main effects. For example, to test the main effect of A, use F A C AC If ( ) ijk and ( ) ij can be eliminated, the model becomes For this model, the analysis of variance is y ijk i j k jk ik ijk Source DF E() A a-1 b bc B b-1 a ac C c-1 a b ab AC (a-1)(c-1) b BC (b-1)(c-1) a Error ( + C) c(a-1)(b-1) Total abc-1 There are exact tests for all effect except C. To test the main effect of C, use the statistic: 13-7
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY F C BC E AC 13.. In Problem 5.8, assume that both machines and operators were chosen randomly. Determine the power of the test for detecting a machine effect such that, where is the variance component for the machine factor. Are two replicates sufficient? an 1 n If, then an estimate of 379., and an estimate of variance table. Then 3 3.79 1. 1.49 3.79 (7.44) 7.44, from the analysis of and the other OC curve parameters are 1 3 and 6. This results in 0 75 approximately, with 005., or 0. 9 with 001.. Two replicates does not seem sufficient.. Cov, 1 13.3. In the two-factor mixed model analysis of variance, show that ii'. a ij i' j for Since a i1 0 (constant) we have 0 ij a V ij, which implies that i1 a a V, 0 ij Cov ij i' j i1 a1 a! a, 0 i' j a!! Cov ij a a a a Cov 1 1, 0 ij i' j 1 Cov, ij i' j a 13.4. Show that the method of analysis of variance always produces unbiased point estimates of the variance component in any random or mixed model. Let g be the vector of mean squares from the analysis of variance, chosen so that E(g) does not contain any fixed effects. Let be the vector of variance components such that E( g) A, where A is a matrix of constants. Now in the analysis of variance method of variance component estimation, we equate observed and expected mean squares, i.e. 13-8
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY -1 g = As s A g ˆ Since -1 A always exists then, E s = E A g A E g = A As s -1-1 -1 Thus is an unbiased estimator of. This and other properties of the analysis of variance method are discussed by Searle (1971a). 13.5. Invoking the usual normality assumptions, find an expression for the probability that a negative estimate of a variance component will be obtained by the analysis of variance method. Using this result, write a statement giving the probability that usefulness of this probability statement. 0 in a one-factor analysis of variance. Comment on the Suppose 1, where i for i=1, are two mean squares and c is a constant. The c probability that (negative) is 0 ö 1 1 E 1 E 1 E 1 P ˆ 0 P1 0 P 1 P P Fuv, E E E where u is the number of degrees of freedom for 1 and v is the number of degrees of freedom for. For the one-way model, this equation reduces to ˆ 0 1 a 1, N a a 1, N a n 1nk P P F P F where k. Using arbitrary values for some of the parameters in this equation will give an experimenter some idea of the probability of obtaining a negative estimate of ˆ 0. 13.6. Analyze the data in Problem 13.1, assuming that the operators are fixed, using both the unrestricted and restricted forms of the mixed models. Compare the results obtained from the two models. The restricted model is as follows: ANOVA: Measurement versus Part, Operator Part random 10 1 3 4 5 6 7 8 9 10 Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part 9 99.017 11.00 7.33 0.000 13-9
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Operator 1 0.417 0.417 0.69 0.47 Part*Operator 9 5.417 0.60 0.40 0.97 Error 40 60.000 1.500 Total 59 164.850 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Part 1.5836 4 (4) + 6(1) Operator 3 (4) + 3(3) + 30Q[] 3 Part*Operator -0.994 4 (4) + 3(3) 4 Error 1.5000 (4) The second approach is the unrestricted mixed model. ANOVA: Measurement versus Part, Operator Part random 10 1 3 4 5 6 7 8 9 10 Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part 9 99.017 11.00 18.8 0.000 Operator 1 0.417 0.417 0.69 0.47 Part*Operator 9 5.417 0.60 0.40 0.97 Error 40 60.000 1.500 Total 59 164.850 Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) 1 Part 1.7333 3 (4) + 3(3) + 6(1) Operator 3 (4) + 3(3) + Q[] 3 Part*Operator -0.994 4 (4) + 3(3) 4 Error 1.5000 (4) Source Sum of Squares DF Mean Square A 99.016667 a-1=9 11.00185 B 0.416667 b-1=1 0.416667 E() F-test F n bn i1 n an b 1 b i F F A 18.8 B 0.69 5.416667 (a-1)(b-1)=9 0.60185 n Error 60.000000 40 1.50000 Total 164.85000 nabc-1=59 F 0.401 E In the unrestricted model, the F-test for A is different. The F-test for A in the unrestricted model should generally be more conservative, since will generally be larger than E. However, this is not the case with this particular experiment. 13.7. Consider the two-factor mixed model. Show that the standard error of the fixed factor mean (e.g. / bn. A) is 1 13-30
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY The standard error is often used in Duncan s Multiple Range test. Duncan s Multiple Range Test requires the variance of the difference in two means, say V y i y.. where rows are fixed and columns are random. Now, assuming all model parameters to be independent, we have the following: 1 b b y i.. ym.. i m ij mj ijk j1 1 b b j1 m.. 1 bn b n j1 k 1 1 bn b n j1 k 1 mjk and Since V yi.. ym.. 1 bn b b n 1 b b estimates, we would use 1 1 bn bn bn n bn bn as the standard error to test the difference. However, the table of ranges for Duncan s Multiple Range test already includes the constant. 13.8. Consider the variance components in the random model from Problem 13.1. (a) Find an exact 95 percent confidence interval on. f f E E E E, f E 1, fe 401.5 401.5 59.34 4.43 1.011.456 (b) Find approximate 95 percent confidence intervals on the other variance components using the Satterthwaite method. is and are negative, and the Satterthwaithe method does not apply. The confidence interval on an B ˆ 11.00185 0.6018519 ˆ 1.7333 3 b 1 a 1b 1 B 11.00185 0.6018519 r 8.0186 B 11.00185 0.6018519 9 1 9 13-31
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY r r ˆ ˆO, r 1, r 8.01861.7333 8.01861.7333 17.5575.18950 0.79157 6.34759 13.9. Use the experiment described in Problem 5.8 and assume that both factors are random. Find an exact 95 percent confidence interval on. Construct approximate 95 percent confidence interval on the other variance components using the Satterthwaite method. ˆ E ˆ 3.79167 f f E E E E, f E 1, fe 13.79167 13.79167 3.34 4.40 1.9494 10.3409 Satterthwaite Method: n E ˆ 7.44444 3.79167 ˆ 1.8639 1 1 E 7.44444 3.79167 r 1.7869 E 7.44444 3.79167 a b df E r ˆ 3 1 r ˆ, r 1, r and were estimated by extrapolating 1.7869 between degrees of freedom of one and two in,r 1,r Microsoft Excel. More precise methods can be used as well. 1.78691.8639 1.78691.8639 5.67991 0.01480 0.41117 158.5817 13-3
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY ˆ 0, this variance component does not have a confidence interval using Satterthwaite s Method. bn A ˆ 80.16667 7.44444 ˆ 9.0908 4 a 1 a 1b 1 A 80.16667 7.44444 r 1.64108 A 80.16667 7.44444 3 r ˆ r ˆ, r 1, r (1.64108)(9.0908) (1.64108)(9.0908) 6.5393 0.0381,r and 1,r were estimated by extrapolating 1.64108 between degrees of freedom of one and two in Microsoft Excel. More precise methods can be used as well..8349 454.61891 13.30. Consider the three-factor experiment in Problem 5.19 and assume that operators were selected at random. Find an approximate 95 percent confidence interval on the operator variance component. acn B E ˆ 130.66667 3.77778 ˆ 7.07716 3 3 B E 130.66667 3.7778 r 1.90085 B E 130.66667 3.7778 b 1 df E 36 r ˆ r ˆ, r 1, r and were estimated by extrapolating 1.90085 between degrees of freedom of one and two in,r 1,r Microsoft Excel. More precise methods can be used as well. 1.900857.07716 1.900857.07716 7.14439 0.04571 1.8896 94.870 13.31. Rework Problem 13.8 using the modified large-sample approach described in Section 13.7.. Compare the two sets of confidence intervals obtained and discuss. 13-33
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY an B ˆ ˆ O 0.05,9, 1 F.95,9 i,.95,9 9 11.00185 0.6018519 ˆ 1.7333 3 O 1 1 G1 1 1 0.46809 F 1.88 1 1 1 H 1 1 1 1.707 0.370 F, f i, f G1 F j, fi, f H j 1 1 3.18 1 0.46809 3.18 1.707 Gij 0.36366 F 3.18, fi, fj V G c H c G c c L 1 1 B 1 11 1 B L 1 1 1 1 VL 6 6 6 6 V 0.8375 0.46809 11.00185 1.707 0.60185 0.36366 11.001850.60185 ˆ 1.7333 0.8375 0.8075 L V L 13.3. Rework Problem 13.8 using the modified large-sample method described in Section 13.7.. Compare this confidence interval with the one obtained previously and discuss. abn C E ˆ 0.05,3, 130.66667 3.77778 ˆ 7.07716 3 3 1 1 G1 1 1 0.61538 F.60 H G 1 F.95,36,.95,36 ij 1 1 1 1 1. 1 0.54493 0.6478 36 F, f i, f G1 F j, fi, f H j 1 1.88 1 0.61538.88 0.54493 0.7454 F.88, fi, f j V G c H c G c c V V L 1 1 B 1 11 1 B L L 1 1 1 1 0.61538 130.66667 0.54493 3.7778 0.7454 130.666673.77 18 18 18 18 0.9511 ˆ 7.07716 0.9511.4999 L V L 13.33 Consider the experiment described in Problem 5.8. Estimate the variance component using the REML method. Compare the CIs to the approximate CIs found in Problem 13.9. The JMP REML analysis below was performed with both factors, Operator and Machine, as random. 13-34
Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY The CIs for the error variance are similar to those found in Problem 13.9. The upper CIs for the other variance components are much larger than those estimated in the JMP REML output below. JMP Output RSquare 0.74044 RSquare Adj 0.74044 Root Mean Square Error 1.947 Mean of Response 11.917 Observations (or Sum Wgts) 4 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 11.9167 1.78978 1.831 6.74 0.0005* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Operator.3974359 9.090778 10.0355-10.5784 8.758958 64.198 Machine -0.144689-0.548611 0.914188 -.336919 1.39697-3.874 Operator*Machine 0.481685 1.863889.841505 -.650464 6.303416 1.898 Residual 3.7916667 1.5479414 1.949717 10.33013 6.778 Total 14.1597 100.000 Covariance Matrix of Variance Component Estimates Random Effect Operator Machine Operator*Machine Residual Operator 100.70575 0.3848594-1.154578.151e-1 Machine 0.3848594 0.835081-1.539438-1.17e-13 Operator*Machine -1.154578-1.539438 5.173434-1.198061 Residual.151e-1-1.17e-13-1.198061.39617 13.34 Consider the experiment described in Problem 13.1. Analyze the data using REML. Compare the CIs to those obtained in Problem 13.8. The JMP REML analysis below was performed with both factors, Part Number and Operator, as random. The CIs for the Operator and Part Number Operator interaction were not calculated in Problem 13.8 due to negative estimates for the corresponding variance components. The error variance estimates and CIs found in the JMP REML output below are the same as those calculated in Problem 13.8. The upper CI for the Part Number variance estimated with the Satterthwaite method in Problem 13.8 is approximately twice the value estimated in the JMP REML analysis. JMP Output RSquare 0.388009 RSquare Adj 0.388009 Root Mean Square Error 1.4745 Mean of Response 49.95 Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept 49.95 0.44591 8.563 117.64 <.0001* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number 1.1555556 1.7333333 0.8656795 0.036636 3.430034 59.03 Operator -0.004115-0.006173 0.018-0.0489 0.0365544-0.11 Part Number*Operator -0.199588-0.99383 0.146437-0.586394-0.01371-10.6 Residual 1.5 0.335410 1.0110933.455691 51.33 Total.977778 100.000 13-35