Chapter 13 Experiments with Random Factors Solutions

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1 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 , 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 (a) Analyze the data from this experiment, assuming both parts and operators are random effects. ANOVA: Impedance versus Inspector, Part Inspecto random Part random Analysis of Variance for Impedanc Source DF SS F P Inspecto Part Inspecto*Part Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Inspecto (4) + 3(3) + 30(1) Part (4) + 3(3) + 9() 3 Inspecto*Part (4) + 3(3) 4 Error (4) (b) Estimate the variance components using the analysis of variance method. 13-1

2 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 ˆ ˆ ˆ 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 The following Minitab output contains the analysis of variance and the variance component estimates: ANOVA: Strength versus Operator, Machine Operator random Machine random Analysis of Variance for Strength Source DF SS F P Operator Machine Operator*Machine Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator (4) + (3) + 8(1) Machine (4) + (3) + 6() 3 Operator*Machine (4) + (3) 4 Error (4) (b) Find point estimates of the variance components using the analysis of variance method. ˆ E ˆ

3 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY n an A ˆ bn E ˆ B ˆ ˆ ˆ 0, assume 3() ˆ () These results agree with the Minitab variance component analysis. ˆ Reconsider the data in Problem Suppose that both factors are random. (a) Analyze the data from this experiment. General Linear Model: Response versus Row, Column Row random Column random Column Factor Row Factor Analysis of Variance for Response, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj F P Row ** Column ** Row*Column ** Error Total ** Denominator of F-test is zero. Expected Mean Squares, using Adjusted SS Source Expected Mean Square for Each Term 1 Row (4) + (3) (1) Column (4) + (3) () 3 Row*Column (4) + (3) 4 Error (4) Error Terms for Tests, using Adjusted SS Source Error DF Error Synthesis of Error 1 Row * (3) Column * (3) 3 Row*Column * * (4) Variance Components, using Adjusted SS Source Estimated Value Row Column Row*Column Error

4 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 ˆ ˆ ˆ 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 Analysis of Variance for Density Temperature ( C) Position Source DF SS F P Position Temperat Position*Temperat Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Position (4) + 9(1) Temperat 3 (4) + 3(3) + 6Q[] 3 Position*Temperat (4) + 3(3) 4 Error (4) ˆ E ˆ

5 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 ˆ 0 assume ˆ ˆ 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 The following analysis assumes a restricted model: ANOVA: Measurement versus Part, Operator Part random Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part Operator Part*Operator Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Part (4) + 6(1) Operator 3 (4) + 3(3) + 30Q[] 3 Part*Operator (4) + 3(3) 4 Error (4) n E ˆ ˆ E ˆ ˆ 0 assume ˆ

6 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY bn A E ˆ ˆ These results agree with the Minitab output 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 ANOVA: Impedance versus Inspector, Part Inspecto fixed Part random Analysis of Variance for Impedanc Source DF SS F P Inspecto Part Inspecto*Part Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Inspecto 3 (4) + 3(3) + 30Q[1] Part (4) + 9() 3 Inspecto*Part (4) + 3(3) 4 Error (4) ˆ ˆ E ˆ 0.73 n ˆ an 33 E ˆ B E ˆ These results agree with the Minitab output. 13-6

7 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 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 (b) Perform the analysis and estimate the model components. The following analysis assumes a restricted model: ANOVA: Strength versus Operator, Machine Operator random Machine fixed Analysis of Variance for Strength Source DF SS F P Operator Machine Operator*Machine Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator (4) + 8(1) Machine 3 (4) + (3) + 6Q[] 3 Operator*Machine (4) + (3) 4 Error 3.79 (4) ˆ n bn E ˆ A E ˆ These results agree with the Minitab output. E ˆ ˆ ˆ Rework Problem 13.5 using the REML method. 13-7

8 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Temperature ( C) Position The JMP REML output is shown below. The variance components are similar to those calculated in Problem JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 18 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept * Temperature[800] * Temperature[85] * REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Position Position*Temperature Residual Total Covariance Matrix of Variance Component Estimates Random Effect Position Position*Temperature Residual Position e-9 Position*Temperature Residual -1.41e Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Temperature * Rework Problem 13.6 using the REML method. 13-8

9 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Table P13.1 Operator 1 Operator Part Measurements Measurements Number The JMP REML output is shown below. The variance components are similar to those calculated in Problem JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept <.0001* Operator[Operator 1] REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number Part Number*Operator Residual Total Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Operator Residual Part Number e-14 Part Number*Operator Residual -1.43e Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Operator 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

10 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY The JMP REML output is shown below. The variance components are similar to those calculated in Problem JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response 35.8 Observations (or Sum Wgts) 90 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept <.0001* Inspector[Inspector 1] * Inspector[Inspector ] * REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number Part Number*Inspector Residual Total Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Inspector Residual Part Number e-13 Part Number*Inspector Residual -.76e Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Inspector * Rework Problem 13.8 using the REML method. Operator Machine The JMP REML output is shown below. The variance components are similar to those calculated in Problem JMP Output 13-10

11 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 4 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept * Machine[1] Machine[] Machine[3] REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Operator Operator*Machine Residual Total Covariance Matrix of Variance Component Estimates Random Effect Operator Operator*Machine Residual Operator e-1 Operator*Machine Residual 1.686e Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Machine 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, Substituting these expressions into the sums of squares yields, 0. j, which imply that 13-11

12 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

13 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Thus E A and E agree with Equation Consider the three-factor factorial design in Example 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 n n ijkl cn These are exact tests for all effects. j ij a 1b Consider the experiment in Example Analyze the data for the case where A, B, and C are random. ANOVA: Drop versus Temp, Operator, Gauge Temp random Operator random Gauge random Analysis of Variance for Drop Source DF SS F P Temp x 13-13

14 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Operator x Gauge x Temp*Operator Temp*Gauge Operator*Gauge Temp*Operator*Gauge Error Total x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Temp * (8) + (7) + 8(5) + 6(4) + 4(1) Operator * (8) + (7) + 6(6) + 6(4) + 18() 3 Gauge * (8) + (7) + 6(6) + 8(5) + 4(3) 4 Temp*Operator (8) + (7) + 6(4) 5 Temp*Gauge (8) + (7) + 8(5) 6 Operator*Gauge (8) + (7) + 6(6) 7 Temp*Operator*Gauge (8) + (7) 8 Error (8) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 Temp (4) + (5) - (7) Operator (4) + (6) - (7) 3 Gauge (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 Derive the expected mean squares shown in Table 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 n n ijkl cn i bcn a 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

15 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 d n C ( ) ijl 0 0 c 0 n D ( ) jkl a n BCD ( ) ikl 0 b 0 0 n ACD ( ) ijkl n CD ( ijkl) h 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 B C D A*B A*C A*D B*C B*D C*D A*B*C

16 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY A*B*D A*C*D B*C*D A*B*C*D Error Total 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 d n CD C ( ) ijl 1 1 c 1 n CD D ( ) jkl a n CD BCD ( ) ikl 1 b 1 1 n CD ACD ( ) ijkl n CD ( ijkl) h 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

17 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 ** B ** C x D ** A*B x A*C x A*D x B*C x B*D x C*D x A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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) + 16(1) B * (16) + (15) + 4(14) + 4(1) + 4(11) + 8(9) + 8(8) + 8(5) + 16() 3 C * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D * (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) + 8(5) 6 A*C * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C (16) + (15) + 4(11) 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + (15) + 4(13) 14 B*C*D (16) + (15) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (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 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B (11) + (1) - (15) 6 A*C (11) + (13) - (15) 13-17

18 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 7 A*D (1) + (13) - (15) 8 B*C (11) + (14) - (15) 9 B*D (1) + (14) - (15) 10 C*D (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 d n CD C ( ) ijl 0 1 c 1 n CD D ( ) jkl a n BCD ( ) ikl 0 b 1 1 n CD ACD ( ) ijkl n CD ( ijkl) h 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

19 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 ** B x C x D x A*B x A*C x A*D x B*C B*D C*D A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 * (16) + 4(14) + 8(9) + 8(8) + 16() 3 C * (16) + 4(14) + 8(10) + 8(8) + 16(3) 4 D * (16) + 4(14) + 8(10) + 8(9) + 16(4) 5 A*B * (16) + (15) + 4(1) + 4(11) + 8(5) 6 A*C * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C (16) + 4(14) + 8(8) 9 B*D (16) + 4(14) + 8(9) 10 C*D (16) + 4(14) + 8(10) 11 A*B*C (16) + (15) + 4(11) 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + (15) + 4(13) 14 B*C*D (16) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (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 (8) + (9) - (14) 3 C (8) + (10) - (14) 4 D (9) + (10) - (14) 5 A*B (11) + (1) - (15) 6 A*C (11) + (13) - (15) 7 A*D (1) + (13) - (15) 13-19

20 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 d n CD C CD 0 0 c 1 n CD D a n 0 b 1 1 n ( ) ijkl n ( ) ikl ( ijkl) h 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 x B x C D A*B x A*C A*D B*C B*D CD D 13-0

21 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY C*D A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 (16) + 8(10) + 16(3) 4 D (16) + 8(10) + 16(4) 5 A*B * (16) + (15) + 4(1) + 4(11) + 8Q[5] 6 A*C (16) + 4(13) + 8(6) 7 A*D (16) + 4(13) + 8(7) 8 B*C (16) + 4(14) + 8(8) 9 B*D (16) + 4(14) + 8(9) 10 C*D (16) + 8(10) 11 A*B*C (16) + (15) + 4(11) 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + 4(13) 14 B*C*D (16) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A (6) + (7) - (13) B (8) + (9) - (14) 5 A*B (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 d n CD C ( ) ijl 0 0 c 1 n D ( ) jkl a n BCD ( ) ikl 0 b 0 1 n ACD ( ) ijkl n CD ( ijkl) h

22 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 B C D A*B A*C A*D B*C B*D C*D A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 (16) + 16(4) 5 A*B 1 (16) + 4(1) + 8Q[5] 6 A*C 13 (16) + 4(13) + 8Q[6] 7 A*D (16) + 8(7) 8 B*C 14 (16) + 4(14) + 8Q[8] 9 B*D (16) + 8(9) 10 C*D (16) + 8(10) 11 A*B*C 15 (16) + (15) + 4Q[11] 1 A*B*D (16) + 4(1) 13 A*C*D (16) + 4(13) 14 B*C*D (16) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (16) Reconsider cases (c), (d) and (e) of Problem 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-

23 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Analysis of Variance for y Source DF SS F P A ** B ** C x D ** A*B x A*C x A*D x B*C x B*D x C*D x A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 * (16) + (15) + 4(14) + 4(1) + 4(11) + 8(9) + 8(8) + 8(5) + 16() 3 C * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D * (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) + 8(5) 6 A*C * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C (16) + (15) + 4(11) 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + (15) + 4(13) 14 B*C*D (16) + (15) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (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 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B (11) + (1) - (15) 6 A*C (11) + (13) - (15) 7 A*D (1) + (13) - (15) 8 B*C (11) + (14) - (15) 9 B*D (1) + (14) - (15) 10 C*D (13) + (14) - (15) A and B are fixed and C and D are random. 13-3

24 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 x B x C x D ** A*B x A*C x A*D x B*C x B*D x C*D x A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 * (16) + (15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D * (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 * (16) + (15) + 4(13) + 4(11) + 8(6) 7 A*D * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C * (16) + (15) + 4(14) + 4(11) + 8(8) 9 B*D * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C (16) + (15) + 4(11) 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + (15) + 4(13) 14 B*C*D (16) + (15) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error Synthesis of Error 1 A (6) + (7) - (13) B (8) + (9) - (14) 3 C (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (1) - (13) - (14) + (15) 5 A*B (11) + (1) - (15) 6 A*C (11) + (13) - (15) 7 A*D (1) + (13) - (15) 8 B*C (11) + (14) - (15) 9 B*D (1) + (14) - (15) 10 C*D (13) + (14) - (15) 13-4

25 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 B C D ** A*B A*C A*D x B*C B*D x C*D x A*B*C A*B*D A*C*D B*C*D A*B*C*D Error Total 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 * (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 * (16) + (15) + 4(13) + 4(1) + 8(7) 8 B*C 14 (16) + (15) + 4(14) + Q[8,11] 9 B*D * (16) + (15) + 4(14) + 4(1) + 8(9) 10 C*D * (16) + (15) + 4(14) + 4(13) + 8(10) 11 A*B*C 15 (16) + (15) + Q[11] 1 A*B*D (16) + (15) + 4(1) 13 A*C*D (16) + (15) + 4(13) 14 B*C*D (16) + (15) + 4(14) 15 A*B*C*D (16) + (15) 16 Error (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 (1) + (13) - (15) 9 B*D (1) + (14) - (15) 10 C*D (13) + (14) - (15) 13-5

26 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY 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 Operator random Temperat fixed Analysis of Variance for Score Source DF SS F P Cycle Ti Operator Temperat Cycle Ti*Operator Cycle Ti*Temperat Operator*Temperat Cycle Ti*Operator*Temperat Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Cycle Ti 4 (8) + 6(4) + 18Q[1] Operator (8) + 18() 3 Temperat 6 (8) + 9(6) + 7Q[3] 4 Cycle Ti*Operator (8) + 6(4) 5 Cycle Ti*Temperat 7 (8) + 3(7) + 9Q[5] 6 Operator*Temperat (8) + 9(6) 7 Cycle Ti*Operator*Temperat (8) + 3(7) 8 Error (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 ˆ ˆ ˆ ˆ 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

27 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 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

28 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 (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 Two replicates does not seem sufficient.. Cov, 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 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

29 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 Thus is an unbiased estimator of. This and other properties of the analysis of variance method are discussed by Searle (1971a) 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 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 Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part

30 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY Operator Part*Operator Error Total Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Part (4) + 6(1) Operator 3 (4) + 3(3) + 30Q[] 3 Part*Operator (4) + 3(3) 4 Error (4) The second approach is the unrestricted mixed model. ANOVA: Measurement versus Part, Operator Part random Operator fixed 1 Analysis of Variance for Measurem Source DF SS F P Part Operator Part*Operator Error Total Source Variance Error Expected Mean Square for Each Term component term (using unrestricted model) 1 Part (4) + 3(3) + 6(1) Operator 3 (4) + 3(3) + Q[] 3 Part*Operator (4) + 3(3) 4 Error (4) Source Sum of Squares DF Mean Square A a-1= B b-1= E() F-test F n bn i1 n an b 1 b i F F A 18.8 B (a-1)(b-1)= n Error Total nabc-1=59 F 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 Consider the two-factor mixed model. Show that the standard error of the fixed factor mean (e.g. / bn. A) is

31 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 Consider the variance components in the random model from Problem (a) Find an exact 95 percent confidence interval on. f f E E E E, f E 1, fe (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 ˆ b 1 a 1b 1 B r B

32 Solutions from Montgomery, D. C. (01) Design and Analysis of Experiments, Wiley, NY r r ˆ ˆO, r 1, r 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 f f E E E E, f E 1, fe Satterthwaite Method: n E ˆ ˆ E r E a b df E r ˆ 3 1 r ˆ, r 1, r and were estimated by extrapolating between degrees of freedom of one and two in,r 1,r Microsoft Excel. More precise methods can be used as well

33 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 ˆ a 1 a 1b 1 A r A r ˆ r ˆ, r 1, r ( )(9.0908) ( )(9.0908) ,r and 1,r were estimated by extrapolating between degrees of freedom of one and two in Microsoft Excel. More precise methods can be used as well 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 ˆ B E r B E b 1 df E 36 r ˆ r ˆ, r 1, r and were estimated by extrapolating between degrees of freedom of one and two in,r 1,r Microsoft Excel. More precise methods can be used as well Rework Problem 13.8 using the modified large-sample approach described in Section Compare the two sets of confidence intervals obtained and discuss

34 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, ˆ O 1 1 G F H F, f i, f G1 F j, fi, f H j Gij F 3.18, fi, fj V G c H c G c c L 1 1 B B L VL V ˆ L V L Rework Problem 13.8 using the modified large-sample method described in Section Compare this confidence interval with the one obtained previously and discuss. abn C E ˆ 0.05,3, ˆ G F.60 H G 1 F.95,36,.95,36 ij F, f i, f G1 F j, fi, f H j F.88, fi, f j V G c H c G c c V V L 1 1 B B L L ˆ L V L 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 The JMP REML analysis below was performed with both factors, Operator and Machine, as random

35 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 The upper CIs for the other variance components are much larger than those estimated in the JMP REML output below. JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 4 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept * REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Operator Machine Operator*Machine Residual Total Covariance Matrix of Variance Component Estimates Random Effect Operator Machine Operator*Machine Residual Operator e-1 Machine e-13 Operator*Machine Residual.151e e Consider the experiment described in Problem Analyze the data using REML. Compare the CIs to those obtained in Problem 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 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 RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob> t Intercept <.0001* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Part Number Operator Part Number*Operator Residual Total

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