Chapter 13 Experiments with Random Factors Solutionssmills/2013-14... · Solutions from Montgomery,...

Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY

13-1

Chapter 13 Experiments with Random Factors

Solutions 13.1. An experiment was performed to investigate the capability of a measurement system. Ten parts were randomly selected, and two randomly selected operators measured each part three times. The tests were made in random order, and the data are shown in Table P13.1.

Table P13.1

Part Number

Operator 1 Operator 2 Measurements Measurements

1 2 3 1 2 3 1 50 49 50 50 48 51 2 52 52 51 51 51 51 3 53 50 50 54 52 51 4 49 51 50 48 50 51 5 48 49 48 48 49 48 6 52 50 50 52 50 50 7 51 51 51 51 50 50 8 52 50 49 53 48 50 9 50 51 50 51 48 49 10 47 46 49 46 47 48

(a) Analyze the data from this experiment. Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 10 1 2 3 4 5 6 7 8 9 10 Operator random 2 1 2 Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.017 11.002 18.28 0.000 Operator 1 0.417 0.417 0.69 0.427 Part*Operator 9 5.417 0.602 0.40 0.927 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 1.73333 3 (4) + 3(3) + 6(1) 2 Operator -0.00617 3 (4) + 3(3) + 30(2) 3 Part*Operator -0.29938 4 (4) + 3(3) 4 Error 1.50000 (4)

(b) Find point estimates of the variance components using the analysis of variance method.

2ˆ EMSσ = 2ˆ 1.5σ =

2ˆ AB EMS MSnτβσ −

= 2 0.6018519 1.5000000ˆ 03τβσ −

= < , assume 2ˆ 0τβσ =


13-2

2ˆ B ABMS MSanβσ−

= ( )

2 11.001852 0.6018519ˆ 1.73332 3βσ−

= =

2ˆ A ABMS MSbnτσ−

= ( )

2 0.416667 0.6018519ˆ 010 3τσ−

= < , assume 2ˆ 0τσ =

All estimates agree with the Minitab output. 13.2. 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. 516-520, 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 P13.2. Table P13.2

Part Number

Inspector 1 Inspector 2 Inspector 3 Test 1 Test 2 Test 3 Test 1 Test 2 Test 3 Test 1 Test 2 Test 3

1 37 38 37 41 41 40 41 42 41 2 42 41 43 42 42 42 43 42 43 3 30 31 31 31 31 31 29 30 28 4 42 43 42 43 43 43 42 42 42 5 28 30 29 29 30 29 31 29 29 6 42 42 43 45 45 45 44 46 45 7 25 26 27 28 28 30 29 27 27 8 40 40 40 43 42 42 43 43 41 9 25 25 25 27 29 28 26 26 26 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. Minitab Output ANOVA: Impedance versus Inspector, Part Factor Type Levels Values Inspecto random 3 1 2 3 Part random 10 1 2 3 4 5 6 7 8 9 10 Analysis of Variance for Impedanc Source DF SS MS F P Inspecto 2 39.27 19.63 7.28 0.005 Part 9 3935.96 437.33 162.27 0.000 Inspecto*Part 18 48.51 2.70 5.27 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) 2 Part 48.2926 3 (4) + 3(3) + 9(2) 3 Inspecto*Part 0.7280 4 (4) + 3(3) 4 Error 0.5111 (4)


13-3

(b) Estimate the variance components using the analysis of variance method.

2ˆ EMSσ = 2ˆ 0.51σ =


= 2 2.70 0.51ˆ 0.733τβσ −

= =


= ( )

2 437.33 2.70ˆ 48.293 3βσ

−= =


= ( )

2 19.63 2.70ˆ 0.5610 3τσ

−= =

All estimates agree with the Minitab output. 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 2 3 4

1 109 110 108 110 110 115 109 108

2 110 110 111 114 112 111 109 112

3 116 112 114 120 114 115 119 117

The following Minitab output contains the analysis of variance and the variance component estimates: Minitab Output ANOVA: Strength versus Operator, Machine Factor Type Levels Values Operator random 3 1 2 3 Machine random 4 1 2 3 4 Analysis of Variance for Strength Source DF SS MS F P Operator 2 160.333 80.167 10.77 0.010 Machine 3 12.458 4.153 0.56 0.662 Operator*Machine 6 44.667 7.444 1.96 0.151 Error 12 45.500 3.792 Total 23 262.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator 9.0903 3 (4) + 2(3) + 8(1) 2 Machine -0.5486 3 (4) + 2(3) + 6(2) 3 Operator*Machine 1.8264 4 (4) + 2(3) 4 Error 3.7917 (4)

(b) Find point estimates of the variance components using the analysis of variance method.

2ˆ EMSσ = 2ˆ 3.79167σ =


13-4


= 2 7.44444 3.79167ˆ 1.826392τβσ −

= =


= 2 4.15278 7.44444ˆ 03(2)βσ−

= < , assume 2ˆ 0βσ =


= 2 80.16667 7.44444ˆ 9.090284(2)τσ−

= =

These results agree with the Minitab variance component analysis. 13.4. Reconsider the data in Problem 5.15. Suppose that both factors are random. (a) Analyze the data from this experiment.

Column Factor Row Factor 1 2 3 4

1 36 39 36 32 2 18 20 22 20 3 30 37 33 34

Minitab Output General Linear Model: Response versus Row, Column Factor Type Levels Values Row random 3 1 2 3 Column random 4 1 2 3 4 Analysis of Variance for Response, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Row 2 580.500 580.500 290.250 60.40 ** Column 3 28.917 28.917 9.639 2.01 ** Row*Column 6 28.833 28.833 4.806 ** Error 0 0.000 0.000 0.000 Total 11 638.250 ** 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) 2 Column (4) + (3) + 3.0000(2) 3 Row*Column (4) + (3) 4 Error (4) Error Terms for Tests, using Adjusted SS Source Error DF Error MS Synthesis of Error MS 1 Row * 4.806 (3) 2 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


13-5

(b) Estimate the variance components. Because the experiment is unreplicated and the interaction term was included in the model, there is no estimate of MSE, and therefore, no estimate of 2σ .


= 2 4.8056 0ˆ 4.80561τβσ −

= =


= ( )

2 9.6389 4.8056ˆ 1.61113 1βσ−

= =


= ( )

2 290.2500 4.8056ˆ 71.36114 1τσ−

= =

These estimates agree with the Minitab output. 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.

Temperature (°C) Position 800 825 850

570 1063 565 1 565 1080 510 583 1043 590 528 988 526

2 547 1026 538 521 1004 532

The following analysis assumes a restricted model: Minitab Output ANOVA: Density versus Position, Temperature Factor Type Levels Values Position random 2 1 2 Temperat fixed 3 800 825 850 Analysis of Variance for Density Source DF SS MS F P Position 1 7160 7160 16.00 0.002 Temperat 2 945342 472671 1155.52 0.001 Position*Temperat 2 818 409 0.91 0.427 Error 12 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) 2 Temperat 3 (4) + 3(3) + 6Q[2] 3 Position*Temperat -12.83 4 (4) + 3(3) 4 Error 447.56 (4)

2ˆ EMSσ = 2ˆ 447.56σ =


13-6


= 2 409 448ˆ 03τβσ −

= < assume 2ˆ 0τβσ =

2ˆ A EMS MSbnτσ−

= ( )

2 7160 448ˆ 745.833 3τσ−

= =

These results agree with the Minitab output. 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

Part Number


1 2 3 1 2 3 1 50 49 50 50 48 51 2 52 52 51 51 51 51 3 53 50 50 54 52 51 4 49 51 50 48 50 51 5 48 49 48 48 49 48 6 52 50 50 52 50 50 7 51 51 51 51 50 50 8 52 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: Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 10 1 2 3 4 5 6 7 8 9 10 Operator fixed 2 1 2 Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.017 11.002 7.33 0.000 Operator 1 0.417 0.417 0.69 0.427 Part*Operator 9 5.417 0.602 0.40 0.927 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) 2 Operator 3 (4) + 3(3) + 30Q[2] 3 Part*Operator -0.2994 4 (4) + 3(3) 4 Error 1.5000 (4)

2ˆ EMSσ = 2ˆ 1.5000σ =


= 2 0.60185 1.5000ˆ 03τβσ −

= < assume 2ˆ 0τβσ =


13-7


= ( )

2 11.00185 1.50000ˆ 1.583642 3τσ−

= =

These results agree with the Minitab output. 13.7. Reanalyze the measurement system experiment in Problem 13.2, assuming that operators are a fixed factor. Estimate the appropriate model components. Table P13.2

Part Number


1 37 38 37 41 41 40 41 42 41 2 42 41 43 42 42 42 43 42 43 3 30 31 31 31 31 31 29 30 28 4 42 43 42 43 43 43 42 42 42 5 28 30 29 29 30 29 31 29 29 6 42 42 43 45 45 45 44 46 45 7 25 26 27 28 28 30 29 27 27 8 40 40 40 43 42 42 43 43 41 9 25 25 25 27 29 28 26 26 26 10 35 34 34 35 35 34 35 34 35

Minitab Output ANOVA: Impedance versus Inspector, Part Factor Type Levels Values Inspecto fixed 3 1 2 3 Part random 10 1 2 3 4 5 6 7 8 9 10 Analysis of Variance for Impedanc Source DF SS MS F P Inspecto 2 39.27 19.63 7.28 0.005 Part 9 3935.96 437.33 855.64 0.000 Inspecto*Part 18 48.51 2.70 5.27 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] 2 Part 48.5353 4 (4) + 9(2) 3 Inspecto*Part 0.7280 4 (4) + 3(3) 4 Error 0.5111 (4)

2ˆ EMSσ = 2ˆ 0.51σ =


= 2 2.70 0.51ˆ 0.733τβσ −

= =

2ˆ B EMS MSanβσ−

= ( )

2 437.33 0.51ˆ 48.543 3βσ

−= =

These results agree with the Minitab output.


13-8

13.8. In problem 5.8, suppose that there are only four machines of interest, but the operators were selected at random.

Operator Machine 1 2 3 4

1 109 110 108 110 110 115 109 108

2 110 110 111 114 112 111 109 112

3 116 112 114 120 114 115 119 117

(a) What type of model is appropriate? A mixed model is appropriate. (b) Perform the analysis and estimate the model components. The following analysis assumes a restricted model: Minitab Output ANOVA: Strength versus Operator, Machine Factor Type Levels Values Operator random 3 1 2 3 Machine fixed 4 1 2 3 4 Analysis of Variance for Strength Source DF SS MS F P Operator 2 160.333 80.167 21.14 0.000 Machine 3 12.458 4.153 0.56 0.662 Operator*Machine 6 44.667 7.444 1.96 0.151 Error 12 45.500 3.792 Total 23 262.958 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Operator 9.547 4 (4) + 8(1) 2 Machine 3 (4) + 2(3) + 6Q[2] 3 Operator*Machine 1.826 4 (4) + 2(3) 4 Error 3.792 (4)

2ˆ EMSσ = 2ˆ 3.792σ =


= 2 7.444 3.792ˆ 1.8262τβσ −

= =


= ( )

2 80.167 3.792ˆ 9.5474 2τσ−

= =

These results agree with the Minitab output. 13.9. Rework Problem 13.5 using the REML method.


13-9

Temperature (°C) Position 800 825 850

570 1063 565 1 565 1080 510 583 1043 590 528 988 526

2 547 1026 538 521 1004 532

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 0.99246 Root Mean Square Error 21.15551 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 2 -23.38 0.0018* Temperature[825] 324.05556 6.741707 2 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 1126.0118 -1456.832 2957.0538 63.309 Position*Temperature -0.028674 -12.83333 149.33585 -305.5262 279.85956 -1.083 Residual 447.55556 182.71379 230.13858 1219.556 37.774 Total 1184.8333 100.000 Covariance Matrix of Variance Component Estimates Random Effect Position Position*Temperature Residual Position 1267902.7 -6197.276 -1.412e-9 Position*Temperature -6197.276 22301.197 -11128.11 Residual -1.412e-9 -11128.11 33384.329 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Temperature 2 2 2 1155.518 0.0009* 13.10. Rework Problem 13.6 using the REML method.


13-10

Table P13.1

Part Number


1 2 3 1 2 3 1 50 49 50 50 48 51 2 52 52 51 51 51 51 3 53 50 50 54 52 51 4 49 51 50 48 50 51 5 48 49 48 48 49 48 6 52 50 50 52 50 50 7 51 51 51 51 50 50 8 52 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 0.420766 RSquare Adj 0.410779 Root Mean Square Error 1.224745 Mean of Response 49.95 Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 49.95 0.42821 9 116.65 <.0001* Operator[Operator 1] 0.0833333 0.100154 9 0.83 0.4269 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 0.0366326 3.430034 59.078 Part Number*Operator -0.199588 -0.299383 0.1464372 -0.586394 -0.012371 -10.204 Residual 1.5 0.3354102 1.0110933 2.4556912 51.126 Total 2.9339506 100.000 Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Operator Residual Part Number 0.749401 -0.004472 -1.43e-14 Part Number*Operator -0.004472 0.0214438 -0.0375 Residual -1.43e-14 -0.0375 0.1125 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Operator 1 1 9 0.6923 0.4269 13.11. Rework Problem 13.7 using the REML method. Table P13.2

Part Number


1 37 38 37 41 41 40 41 42 41 2 42 41 43 42 42 42 43 42 43 3 30 31 31 31 31 31 29 30 28 4 42 43 42 43 43 43 42 42 42


13-11

5 28 30 29 29 30 29 31 29 29 6 42 42 43 45 45 45 44 46 45 7 25 26 27 28 28 30 29 27 27 8 40 40 40 43 42 42 43 43 41 9 25 25 25 27 29 28 26 26 26 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 0.992005 RSquare Adj 0.991821 Root Mean Square Error 0.71492 Mean of Response 35.8 Observations (or Sum Wgts) 90 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 35.8 2.20436 9 16.24 <.0001* Inspector[Inspector 1] -0.9 0.244725 18 -3.68 0.0017* Inspector[Inspector 2] 0.6666667 0.244725 18 2.72 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 48.292593 22.906727 3.3962334 93.188952 97.498 Part Number*Inspector 1.4243156 0.7279835 0.3010625 0.1379119 1.3180552 1.470 Residual 0.5111111 0.0933157 0.3681575 0.757543 1.032 Total 49.531687 100.000 Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Inspector Residual Part Number 524.71813 -0.02989 -2.76e-13 Part Number*Inspector -0.02989 0.0906386 -0.002903 Residual -2.76e-13 -0.002903 0.0087078 Fixed Effect Tests Source Nparm DF DFDen F Ratio Prob > F Inspector 2 2 18 7.2849 0.0048* 13.12. Rework Problem 13.8 using the REML method.

Operator Machine 1 2 3 4

1 109 110 108 110 110 115 109 108

2 110 110 111 114 112 111 109 112

3 116 112 114 120 114 115 119 117

The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.8. JMP Output


13-12

RSquare 0.78154 RSquare Adj 0.748771 Root Mean Square Error 1.94722 Mean of Response 112.2917 Observations (or Sum Wgts) 24 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 112.29167 1.827643 2 61.44 0.0003* Machine[1] -0.458333 0.964653 6 -0.48 0.6515 Machine[2] -0.125 0.964653 6 -0.13 0.9011 Machine[3] -0.625 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 2.3974359 9.0902778 10.035225 -10.5784 28.758958 61.804 Operator*Machine 0.481685 1.8263889 2.2841505 -2.650464 6.3032416 12.417 Residual 3.7916667 1.5479414 1.9497217 10.332013 25.779 Total 14.708333 100.000 Covariance Matrix of Variance Component Estimates Random Effect Operator Operator*Machine Residual Operator 100.70575 -1.154578 1.686e-12 Operator*Machine -1.154578 5.2173434 -1.198061 Residual 1.686e-12 -1.198061 2.3961227 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

( )∑=

−=a

i.....iA yybnSS

1

2 , ( )∑=

−=b

j....j.B yyanSS

1

2

( )∑∑= =

+−−=a

i

b

j....j...i.ijAB yyyynSS

1 1

2 , ( )∑∑∑= = =

−=a

i

b

j

n

k...ijkE yySS

1 1 1

2

Using the model ( ) ijkijjiijky ετββτµ ++++= , we may find that

( )

( )

.. ...

. . . .

. .

... . ...

i i ii

j j j

ij i j ijij

y

y

y

y

µ τ τβ ε

µ β ε

µ τ β τβ ε

µ β ε

= + + +

= + +

= + + + +

= + +

Using the assumptions for the restricted form of the mixed model, . 0τ = , ( ) 0=j.τβ , which imply that

( ) 0=..τβ . Substituting these expressions into the sums of squares yields


13-13

( )( )

( )

( ) ( )( )

( )

2

.. ....1

2

. . ...1

2

. .. . . ....1 1

2

.1 1 1

)

a

A i iiib

B j jj

a b

AB ij ij i jii j

a b n

E ijk iji j k

SS bn

SS an

SS n

SS

τ τβ ε ε

β ε ε

τβ τβ ε ε ε ε

ε ε

=

=

= =

= = =

= + + −

= + −

= − + − − +

= −

∑

∑

∑∑

∑∑∑

Using the assumption that ( ) 0=ijkE ε , V ijk( )ε = 0 , and ( ) 0=⋅ 'k'j'iijkE εε , we may divide each sum of squares by its degrees of freedom and take the expectation to produce

( ) ( ) ( )( )

( ) ( )

( ) ( )( ) ( ) ( )( )( )

22.

1

2 2

1

22

.1 1

2

1

1

1 1

a

A i ii

b

B jj

a b

AB ij ii j

E

bnE MS Ea

anE MSb

nE MS Ea b

E MS

σ τ τβ

σ β

σ τβ τβ

σ

=

=

= =

= + +

−

= + −

= + −

− − =

∑

∑

∑∑

Note that ( )BMSE and ( )EMSE are the results given in Table 8-3. We need to simplify ( )AMSE and ( )ABMSE . Consider ( )AMSE

( ) ( ) ( ) ( )

( )( )

( ) ∑

∑

∑ ∑

=

=

= =

−++=

−

+−

+=

=++

−+=

a

iiA

a

iiA

a

i

a

i.iiA

abnnMSE

ba

a

aabnMSE

ctscrossproduEEabnMSE

1

222

2

1

22

1 1

222

1

1

1

01

τσσ

στσ

τβτσ

τβ

τβ

since ( )ijτβ is

− 210 τβσ

aa,NID . Consider ( )ABMSE

( ) ( )( ) ( ) ( )( )

( ) ( )( )( )

22

.1 1

2 2

1 1

2 2

1 1

1 11 1

a b

AB ij ii j

a b

ABi j

AB

nE MS Ea b

n b aE MSa b b a

E MS n

τβ

τβ

σ τβ τβ

σ σ

σ σ

= =

= =

= + −− −

− − = + − −

= +

∑∑

∑∑


13-14

Thus ( )AMSE and ( )ABMSE 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:

BCAC

ABCC

BCAB

ABCB

ACAB

ABCA

MSMSMSMSF:C

MSMSMSMSF:B

MSMSMSMSF:A

++

=

++

=

++

=

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(MS)

τ i 0 b c n ( )∑ −++

1

222

abcnbn iτσσ τγ

β j a 0 c n ( )∑ −++

1

222

bacnan jβ

σσ βγ

γ k a b 1 n σ σγ2 2+ abn

( )ijτβ 0 0 c n ( )

( )( )

2

2 2

1 1ijn cn

a bτβγ

τβσ σ+ +

− −∑∑

( )ikτγ 0 b 1 n σ στγ2 2+ bn

( )jkβγ a 0 1 n σ σβγ2 2+ an

( )ijkτβγ 0 0 1 n σ στβγ2 2+ n

( )lijkε 1 1 1 1 σ 2

These are exact tests for all effects. 13.15. Consider the experiment in Example 13.6. Analyze the data for the case where A, B, and C are random. Minitab Output ANOVA: Drop versus Temp, Operator, Gauge Factor Type Levels Values Temp random 3 60 75 90 Operator random 4 1 2 3 4 Gauge random 3 1 2 3 Analysis of Variance for Drop Source DF SS MS F P Temp 2 1023.36 511.68 2.30 0.171 x


13-15

Operator 3 423.82 141.27 0.63 0.616 x Gauge 2 7.19 3.60 0.06 0.938 x Temp*Operator 6 1211.97 202.00 14.59 0.000 Temp*Gauge 4 137.89 34.47 2.49 0.099 Operator*Gauge 6 209.47 34.91 2.52 0.081 Temp*Operator*Gauge 12 166.11 13.84 0.65 0.788 Error 36 770.50 21.40 Total 71 3950.32 x Not an exact F-test. Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 Temp 12.044 * (8) + 2(7) + 8(5) + 6(4) + 24(1) 2 Operator -4.544 * (8) + 2(7) + 6(6) + 6(4) + 18(2) 3 Gauge -2.164 * (8) + 2(7) + 6(6) + 8(5) + 24(3) 4 Temp*Operator 31.359 7 (8) + 2(7) + 6(4) 5 Temp*Gauge 2.579 7 (8) + 2(7) + 8(5) 6 Operator*Gauge 3.512 7 (8) + 2(7) + 6(6) 7 Temp*Operator*Gauge -3.780 8 (8) + 2(7) 8 Error 21.403 (8) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 1 Temp 6.97 222.63 (4) + (5) - (7) 2 Operator 7.09 223.06 (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(MS)

τ i 0 b c n ( )∑ −++++

1

22222

abcncnbnn iτσσσσ τβτγτβγ

β j a 1 c n σ σ σβγ β2 2 2+ +an acn

γ k a b 1 n σ σ σβγ γ2 2 2+ +an abn

( )ijτβ 0 1 c n σ σ στβγ τβ2 2 2+ +n cn

( )ikτγ 0 b 1 n σ σ στβγ τγ2 2 2+ +n bn

( )jkβγ a 1 1 n σ σβγ2 2+ an

( )ijkτβγ 0 1 1 n σ στβγ2 2+ n

ε ijkl 1 1 1 1 σ 2 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:


13-16

( ) ( ) ( ) ( ) ( ) ( )ijklh i j k l ij ik il jk jl kl

y µ τ β γ δ τβ τγ τδ βγ βδ γδ= + + + + + + + + + + +

( ) ( ) ( ) ( ) ( ) ijklhijk ijl jkl ikl ijklτβγ τβδ βγδ τγδ τβγδ ε+ + + + +

To simplify the expected mean square derivations, let capital Latin letters represent the factor effects or

variance components. For example, Abcdn

ai

=−∑τ 2

1, or B acdn= σβ

2 .

(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(MS) τ i 0 b c d n σ 2 + A β j a 0 c d n σ 2 + B γ k a b 0 d n σ 2 + C δl a b c 0 n σ 2 + D ( )τβ ij 0 0 c d n σ 2 + AB ( )τγ ik 0 b 0 d n σ 2 + AC ( )τδ il 0 b c 0 n σ 2 + AD ( )βγ jk a 0 0 d n σ 2 + BC ( )βδ jl a 0 c 0 n σ 2 + BD ( )γδ kl a b 0 0 n σ 2 + CD ( )τβγ ijk 0 0 0 d n σ 2 + ABC ( )τβδ ijl 0 0 c 0 n σ 2 + ABD ( )βγδ jkl a 0 0 0 n σ 2 + BCD ( )τγδ ikl 0 b 0 0 n σ 2 + ACD ( )τβγδ ijkl 0 0 0 0 n σ 2 + ABCD ε ( )ijkl h 1 1 1 1 1 σ 2

There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B fixed 2 H L C fixed 2 H L D fixed 2 H L Analysis of Variance for y Source DF SS MS F P A 1 6.13 6.13 0.49 0.492 B 1 0.13 0.13 0.01 0.921 C 1 1.13 1.13 0.09 0.767 D 1 0.13 0.13 0.01 0.921 A*B 1 3.13 3.13 0.25 0.622 A*C 1 3.13 3.13 0.25 0.622 A*D 1 3.13 3.13 0.25 0.622 B*C 1 3.13 3.13 0.25 0.622 B*D 1 3.13 3.13 0.25 0.622 C*D 1 3.13 3.13 0.25 0.622 A*B*C 1 3.13 3.13 0.25 0.622 A*B*D 1 28.13 28.13 2.27 0.151


13-17

A*C*D 1 3.13 3.13 0.25 0.622 B*C*D 1 3.13 3.13 0.25 0.622 A*B*C*D 1 3.13 3.13 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A 16 (16) + 16Q[1] 2 B 16 (16) + 16Q[2] 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] 12 A*B*D 16 (16) + 4Q[12] 13 A*C*D 16 (16) + 4Q[13] 14 B*C*D 16 (16) + 4Q[14] 15 A*B*C*D 16 (16) + 2Q[15] 16 Error 12.38 (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(MS) τ i 1 b c d n σ 2 + + + + + + + +ABCD ACD ABD ABC AD AC AB A β j a 1 c d n σ 2 + + + + + + + +ABCD BCD ABD ABC BD BC AB B γ k a b 1 d n 2 ABCD ACD BCD ABC AC BC CD Cσ + + + + + + + + δl a b c 1 n σ 2 + + + + + + + +ABCD ACD BCD ABD BD AD CD D ( )τβ ij 1 1 c d n σ 2 + + + +ABCD ABC ABD AB ( )τγ ik 1 b 1 d n σ 2 + + + +ABCD ABC ACD AC ( )τδ il 1 b c 1 n σ 2 + + + +ABCD ABD ACD AD ( )βγ jk a 1 1 d n σ 2 + + + +ABCD ABC BCD BC ( )βδ jl a 1 c 1 n σ 2 + + + +ABCD ABD BCD BD ( )γδ kl a b 1 1 n σ 2 + + + +ABCD ACD BCD CD ( )τβγ ijk 1 1 1 d n σ 2 + +ABCD ABC ( )τβδ ijl 1 1 c 1 n σ 2 + +ABCD ABD ( )βγδ jkl a 1 1 1 n σ 2 + +ABCD BCD ( )τγδ ikl 1 b 1 1 n σ 2 + +ABCD ACD ( )τβγδ ijkl 1 1 1 1 n σ 2 + ABCD ε ( )ijkl h 1 1 1 1 1 σ 2

No exact tests exist on main effects or two-factor interactions. For main effects use statistics such as:

A FMS MS MS MSMS MS MS MS

A ABC ABD ACD

AB AC AD ABCD: =

+ + ++ + +

For testing two-factor interactions use statistics such as: AB FMS MSMS MS

AB ABCD

ABC ABD: =

++

The results can also be generated in Minitab as follows:


13-18

Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A random 2 H L B random 2 H L C random 2 H L D random 2 H L Analysis of Variance for y Source DF SS MS 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 28.13 28.13 9.00 0.205 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 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 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) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6) + 8(5) + 16(1) 2 B 1.3750 * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8) + 8(5) + 16(2) 3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6) 7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7) 8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8) 9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9) 10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + 2(15) + 4(11) 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D 0.0000 15 (16) + 2(15) + 4(13) 14 B*C*D 0.0000 15 (16) + 2(15) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15) 2 B 0.56 * (5) + (8) + (9) - (11) - (12) - (14) + (15) 3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15) 5 A*B 0.98 28.13 (11) + (12) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 28.13 (12) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15)


13-19

9 B*D 0.98 28.13 (12) + (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(MS) τ i 0 b c d n σ 2 + + + + + + + +ABCD ACD ABD ABC AD AC AB A β j a 1 c d n 2 BCD BD BC Bσ + + + + γ k a b 1 d n σ 2 + + + +BCD BC CD C δl a b c 1 n σ 2 + + + +BCD BD CD D ( )τβ ij 0 1 c d n σ 2 + + + +ABCD ABC ABD AB ( )τγ ik 0 b 1 d n σ 2 + + + +ABCD ABC ACD AC ( )τδ il 0 b c 1 n σ 2 + + + +ABCD ABD ACD AD ( )βγ jk a 1 1 d n σ 2 + +BCD BC ( )βδ jl a 1 c 1 n σ 2 + +BCD BD ( )γδ kl a b 1 1 n σ 2 + +BCD CD ( )τβγ ijk 0 1 1 d n σ 2 + +ABCD ABC ( )τβδ ijl 0 1 c 1 n σ 2 + +ABCD ABD ( )βγδ jkl a 1 1 1 n σ 2 + BCD ( )τγδ ikl 0 b 1 1 n σ 2 + +ABCD ACD ( )τβγδ ijkl 0 1 1 1 n σ 2 + ABCD ε ( )ijkl h 1 1 1 1 1 σ 2

No exact tests exist on main effects or two-factor interactions involving the fixed factor A. To test the fixed factor A use

A FMS MS MS MSMS MS MS MS

A ABC ABD ACD

AB AC AD ABCD: =

+ + ++ + +

Random main effects could be tested by, for example: : D BCD

BD CD

MS MSD FMS MS

+=

+

For testing two-factor interactions involving A use: AB FMS MSMS MS

AB ABCD

ABC ABD: =

++

The results can also be generated in Minitab as follows:


13-20

Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B random 2 H L C random 2 H L D random 2 H L Analysis of Variance for y Source DF SS MS 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 28.13 28.13 9.00 0.205 A*C*D 1 3.13 3.13 1.00 0.500 B*C*D 1 3.13 3.13 0.25 0.622 A*B*C*D 1 3.13 3.13 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 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) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6) + 8(5) + 16Q[1] 2 B -0.1875 * (16) + 4(14) + 8(9) + 8(8) + 16(2) 3 C -0.1250 * (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 -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6) 7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 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) + 2(15) + 4(11) 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D 0.0000 15 (16) + 2(15) + 4(13) 14 B*C*D -2.3125 16 (16) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15) 2 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 28.13 (11) + (12) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 28.13 (12) + (13) - (15)


13-21

(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(MS) τ i 0 b c d n σ 2 + + + +ACD AD AC A β j a 0 c d n σ 2 + + + +BCD BC BD B γ k a b 1 d n σ 2 + +CD C δl a b c 1 n σ 2 + +CD D ( )τβ ij 0 0 c d n σ 2 + + + +ABCD ABC ABD AB ( )τγ ik 0 b 1 d n σ 2 + +ACD AC ( )τδ il 0 b c 1 n σ 2 + +ACD AD ( )βγ jk a 0 1 d n σ 2 + +BCD BC ( )βδ jl a 0 c 1 n σ 2 + +BCD BD ( )γδ kl a b 1 1 n σ 2 + CD ( )τβγ ijk 0 0 1 d n σ 2 + +ABCD ABC ( )τβδ ijl 0 0 c 1 n σ 2 + +ABCD ABD ( )βγδ jkl a 0 1 1 n σ 2 + BCD ( )τγδ ikl 0 b 1 1 n σ 2 + ACD ( )τβγδ ijkl 0 0 1 1 n σ 2 + ABCD ε ( )ijkl h 1 1 1 1 1 σ 2

There are no exact tests on the fixed factors A and B, or their two-factor interaction AB. The appropriate test statistics are:

A FMS MSMS MS

B FMS MSMS MS

A ACD

AC AD

B BCD

BC BD

:

:

=++

=++

AB FMS MSMS MS

AB ABCD

ABC ABD: =

++

The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B fixed 2 H L C random 2 H L D random 2 H L Analysis of Variance for y Source DF SS MS 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


13-22

C*D 1 3.13 3.13 0.25 0.622 A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 28.13 28.13 9.00 0.205 A*C*D 1 3.13 3.13 0.25 0.622 B*C*D 1 3.13 3.13 0.25 0.622 A*B*C*D 1 3.13 3.13 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 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] 2 B * (16) + 4(14) + 8(9) + 8(8) + 16Q[2] 3 C -0.1250 10 (16) + 8(10) + 16(3) 4 D -0.1875 10 (16) + 8(10) + 16(4) 5 A*B * (16) + 2(15) + 4(12) + 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) + 2(15) + 4(11) 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D -2.3125 16 (16) + 4(13) 14 B*C*D -2.3125 16 (16) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 1 A 0.33 3.13 (6) + (7) - (13) 2 B 0.33 3.13 (8) + (9) - (14) 5 A*B 0.98 28.13 (11) + (12) - (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(MS) τ i 0 b c d n σ 2 + +AD A β j a 0 c d n σ 2 + +BD B γ k a b 0 d n σ 2 + +CD C δl a b c 1 n σ 2 + D ( )τβ ij 0 0 c d n σ 2 + +ABD AB ( )τγ ik 0 b 0 d n σ 2 + +ACD AC ( )τδ il 0 b c 1 n σ 2 + AD ( )βγ jk a 0 0 d n σ 2 + +BCD BC ( )βδ jl a 0 c 1 n σ 2 + BD ( )γδ kl a b 0 1 n σ 2 + CD ( )τβγ ijk 0 0 0 d n σ 2 + +ABCD ABC ( )τβδ ijl 0 0 c 1 n σ 2 + ABD ( )βγδ jkl a 0 0 1 n σ 2 + BCD ( )τγδ ikl 0 b 0 1 n σ 2 + ACD ( )τβγδ ijkl 0 0 0 1 n σ 2 + ABCD ε ( )ijkl h 1 1 1 1 1 σ 2


13-23

There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B fixed 2 H L C fixed 2 H L D random 2 H L Analysis of Variance for y Source DF SS MS 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 0.921 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 0.25 0.622 B*C 1 3.13 3.13 1.00 0.500 B*D 1 3.13 3.13 0.25 0.622 C*D 1 3.13 3.13 0.25 0.622 A*B*C 1 3.13 3.13 1.00 0.500 A*B*D 1 28.13 28.13 2.27 0.151 A*C*D 1 3.13 3.13 0.25 0.622 B*C*D 1 3.13 3.13 0.25 0.622 A*B*C*D 1 3.13 3.13 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 Source Variance Error Expected Mean Square for Each Term component term (using restricted model) 1 A 7 (16) + 8(7) + 16Q[1] 2 B 9 (16) + 8(9) + 16Q[2] 3 C 10 (16) + 8(10) + 16Q[3] 4 D -0.7656 16 (16) + 16(4) 5 A*B 12 (16) + 4(12) + 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) + 2(15) + 4Q[11] 12 A*B*D 3.9375 16 (16) + 4(12) 13 A*C*D -2.3125 16 (16) + 4(13) 14 B*C*D -2.3125 16 (16) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (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. Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B random 2 H L C random 2 H L D random 2 H L


13-24

Analysis of Variance for y Source DF SS MS 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 28.13 28.13 9.00 0.205 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 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 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) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6) + 8(5) + Q[1] 2 B 1.3750 * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8) + 8(5) + 16(2) 3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B -3.1250 * (16) + 2(15) + 4(12) + 4(11) + 8(5) 6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6) 7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7) 8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8) 9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9) 10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + 2(15) + 4(11) 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D 0.0000 15 (16) + 2(15) + 4(13) 14 B*C*D 0.0000 15 (16) + 2(15) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 1 A 0.56 * (5) + (6) + (7) - (11) - (12) - (13) + (15) 2 B 0.56 * (5) + (8) + (9) - (11) - (12) - (14) + (15) 3 C 0.14 3.13 (6) + (8) + (10) - (11) - (13) - (14) + (15) 4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15) 5 A*B 0.98 28.13 (11) + (12) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 28.13 (12) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15) 9 B*D 0.98 28.13 (12) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15)

A and B are fixed and C and D are random.


13-25

Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B fixed 2 H L C random 2 H L D random 2 H L

Analysis of Variance for y

Source DF SS MS 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 28.13 28.13 9.00 0.205 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 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88

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) + 2(15) + 4(13) + 4(12) + 4(11) + 8(7) + 8(6) + Q[1,5] 2 B * (16) + 2(15) + 4(14) + 4(12) + 4(11) + 8(9) + 8(8) + Q[2,5] 3 C -0.1250 * (16) + 2(15) + 4(14) + 4(13) + 4(11) + 8(10) + 8(8) + 8(6) + 16(3) 4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B * (16) + 2(15) + 4(12) + 4(11) + Q[5] 6 A*C 0.0000 * (16) + 2(15) + 4(13) + 4(11) + 8(6) 7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7) 8 B*C 0.0000 * (16) + 2(15) + 4(14) + 4(11) + 8(8) 9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9) 10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10) 11 A*B*C 0.0000 15 (16) + 2(15) + 4(11) 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D 0.0000 15 (16) + 2(15) + 4(13) 14 B*C*D 0.0000 15 (16) + 2(15) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16)

* Synthesized Test.

Error Terms for Synthesized Tests

Source Error DF Error MS Synthesis of Error MS 1 A 0.33 3.13 (6) + (7) - (13) 2 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) - (12) - (13) - (14) + (15) 5 A*B 0.98 28.13 (11) + (12) - (15) 6 A*C 0.33 3.13 (11) + (13) - (15) 7 A*D 0.98 28.13 (12) + (13) - (15) 8 B*C 0.33 3.13 (11) + (14) - (15) 9 B*D 0.98 28.13 (12) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15)


13-26

(e) A, B and C are fixed and D is random. Minitab Output ANOVA: y versus A, B, C, D Factor Type Levels Values A fixed 2 H L B fixed 2 H L C fixed 2 H L D random 2 H L Analysis of Variance for y Source DF SS MS 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 28.13 28.13 9.00 0.205 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 0.25 0.622 Error 16 198.00 12.38 Total 31 264.88 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) + 2(15) + 4(13) + 4(12) + 8(7) + Q[1,5,6,11] 2 B 9 (16) + 2(15) + 4(14) + 4(12) + 8(9) + Q[2,5,8,11] 3 C 10 (16) + 2(15) + 4(14) + 4(13) + 8(10) + Q[3,6,8,11] 4 D 1.3750 * (16) + 2(15) + 4(14) + 4(13) + 4(12) + 8(10) + 8(9) + 8(7) + 16(4) 5 A*B 12 (16) + 2(15) + 4(12) + Q[5,11] 6 A*C 13 (16) + 2(15) + 4(13) + Q[6,11] 7 A*D -3.1250 * (16) + 2(15) + 4(13) + 4(12) + 8(7) 8 B*C 14 (16) + 2(15) + 4(14) + Q[8,11] 9 B*D -3.1250 * (16) + 2(15) + 4(14) + 4(12) + 8(9) 10 C*D 0.0000 * (16) + 2(15) + 4(14) + 4(13) + 8(10) 11 A*B*C 15 (16) + 2(15) + Q[11] 12 A*B*D 6.2500 15 (16) + 2(15) + 4(12) 13 A*C*D 0.0000 15 (16) + 2(15) + 4(13) 14 B*C*D 0.0000 15 (16) + 2(15) + 4(14) 15 A*B*C*D -4.6250 16 (16) + 2(15) 16 Error 12.3750 (16) * Synthesized Test. Error Terms for Synthesized Tests Source Error DF Error MS Synthesis of Error MS 4 D 0.56 * (7) + (9) + (10) - (12) - (13) - (14) + (15) 7 A*D 0.98 28.13 (12) + (13) - (15) 9 B*D 0.98 28.13 (12) + (14) - (15) 10 C*D 0.33 3.13 (13) + (14) - (15)


13-27

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. Minitab Output ANOVA: Score versus Cycle Time, Operator, Temperature Factor Type Levels Values Cycle Ti fixed 3 40 50 60 Operator random 3 1 2 3 Temperat fixed 2 300 350 Analysis of Variance for Score Source DF SS MS F P Cycle Ti 2 436.000 218.000 2.45 0.202 Operator 2 261.333 130.667 39.86 0.000 Temperat 1 50.074 50.074 8.89 0.096 Cycle Ti*Operator 4 355.667 88.917 27.13 0.000 Cycle Ti*Temperat 2 78.815 39.407 3.41 0.137 Operator*Temperat 2 11.259 5.630 1.72 0.194 Cycle Ti*Operator*Temperat 4 46.185 11.546 3.52 0.016 Error 36 118.000 3.278 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] 2 Operator 7.0772 8 (8) + 18(2) 3 Temperat 6 (8) + 9(6) + 27Q[3] 4 Cycle Ti*Operator 14.2731 8 (8) + 6(4) 5 Cycle Ti*Temperat 7 (8) + 3(7) + 9Q[5] 6 Operator*Temperat 0.2613 8 (8) + 9(6) 7 Cycle Ti*Operator*Temperat 2.7562 8 (8) + 3(7) 8 Error 3.2778 (8)

The following calculations agree with the Minitab results:

2ˆ EMSσ = 2ˆ 3.27778σ =

2ˆ ABC EMS MSnτβγσ −

= 2 11.546296 3.277778ˆ 2.75623τβγσ −

= =

2ˆ AB EMS MScnτβσ −

= ( )

2 88.91667 3.277778ˆ 14.273152 3τβσ −

= =

2ˆ BC EMS MSanβγσ−

= ( )

2 5.629630 3.277778ˆ 0.261323 3τγσ −

= =

2ˆ B EMS MSacnβσ−

= ( )( )

2 130.66667 3.277778ˆ 7.077163 2 3γσ

−= =

13.20. Consider the three-factor factorial model

( ) ( )ijk i j k ijkij jky µ τ β γ τβ βγ ε= + + + + + +

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.


13-28

Source DF E(MS) A a-1 σ σ στβ τ

2 2 2+ +c bc

B b-1 σ σ σ στβ βγ β2 2 2 2+ + +c a ac

C c-1 σ σ σβγ γ2 2 2+ +a ab

AB (a-1)(b-1) σ στβ2 2+ c

BC (b-1)(c-1) σ σβγ2 2+ a

Error (AC + ABC) b(a-1)(c-1) σ 2 Total abc-1

There are exact tests for all effects except B. To test B, use the statistic FMS MS

MS MSB E

AB BC=

++

13.21. The three-factor factorial model for a single replicate is

( ) ( ) ( ) ( )ijk i j k ij jk ik ijk ijky µ τ β γ τβ βγ τγ τβγ ε= + + + + + + + + If all the factors are random, can any effects be tested? If the three-factor interaction and the ( )τβ ij interaction do not exist, can all the remaining effects be tested? The expected mean squares are found by referring to Table 13.9, deleting the line for the error term ε( )ijk l and setting n=1. The three-factor interaction now cannot be tested; however, exact tests exist for the two-factor interactions and approximate F tests can be conducted for the main effects. For example, to test the main effect of A, use

FMS MSMS MS

A ABC

AB AC=

++

If ( )τβγ ijk and ( )τβ ij can be eliminated, the model becomes

( ) ( )ijk i j k ijkjk iky µ τ β γ βγ τγ ε= + + + + + +

For this model, the analysis of variance is

Source DF E(MS) A a-1 σ σ στγ τ

2 2 2+ +b bc

B b-1 σ σ σβγ β2 2 2+ +a ac

C c-1 σ σ σ σβγ τγ γ2 2 2 2+ + +a b ab

AC (a-1)(c-1) σ στγ2 2+ b

BC (b-1)(c-1) σ σβγ2 2+ a

Error (AB + ABC) c(a-1)(b-1) σ 2 Total abc-1

There are exact tests for all effect except C. To test the main effect of C, use the statistic:


13-29

FMS MS

MS MSC E

BC AC=

++

13.22. 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 σ σβ

2 2= , where σβ2 is the variance component

for the machine factor. Are two replicates sufficient?

λσ

σ σβ

τβ= +

+1

2

2 2

an

n

If σ σβ2 2= , then an estimate of σ σβ

2 2 379= = . , and an estimate of 2 7.44τβσ = , from the analysis of variance table. Then

( )( )( )3 2 3.791 2.22 1.49

3.79 2(7.44)λ = + = =

+

and the other OC curve parameters are υ1 3= and υ2 6= . This results in β ≈ 0 75. approximately, with α = 0 05. , or β ≈ 0 9. with α = 0 01. . Two replicates does not seem sufficient. 13.23. In the two-factor mixed model analysis of variance, show that ( ) ( ) ( )2

'Cov , 1

ij i ja

τβστβ τβ = −

for

i≠i'.

Since ( )∑=

=a

iij

1

0τβ (constant) we have ( ) 01

=

∑=

a

iijV τβ , which implies that

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

'1

2'

2'

2'

2 , 02

1 ! 2 , 02! 2 !

1 1 , 0

1,

a

ij ij i ji

ij i j

ij i j

ij i j

aV Cov

a aa Cova a

a a a Cov

Cova

τβ

τβ

τβ

τβ τβ τβ

σ τβ τβ

σ τβ τ β

τ β τβ σ

=

+ = − + = −

− + − = = −

∑

13.24. 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 σ 2 be the vector of variance components such that E( )g A= σ 2 , 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.


13-30

2 2 -1ˆ⇒ =g = As s A g

Since -1A always exists then,

( ) ( ) ( ) ( )2 -1 -1 -1 2 2E E E= =s = A g A g = A As s

Thus σ 2 is an unbiased estimator of σ 2 . This and other properties of the analysis of variance method are discussed by Searle (1971a). 13.25. 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 στ

2 0< in a one-factor analysis of variance. Comment on the usefulness of this probability statement.

Suppose σ 2 1 2=−MS MSc

, where MSi for i=1,2 are two mean squares and c is a constant. The

probability that 02 <τσö (negative) is

{ } { } ( )

( )

( )( )

( )( )

1

1 1 12 11 2 ,

22 2 2

2

ˆ 0 0 1 u v

MSE MS E MS E MSMSP P MS MS P P P FMSMS E MS E MSE MS

σ

< = − < = < = < = <

where u is the number of degrees of freedom for MS1 and v is the number of degrees of freedom for MS2 . For the one-way model, this equation reduces to

{ }2

21, 1,2 2

1ˆ 01a N a a N aP P F P F

n nkτ

σσσ σ− − − −

< = < = < + +

where 2

2

σ

στ=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 2ˆ 0τσ < . 13.26. 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: Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 10 1 2 3 4 5 6 7 8 9 10 Operator fixed 2 1 2 Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.017 11.002 7.33 0.000


13-31

Operator 1 0.417 0.417 0.69 0.427 Part*Operator 9 5.417 0.602 0.40 0.927 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) 2 Operator 3 (4) + 3(3) + 30Q[2] 3 Part*Operator -0.2994 4 (4) + 3(3) 4 Error 1.5000 (4)

The second approach is the unrestricted mixed model. Minitab Output ANOVA: Measurement versus Part, Operator Factor Type Levels Values Part random 10 1 2 3 4 5 6 7 8 9 10 Operator fixed 2 1 2 Analysis of Variance for Measurem Source DF SS MS F P Part 9 99.017 11.002 18.28 0.000 Operator 1 0.417 0.417 0.69 0.427 Part*Operator 9 5.417 0.602 0.40 0.927 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) 2 Operator 3 (4) + 3(3) + Q[2] 3 Part*Operator -0.2994 4 (4) + 3(3) 4 Error 1.5000 (4)

Source Sum of Squares

DF Mean Square

E(MS) F-test F

A 99.016667 a-1=9 11.00185 2 2 2n bnτβ τσ σ σ+ + A

AB

MSFMS

= 18.28

B 0.416667 b-1=1 0.416667

2

2 2 1

1

b

iin anbτβ

βσ σ =+ +

−

∑

B

AB

MSFMS

= 0.692

AB 5.416667 (a-1)(b-1)=9 0.60185 σ στβ2 2+ n

E

AB

MSMS

F = 0.401

Error 60.000000 40 1.50000 σ 2 Total 164.85000 nabc-1=59

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 MSAB will generally be larger than MSE. However, this is not the case with this particular experiment. 13.27. Consider the two-factor mixed model. Show that the standard error of the fixed factor mean (e.g.

A) is [ ]1 2/ABMS bn .


13-32

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

( )..m..i yyV − where rows are fixed and columns are random. Now, assuming all model parameters to be independent, we have the following:

( ) ( ) ( ) ∑∑∑∑∑∑= == ===

−+−+−=−b

j

n

kmjk

b

j

n

kijk

b

jmj

b

jijmi..m..i bnbnbb

yy1 11 111

1111 εετβτβττ

and

( ) ( )bn

nbn

bnbn

bnb

bb

byyV ..m..i

222

22

22

22

2 21111 τβτβτβ

σσσσσσ

+=

+

+

+

=−

Since MSAB estimates σ στβ

2 2+ n , we would use

2 MSbn

AB

as the standard error to test the difference. However, the table of ranges for Duncan’s Multiple Range test already includes the constant 2. 13.28. Consider the variance components in the random model from Problem 13.1. (a) Find an exact 95 percent confidence interval on σ2.

22 2

2, 1 2,E E

E E E E

f f

f MS f MS

α α

σχ χ −

≤ ≤

( )( ) ( )( )240 1.5 40 1.559.34 24.43

σ≤ ≤

21.011 2.456σ≤ ≤ (b) Find approximate 95 percent confidence intervals on the other variance components using the

Satterthwaite method. στβ

2 and στ2 are negative, and the Satterthwaithe method does not apply. The confidence interval on σβ

2 is


= ( )

2 11.001852 0.6018519ˆ 1.73332 3βσ−

= =

( )

( ) ( )( )

( )

( ) ( )( )

2 2

2 2 2 2

11.001852 0.60185198.01826

11.001852 0.60185199 1 91 1 1

B AB

B AB

MS MSr

MS MSb a b

− −= = =

++− − −


13-33

22

22 2

2, 1 2,

ˆˆO

r r

rr ββ

α α

σσ σχ χ −

≤ ≤

( )( ) ( )( )28.01826 1.7333 8.01826 1.7333

17.55752 2.18950βσ≤ ≤

20.79157 6.34759βσ≤ ≤

13.29. Use the experiment described in Problem 5.8 and assume that both factors are random. Find an exact 95 percent confidence interval on σ2. Construct approximate 95 percent confidence interval on the other variance components using the Satterthwaite method.

2ˆ EMSσ = 2ˆ 3.79167σ = 2

2 22, 1 2,E E

E E E E

f f

f MS f MS

α α

σχ χ −

≤ ≤

( )( ) ( )( )212 3.79167 12 3.79167

23.34 4.40σ≤ ≤

21.9494 10.3409σ≤ ≤

Satterthwaite Method:


= 2 7.44444 3.79167ˆ 1.826392τβσ −

= =

( )

( )( )

( )

( )( ) ( )

2 2

2 2 2 2

7.44444 3.791671.27869

7.44444 3.791672 3 121 1

AB E

AB E

E

MS MSr

MS MSa b df

− −= = =

++− −

2 2

22 2

2, 1 2,

ˆ ˆ

r r

r rβ ββ

α α

σ σσ

χ χ −

≤ ≤

2

2,rαχ and 21 2,rαχ − were estimated by extrapolating 1.27869 between degrees of freedom of one and two in

Microsoft Excel. More precise methods can be used as well.

( )( ) ( )( )21.27869 1.82639 1.27869 1.826395.67991 0.014820βσ≤ ≤

20.41117 158.58172βσ≤ ≤


13-34

2ˆ 0βσ < , this variance component does not have a confidence interval using Satterthwaite’s Method.


= ( )

2 80.16667 7.44444ˆ 9.090284 2τσ−

= =

( )

( ) ( )( )

( )

( ) ( )( )

2 2

2 2 2 2

80.16667 7.444441.64108

80.16667 7.444442 2 31 1 1

A AB

A AB

MS MSr

MS MSa a b

− −= = =

++− − −

2 2

22 2

2, 1 2,

ˆ ˆ

r r

r rτ ττ

α α

σ σσχ χ −

≤ ≤

2(1.64108)(9.09028) (1.64108)(9.09028)

6.53293 0.03281τσ≤ ≤

2



22.28349 454.61891τσ≤ ≤

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.

2ˆ B EMS MSacnβσ−

= ( )( )

2 130.66667 3.277778ˆ 7.077162 3 3βσ

−= =

( )

( )

( )

( ) ( )

2 2

2 2 2 2

130.66667 3.277781.90085

130.66667 3.277782 361

B E

B E

E

MS MSr

MS MSb df

− −= = =

++−

2 2

22 2

2, 1 2,

ˆ ˆ

r r

r rβ ββ

α α

σ σσ

χ χ −

≤ ≤

2



( )( ) ( )( )21.90085 7.07716 1.90085 7.077167.14439 0.04571βσ≤ ≤

21.88296 294.28720βσ≤ ≤

13.31. Rework Problem 13.28 using the modified large-sample approach described in Section 13.7.2. Compare the two sets of confidence intervals obtained and discuss.


13-35

2 2ˆ ˆ B ABO

MS MSanβσ σ −

= = ( )

2 11.001852 0.6018519ˆ 1.73332 3Oσ−

= =

( ) ( ) ( ) ( )

10.05,9,

1 2.95,9.95,9 ,

22 2 2 2 2

, , 1 , , 1

, ,

1 11 1 0.468091.88

1 1 11 1 1 1.70270.370

9

1 3.18 1 0.46809 3.18 1.70270.36366

3.18

i

i j i j

i j

f f f f

ijf f

GF

HF

F G F HG

Fα α

α

χ

∞

∞

= − = − =

= − = − = − =

− − − − − −= = =

( ) ( ) ( ) ( ) ( ) ( )( )

2 2 2 2 2 21 1 1 2 11 1 2

2 22 2 2 21 1 1 10.46809 11.00185 1.7027 0.60185 0.36366 11.00185 0.60185

6 6 6 60.83275

L B AB B AB

L

L

V G c MS H c MS G c c MS MS

V

V

= + +

= + +

=

2ˆ 1.7333 0.83275 0.82075LL Vβσ= − = − =

13.32. Rework Problem 13.28 using the modified large-sample method described in Section 13.7.2. Compare this confidence interval with the one obtained previously and discuss.

2ˆ C EMS MSabnγσ−

= ( )( )

2 130.66667 3.277778ˆ 7.077162 3 3γσ

−= =

( ) ( ) ( ) ( )

10.05,3,

1 2.95,36.95,36,

22 2 2 2 2

, , 1 , , 1

, ,

1 11 1 0.615382.60

1 1 11 1 . 1 0.544930.64728

36

1 2.88 1 0.61538 2.88 0.544930.74542

2.88i j i j

i j

f f f f

ijf f

GF

HF

F G F HG

Fα α

α

χ

∞

∞

= − = − =

= − = − = − =

− − − − − −= = =

( ) ( ) ( ) ( ) ( ) ( )(

2 2 2 2 2 21 1 1 2 11 1 2

2 22 2 2 21 1 1 10.61538 130.66667 0.54493 3.27778 0.74542 130.66667 3.277

18 18 18 1820.95112

L B AB B AB

L

L

V G c MS H c MS G c c MS MS

V

V

= + +

= + +

=

2ˆ 7.07716 20.95112 2.49992LL Vγσ= − = − = 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 13.29. The JMP REML analysis below was performed with both factors, Operator and Machine, as random.


13-36

The CIs for the error variance are similar to those found in Problem 13.29. The upper CIs for the other variance components are much larger than those estimated in the JMP REML output below. JMP Output RSquare 0.742044 RSquare Adj 0.742044 Root Mean Square Error 1.94722 Mean of Response 112.2917 Observations (or Sum Wgts) 24 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 112.29167 1.789728 1.831 62.74 0.0005* REML Variance Component Estimates Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Operator 2.3974359 9.0902778 10.035225 -10.5784 28.758958 64.198 Machine -0.144689 -0.548611 0.9124188 -2.336919 1.239697 -3.874 Operator*Machine 0.481685 1.8263889 2.2841505 -2.650464 6.3032416 12.898 Residual 3.7916667 1.5479414 1.9497217 10.332013 26.778 Total 14.159722 100.000 Covariance Matrix of Variance Component Estimates Random Effect Operator Machine Operator*Machine Residual Operator 100.70575 0.3848594 -1.154578 2.151e-12 Machine 0.3848594 0.8325081 -1.539438 -1.17e-13 Operator*Machine -1.154578 -1.539438 5.2173434 -1.198061 Residual 2.151e-12 -1.17e-13 -1.198061 2.3961227 13.34 Consider the experiment described in Problem 13.1. Analyze the data using REML. Compare the CIs to those obtained in Problem 13.28. 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 13.28 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 13.28. The upper CI for the Part Number variance estimated with the Satterthwaite method in Problem 13.28 is approximately twice the value estimated in the JMP REML analysis. JMP Output RSquare 0.388009 RSquare Adj 0.388009 Root Mean Square Error 1.224745 Mean of Response 49.95 Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 49.95 0.424591 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 0.0366326 3.430034 59.203 Operator -0.004115 -0.006173 0.0218 -0.0489 0.0365544 -0.211 Part Number*Operator -0.199588 -0.299383 0.1464372 -0.586394 -0.012371 -10.226 Residual 1.5 0.3354102 1.0110933 2.4556912 51.233 Total 2.9277778 100.000


13-37

Covariance Matrix of Variance Component Estimates Random Effect Part Number Operator Part Number*Operator Residual Part Number 0.749401 0.0004472 -0.004472 9.256e-14 Operator 0.0004472 0.0004752 -0.000894 -3.32e-16 Part Number*Operator -0.004472 -0.000894 0.0214438 -0.0375 Residual 9.256e-14 -3.32e-16 -0.0375 0.1125 13.35 Rework Problem 13.31 using REML. Compare all sets of CIs for the variance components. JMP uses the unrestricted approach for estimating variance components with mixed models. The JMP REML output is shown below. The lower confidence interval on 2

βσ (Operator) comparisons between the Satterthwaite method in Problem 13.28, the modified large sample approach in Problem 13.31, and the JMP REML output below are:

Satterthwaite 20.79157 6.34759βσ≤ ≤

Modified 20.82075 βσ≤

REML 20.03663 3.43003βσ≤ ≤ JMP Output RSquare 0.388009 RSquare Adj 0.388009 Root Mean Square Error 1.224745 Mean of Response 49.95 Observations (or Sum Wgts) 60 Parameter Estimates Term Estimate Std Error DFDen t Ratio Prob>|t| Intercept 49.95 0.424591 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 0.0366326 3.430034 59.203 Operator -0.004115 -0.006173 0.0218 -0.0489 0.0365544 -0.211 Part Number*Operator -0.199588 -0.299383 0.1464372 -0.586394 -0.012371 -10.226 Residual 1.5 0.3354102 1.0110933 2.4556912 51.233 Total 2.9277778 100.000 Covariance Matrix of Variance Component Estimates Random Effect Part Number Operator Part Number*Operator Residual Part Number 0.749401 0.0004472 -0.004472 9.256e-14 Operator 0.0004472 0.0004752 -0.000894 -3.32e-16 Part Number*Operator -0.004472 -0.000894 0.0214438 -0.0375 Residual 9.256e-14 -3.32e-16 -0.0375 0.1125

Chapter 13 Experiments with Random Factors Solutionssmills/2013-14... · Solutions from Montgomery,...

Documents

Transcript of Chapter 13 Experiments with Random Factors Solutionssmills/2013-14... · Solutions from Montgomery,...