Experiments, Wiley, NY Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments
Chapter 13 Experiments with Random Factors
Solutions 13.1. An experiment was performed to investigate the capability of a measurement 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 1 2 3 4 5 6 7 8 9 10
1 50 52 53 49 48 52 51 52 50 47
Operator 1 Measurements 2 49 52 50 51 49 50 51 50 51 46
3 50 51 50 50 48 50 51 49 50 49
1 50 51 54 48 48 52 51 53 51 46
Operator 2 Measurements 2 48 51 52 50 49 50 50 48 48 47
3 51 51 51 51 48 50 50 50 49 48
(a) Analyze the data from this experiment. Minitab Output ANOVA: M easur ement vers us Par t, Oper ator
Fact or Part
Type Type Level Level s Val ues ues r ando andom m 10 1 8 Oper per at or r ando andom m 2 1
2 9 2
3 10
4
5
6
7
Anal ysi s of Vari ance ance f or Measurem easurem Sour ce Part Oper per at or Par Par t *Ope *Operr at or Err or Tot al
DF 9 1 9 40 59
SS 99. 99. 017 017 0. 417 417 5. 417 417 60. 60. 000 000 164. 164. 850
MS 11. 11. 002 002 0. 417 417 0. 602 602 1. 500 500
F 18. 18. 28 0. 69 0. 40
P 0. 000 000 0. 427 427 0. 927 927
Sour Sour ce 1 2 3 4
Var i ance ance Er r or Expect Expect ed Mean Squa Squarr e f or Each Each Term com componen onentt t erm ( usi ng r est r i ct ed model odel ) Part 1. 73333 3 ( 4) + 3( 3) + 6( 1) Oper ator - 0. 00617 3 ( 4) + 3( 3) + 30( 2) Par Par t *Op *Operator - 0. 29938 4 ( 4) + 3( 3) Er r or 1. 500 50000 ( 4)
(b) Find point estimates of the variance components using using the analysis of variance variance method.
ˆ2 σ 2 ˆτβ σ
=
MS AB
− MS E n
2 ˆτβ = σ
= MS E
ˆ2 σ
= 1.5
0.6018 0.6018519 519 − 1.50000 1.5000000 00 3
13-1
< 0 , assume σ ˆτβ 2 = 0
Experiments, Wiley, NY Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments
ˆ β 2 σ ˆτ 2 σ
=
=
MS B
− MS AB
ˆ β 2 σ
an
MS A − MS AB bn
ˆτ 2 σ
=
11.00185 11.001852 2 − 0.601851 0.6018519 9 2 ( 3)
0.416667 0.416667 − 0.60185 0.6018519 19 10 ( 3 )
=
= 1.7333
< 0 , assume σ ˆτ 2 = 0
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 shown in Table P13.2. Table P13.2
Part Number 1 2 3 4 5 6 7 8 9 10
Inspector 1 Test 1 Test 2 Test 3 37 38 37 42 41 43 30 31 31 42 43 42 28 30 29 42 42 43 25 26 27 40 40 40 25 25 25 35 34 34
Inspector 2 Test 1 Test 2 Test 3 41 41 40 42 42 42 31 31 31 43 43 43 29 30 29 45 45 45 28 28 30 43 42 42 27 29 28 35 35 34
Inspector 3 Test 1 Test 2 Test 3 41 42 41 43 42 43 29 30 28 42 42 42 31 29 29 44 46 45 29 27 27 43 43 41 26 26 26 35 34 35
(a) Analyze the data from this experiment, assuming both parts and and operators are random effects. Minitab Output ANOVA: Im ped anc e vers us In spec to r, Part
Fact or Type Type Level Level s Val ues ues I nspect nspect o r ando andom m 3 1 Part r ando andom m 10 1 8
2 2 9
3 3 10
4
5
6
7
Anal ysi s of Vari ance ance f or I mpedan edanc c Sour ce I nspect nspect o Part I nspect nspect o*Par o*Par t Err or Tot al
DF 2 9 18 60 89
SS 39. 39. 27 3935 3935.. 96 48. 48. 51 30. 30. 67 4054. 4054. 40
MS 19. 19. 63 437. 437. 33 2. 70 0. 51
F 7. 28 162. 162. 27 5. 27
Sour Sour ce 1 2 3 4
P 0. 005 005 0. 000 000 0. 000 000
Var i ance ance Er r or Expect Expect ed Mean Squa Squarr e f or Each Each Term com componen onentt t erm ( usi ng r est r i ct ed model odel ) I nspec spectt o 0. 5646 3 (4) + 3( 3) + 30( 1) Part 48. 2926 3 ( 4) + 3( 3) + 9( 2) I nspec spectt o*Pa o*Par t 0. 7280 4 ( 4) + 3( 3) Er r or 0. 5111 ( 4)
13-2
Experiments, Wiley, NY Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments
ˆ β 2 σ ˆτ 2 σ
=
=
MS B
− MS AB
ˆ β 2 σ
an
MS A − MS AB bn
ˆτ 2 σ
=
11.00185 11.001852 2 − 0.601851 0.6018519 9 2 ( 3)
0.416667 0.416667 − 0.60185 0.6018519 19 10 ( 3 )
=
= 1.7333
< 0 , assume σ ˆτ 2 = 0
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 shown in Table P13.2. Table P13.2
Part Number 1 2 3 4 5 6 7 8 9 10
Inspector 1 Test 1 Test 2 Test 3 37 38 37 42 41 43 30 31 31 42 43 42 28 30 29 42 42 43 25 26 27 40 40 40 25 25 25 35 34 34
Inspector 2 Test 1 Test 2 Test 3 41 41 40 42 42 42 31 31 31 43 43 43 29 30 29 45 45 45 28 28 30 43 42 42 27 29 28 35 35 34
Inspector 3 Test 1 Test 2 Test 3 41 42 41 43 42 43 29 30 28 42 42 42 31 29 29 44 46 45 29 27 27 43 43 41 26 26 26 35 34 35
(a) Analyze the data from this experiment, assuming both parts and and operators are random effects. Minitab Output ANOVA: Im ped anc e vers us In spec to r, Part
Fact or Type Type Level Level s Val ues ues I nspect nspect o r ando andom m 3 1 Part r ando andom m 10 1 8
2 2 9
3 3 10
4
5
6
7
Anal ysi s of Vari ance ance f or I mpedan edanc c Sour ce I nspect nspect o Part I nspect nspect o*Par o*Par t Err or Tot al
DF 2 9 18 60 89
SS 39. 39. 27 3935 3935.. 96 48. 48. 51 30. 30. 67 4054. 4054. 40
MS 19. 19. 63 437. 437. 33 2. 70 0. 51
F 7. 28 162. 162. 27 5. 27
Sour Sour ce 1 2 3 4
P 0. 005 005 0. 000 000 0. 000 000
Var i ance ance Er r or Expect Expect ed Mean Squa Squarr e f or Each Each Term com componen onentt t erm ( usi ng r est r i ct ed model odel ) I nspec spectt o 0. 5646 3 (4) + 3( 3) + 30( 1) Part 48. 2926 3 ( 4) + 3( 3) + 9( 2) I nspec spectt o*Pa o*Par t 0. 7280 4 ( 4) + 3( 3) Er r or 0. 5111 ( 4)
13-2
Experiments, Wiley, NY Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments
(b) Estimate the variance components using the the analysis of variance method.
ˆ2 σ
=
2 ˆτβ σ
ˆ β 2 σ
=
ˆτ 2 σ
= MS E
− MS E
MS AB
n MS B − MS AB
=
ˆ 2 = 0.51 σ
an
2 ˆτβ σ
ˆ β 2 σ
MS A − MS AB
=
ˆτ 2 σ
bn
2.70 2.70 − 0.51 0.51 = 0.73 3 437. 437.33 33 − 2.70 2.70 = 48.29 3 ( 3)
=
19.63 19.63 − 2.70 2.70 = 0.56 10 ( 3)
=
All estimates agree with the Minitab output.
13.3. Reconsider the data in Problem 5.8. Suppose that both factors, machines machines and and operators, are chosen at random.
(a) Analyze the data from this experiment. Machine Operator 1
1 109 110
2 110 115
3 108 109
4 110 108
2
110 112
110 111
111 109
114 112
3
116 114
112 115
114 119
120 117
The following Minitab output contains the analysis of variance and the variance component estimates: Minitab Output ANOVA: St ren gt h v ersu s Oper ato r, Mach in e
Fact or Type Type Level Level s Val ues ues Oper per at or r ando andom m 3 1 Machi ne r andom andom 4 1
2 2
3 3
4
Anal ysi s of Var i ance for St r eng engt h Sour ce Oper per at or Machi ne Oper per at or *Machi ne Err or Tot al
DF 2 3 6 12 23
SS 160. 160. 333 333 12. 12. 458 458 44. 44. 667 667 45. 45. 500 500 262. 262. 958
MS 80. 80. 167 167 4. 153 153 7. 444 444 3. 792 792
F 10. 10. 77 0. 56 1. 96
P 0. 010 010 0. 662 662 0. 151 151
Sour Sour ce 1 2 3 4
Var i ance ance Err or Expect Expect ed Mean Squa Squarr e f or Each Each Term componen onentt t erm ( usi ng r est r i ct ed model odel ) Oper ator 9. 0903 3 ( 4) + 2( 3) + 8( 1) Mach achi ne - 0. 5486 3 ( 4) + 2( 3) + 6( 2) Oper per at or* Machi achi ne 1. 8264 8264 4 ( 4) + 2( 3) Er r or 3. 7917 ( 4)
(b) Find point estimates of the variance components using using the analysis of variance variance method.
ˆ2 σ
= MS E
ˆ 2 = 3.79167 σ
13-3
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
− MS E
7.44444 − 3.79167 = 1.82639 2 n MS B − MS AB 4.15278 − 7.44444 ˆ β 2 = σ < 0 , assume σ ˆ β 2 = 0 σ ˆ β 2 = an 3(2) MS A − MS AB 80.16667 − 7.44444 ˆτ 2 = ˆτ 2 = σ = 9.09028 σ bn 4(2) 2 ˆτβ σ =
MS AB
2 ˆτβ σ
=
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.
Row Factor 1 2 3
Column 2 39 20 37
1 36 18 30
Factor 3 36 22 33
4 32 20 34
Minitab Output General Lin ear Model: Response versus Row, Column
Fact or Row Col umn
Type Level s Val ues r andom 3 1 2 3 r andom 4 1 2 3 4
Anal ysi s of Vari ance f or Response, usi ng Adj ust ed SS f or Test s Sour ce Row Col umn Row*Col umn Err or Tot al
DF 2 3 6 0 11
Seq SS 580. 500 28. 917 28. 833 0. 000 638. 250
Adj SS 580. 500 28. 917 28. 833 0. 000
Adj MS 290. 250 9. 639 4. 806 0. 000
F 60. 40 2. 01 **
** Denomi nator of F- t est i s zero. Expect ed Mean Squar es, usi ng Adj ust ed SS Sour ce 1 Row 2 Col umn 3 Row*Col umn 4 Er ror
Expect ed Mean Squar e f or Each Ter m ( 4) + ( 3) + 4. 0000( 1) ( 4) + ( 3) + 3. 0000( 2) ( 4) + ( 3) ( 4)
Er r or Ter ms f or Test s, usi ng Adj ust ed SS Source 1 Row 2 Col umn 3 Row*Col umn
Err or DF * * *
Err or MS 4. 806 4. 806 *
Synt hesi s of Err or MS ( 3) ( 3) ( 4)
Var i ance Component s, usi ng Adj ust ed SS Sour ce Row Col umn Row*Col umn Er r or
Est i mat ed Val ue 71. 3611 1. 6111 4. 8056 0. 0000
13-4
P ** **
Solutions from Montgomery, D. C. (2012) 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 MS E , and therefore, no estimate of σ 2 . 2 ˆτβ σ
ˆ β 2 σ ˆτ 2 σ
=
=
=
MS AB
− MS E
n MS B − MS AB
ˆ β 2 σ
an
MS A − MS AB
ˆτ 2 σ
bn
4.8056 − 0 = 4.8056 1 9.6389 − 4.8056 = 1.6111 3 (1)
2 ˆτβ σ
=
=
=
290.2500 − 4.8056 4 (1)
= 71.3611
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.
Position 1
2
800 570 565 583
Temperature (°C) 825 1063 1080 1043
850 565 510 590
528 547 521
988 1026 1004
526 538 532
The following analysis assumes a restricted model: Minitab Output ANOVA: Den si ty v ersu s Pos it io n, Temp eratu re
Fact or Type Level s Val ues Posi t i on r andom 2 1 Temper at f i xed 3 800
2 825
850
Anal ysi s of Var i ance f or Densi t y Sour ce Posi t i on Temper at Posi t i on*Temper at Er r or Tot al
DF 1 2 2 12 17
SS 7160 945342 818 5371 958691
MS F 7160 16. 00 472671 1155. 52 409 0. 91 448
P 0. 002 0. 001 0. 427
Sour ce 1 2 3 4
Var i ance Er r or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) Posi t i on 745. 83 4 ( 4) + 9( 1) Temperat 3 ( 4) + 3(3) + 6Q[ 2] Posi t i on*Temper at - 12. 83 4 ( 4) + 3( 3) Er r or 447. 56 ( 4)
ˆ2 σ
= MS E
ˆ 2 = 447.56 σ
13-5
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
2 ˆτβ σ
=
MS AB
ˆτ 2 σ
=
− MS E
n MS A − MS E
=
2 ˆτβ σ
409 − 448
ˆτ 2 σ
bn
=
< 0 assume σ ˆτβ 2 = 0
3 7160 − 448 3 (3)
= 745.83
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 4 5 6 7 8 9 10
1 50 52 53 49 48 52 51 52 50 47
Operator 1 Measurements 2 49 52 50 51 49 50 51 50 51 46
3 50 51 50 50 48 50 51 49 50 49
1 50 51 54 48 48 52 51 53 51 46
Operator 2 Measurements 2 48 51 52 50 49 50 50 48 48 47
3 51 51 51 51 48 50 50 50 49 48
The following analysis assumes a restricted model: Minitab Output ANOVA: M easur ement vers us Par t, Oper ator
Fact or Part
Type Level s Val ues r andom 10 1 8 Oper at or f i xed 2 1
2 9 2
3 10
4
5
F 7. 33 0. 69 0. 40
P 0. 000 0. 427 0. 927
6
7
Anal ysi s of Vari ance f or Measurem Sour ce Par t Oper at or Par t *Oper at or Err or Tot al
DF 9 1 9 40 59
SS 99. 017 0. 417 5. 417 60. 000 164. 850
MS 11. 002 0. 417 0. 602 1. 500
Sour ce 1 2 3 4
Var i ance Er r or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) Par t 1. 5836 4 ( 4) + 6( 1) Oper ator 3 ( 4) + 3( 3) + 30Q[ 2] Par t *Operator - 0. 2994 4 ( 4) + 3( 3) Er r or 1. 5000 ( 4)
ˆ2 σ 2 ˆτβ σ
=
MS AB
− MS E n
= MS E 2 ˆτβ σ
=
ˆ 2 = 1.5000 σ
0.60185 −1.5000 3
13-6
< 0 assume σ ˆτβ 2 = 0
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
σ ˆτ 2
=
MS A − MS E
ˆτ 2 σ
bn
=
11.00185 −1.50000 2 ( 3)
= 1.58364
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 2 3 4 5 6 7 8 9 10
Inspector 1 Test 1 Test 2 Test 3 37 38 37 42 41 43 30 31 31 42 43 42 28 30 29 42 42 43 25 26 27 40 40 40 25 25 25 35 34 34
Inspector 2 Test 1 Test 2 Test 3 41 41 40 42 42 42 31 31 31 43 43 43 29 30 29 45 45 45 28 28 30 43 42 42 27 29 28 35 35 34
Inspector 3 Test 1 Test 2 Test 3 41 42 41 43 42 43 29 30 28 42 42 42 31 29 29 44 46 45 29 27 27 43 43 41 26 26 26 35 34 35
Minitab Output ANOVA: Im ped anc e vers us In spec to r, Part
Fact or Type Level s Val ues I nspect o f i xed 3 1 Part r andom 10 1 8
2 2 9
3 3 10
4
5
6
7
Anal ysi s of Vari ance f or I mpedanc Sour ce I nspect o Part I nspect o*Par t Err or Tot al
DF 2 9 18 60 89
SS 39. 27 3935. 96 48. 51 30. 67 4054. 40
MS 19. 63 437. 33 2. 70 0. 51
F 7. 28 855. 64 5. 27
P 0. 005 0. 000 0. 000
Sour ce 1 2 3 4
Var i ance Er r or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) I nspect o 3 ( 4) + 3( 3) + 30Q[ 1] Par t 48. 5353 4 ( 4) + 9( 2) I nspect o*Par t 0. 7280 4 ( 4) + 3( 3) Er r or 0. 5111 ( 4)
ˆ2 σ 2 ˆτβ σ
ˆ β 2 σ
=
=
MS AB
MS B
= MS E
− MS E
an
2.70 − 0.51 = 0.73 3 437.33 − 0.51 = 48.54 3 ( 3)
2 ˆτβ σ
n
− MS E
ˆ 2 = 0.51 σ
ˆ β 2 σ
=
These results agree with the Minitab output.
13-7
=
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
13.8. In problem 5.8, suppose that there are only four machines of interest, but the operators were selected at random.
Operator 1 109 110
Machine 2 3 110 108 115 109
4 110 108
2
110 112
110 111
111 109
114 112
3
116 114
112 115
114 119
120 117
1
(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: St ren gt h v ersu s Oper ato r, Mach in e
Fact or Type Level s Val ues Oper at or r andom 3 1 Machi ne f i xed 4 1
2 2
3 3
4
Anal ysi s of Var i ance for St r engt h Sour ce Oper at or Machi ne Oper at or *Machi ne Err or Tot al
DF 2 3 6 12 23
SS 160. 333 12. 458 44. 667 45. 500 262. 958
MS 80. 167 4. 153 7. 444 3. 792
F 21. 14 0. 56 1. 96
P 0. 000 0. 662 0. 151
Sour ce 1 2 3 4
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) Oper ator 9. 547 4 ( 4) + 8( 1) Machi ne 3 ( 4) + 2( 3) + 6Q[ 2] Oper ator *Machi ne 1. 826 4 ( 4) + 2( 3) Er r or 3. 792 ( 4)
ˆ2 σ 2 ˆτβ σ
=
ˆτ 2 σ
=
MS AB
= MS E
− MS E
n MS A − MS E bn
ˆ 2 = 3.792 σ
7.444 − 3.792 = 1.826 2 80.167 − 3.792 ˆτ 2 = σ = 9.547 4 ( 2) 2 ˆτβ σ
=
These results agree with the Minitab output.
13.9.
Rework Problem 13.5 using the REML method.
13-8
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Position 1
2
800 570 565 583
Temperature (°C) 825 1063 1080 1043
850 565 510 590
528 547 521
988 1026 1004
526 538 532
The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.5. JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Parameter Estimates Term Intercept Temperature[800] Temperature[825]
0.993347 0.99246 21.15551 709.9444 18
Estimate 709.94444 -157.6111 324.05556
Std Error 19.94444 6.741707 6.741707
REML Variance Component Estimates Random Effect Var Ratio Var Component Positi on 1.6760179 750.11111 Position*Temperature -0.028674 -12.83333 Residual 447.55556 Total 1184.8333
DFDen 1 2 2
Std Error 1126.0118 149.33585 182.71379
95% Lower -1456.832 -305.5262 230.13858
Covariance Matrix of Variance Component Estimates Random Effect Position Position*Temperature Position 1267902.7 -6197.276 Position*Temperature -6197.276 22301.197 Residual -1.412e-9 -11128.11 Fixed Effect Tests Source Temperature
13.10.
Nparm 2
DF 2
DFDen 2
F Ratio 1155.518
Rework Problem 13.6 using the REML method.
13-9
t Ratio 35.60 -23.38 48.07
Prob>|t| 0.0179* 0.0018* 0.0004*
95% Upper 2957.0538 279.85956 1219.556
Residual -1.412e-9 -11128.11 33384.329
Prob > F 0.0009*
Pct of Total 63.309 -1.083 37.774 100.000
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Table P13.1
Part Number 1 2 3 4 5 6 7 8 9 10
1 50 52 53 49 48 52 51 52 50 47
Operator 1 Measurements 2 49 52 50 51 49 50 51 50 51 46
3 50 51 50 50 48 50 51 49 50 49
1 50 51 54 48 48 52 51 53 51 46
Operator 2 Measurements 2 48 51 52 50 49 50 50 48 48 47
3 51 51 51 51 48 50 50 50 49 48
The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.6. JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Parameter Estimates Term Intercept Operator[Operator 1]
0.420766 0.410779 1.224745 49.95 60
Estimate 49.95 0.0833333
Std Error 0.42821 0.100154
REML Variance Component Estimates Random Effect Var Ratio Var Component Part Number 1.1555556 1.7333333 Part Number*Operator -0.199588 -0.299383 Residual 1.5 Total 2.9339506
DFDen 9 9
Std Error 0.8656795 0.1464372 0.3354102
t Ratio 116.65 0.83
95% Lower 0.0366326 -0.586394 1.0110933
Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number*Operator Part Number 0.749401 -0.004472 Part Number*Operator -0.004472 0.0214438 Residual -1.43e-14 -0.0375 Fixed Effect Tests Source Nparm Operator 1
13.11.
DF 1
DFDen 9
F Ratio 0.6923
Prob>|t| <.0001* 0.4269
95% Upper 3.430034 -0.012371 2.4556912
Pct of Total 59.078 -10.204 51.126 100.000
Residual -1.43e-14 -0.0375 0.1125
Prob > F 0.4269
Rework Problem 13.7 using the REML method.
Table P13.2
Part Number 1 2 3 4
Inspector 1 Test 1 Test 2 Test 3 37 38 37 42 41 43 30 31 31 42 43 42
Inspector 2 Test 1 Test 2 Test 3 41 41 40 42 42 42 31 31 31 43 43 43
13-10
Inspector 3 Test 1 Test 2 Test 3 41 42 41 43 42 43 29 30 28 42 42 42
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
5 6 7 8 9 10
28 42 25 40 25 35
30 42 26 40 25 34
29 43 27 40 25 34
29 45 28 43 27 35
30 45 28 42 29 35
29 45 30 42 28 34
31 44 29 43 26 35
29 46 27 43 26 34
29 45 27 41 26 35
The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.7. JMP Output RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Parameter Estimates Term Intercept Inspector[Inspector 1] Inspector[Inspector 2]
0.992005 0.991821 0.71492 35.8 90
Estimate 35.8 -0.9 0.6666667
Std Error 2.20436 0.244725 0.244725
REML Variance Component Estimates Random Effect Var Ratio Var Component Part Number 94.485507 48.292593 Part Number*Inspector 1.4243156 0.7279835 Residual 0.5111111 Total 49.531687 Covariance Matrix of Variance Component Estimates Random Effect Part Number Part Number 524.71813 Part Number*Inspector -0.02989 Residual -2.76e-13 Fixed Effect Tests Source Nparm Inspector 2
13.12.
DF 2
DFDen 9 18 18
Std Error 22.906727 0.3010625 0.0933157
t Ratio 16.24 -3.68 2.72
95% Lower 3.3962334 0.1379119 0.3681575
Part Number*Inspector -0.02989 0.0906386 -0.002903
DFDen 18
F Ratio 7.2849
Prob>|t| <.0001* 0.0017* 0.0139*
95% Upper 93.188952 1.3180552 0.757543
Pct of Total 97.498 1.470 1.032 100.000
Residual -2.76e-13 -0.002903 0.0087078
Prob > F 0.0048*
Rework Problem 13.8 using the REML method. Operator 1 109 110
Machine 2 3 110 108 115 109
4 110 108
2
110 112
110 111
111 109
114 112
3
116 114
112 115
114 119
120 117
1
The JMP REML output is shown below. The variance components are similar to those calculated in Problem 13.8. JMP Output
13-11
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts)
0.78154 0.748771 1.94722 112.2917 24
Parameter Estimates Term Estimate Intercept 112.29167 Machine[1] -0.458333 Machine[2] -0.125 Machine[3] -0.625
Std Error 1.827643 0.964653 0.964653 0.964653
DFDen 2 6 6 6
REML Variance Component Estimates Random Effect Var Ratio Var Component Operator 2.3974359 9.0902778 Operator*Machine 0.481685 1.8263889 Residual 3.7916667 Total 14.708333 Covariance Matrix of Variance Component Estimates Random Effect Operator Operator 100.70575 Operator*Machine -1.154578 Residual 1.686e-12 Fixed Effect Tests Source Nparm Machine 3
DF 3
Std Error 10.035225 2.2841505 1.5479414
t Ratio 61.44 -0.48 -0.13 -0.65
Prob>|t| 0.0003* 0.6515 0.9011 0.5410
95% Lower -10.5784 -2.650464 1.9497217
95% Upper 28.758958 6.3032416 10.332013
Operator*Machine -1.154578 5.2173434 -1.198061
DFDen 6
Pct of Total 61.804 12.417 25.779 100.000
Residual 1.686e-12 -1.198061 2.3961227
F Ratio 0.5578
Prob > F 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
SS A
= bn
∑ ( y
i ..
− y... )
b
2
SS B
,
= an
SS AB
=n
b
∑∑ ( y
ij .
. j .
− y i.. − y . j . + y ...
), 2
a
SS E =
i =1 j =1
Using the model y ijk
)
− y... 2
j =1
i =1
a
∑ ( y
b
n
∑∑∑ ( y
ijk
)
− y... 2
i =1 j =1 k =1
= µ + τ i + β j + (τβ )ij + ε ijk , we may find that
= µ + τ i + (τβ )i. + ε i.. y. j. = µ + β j + ε . j. yi..
yij .
= µ + τ i + β j + (τβ )ij + ε ij.
y...
= µ + β. + ε ...
Using the assumptions for the restricted form of the mixed model, τ .
(τβ ).. = 0 .
=0,
Substituting these expressions into the sums of squares yields
13-12
(τβ ). j = 0 , which imply that
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
a
SS A
= bn∑ i =1
(τ + (τβ ) + ε i
)
i .. − ε ...
i.
b
SS B
2
= an∑ ( β j + ε. j . − ε ... )
2
j =1 a
SS AB
b
= n∑∑ i =1 j =1
a
SS E
b
((τβ )) − (τβ ) + ε ij
i.
n
)
ij . − εi .. − ε . j . + ε ...
2
= ∑∑∑ (εijk − ε ij . )
2
i =1 j =1 k =1
= 0 , V (ε ijk ) = 0 , and E ε ijk ⋅ ε i' j' k ' = 0 , we may divide each sum of
Using the assumption that E ε ijk
squares by its degrees of freedom and take the expectation to produce
bn a 2 E (MS A ) = σ + E ∑ τ i + (τβ )i. (a − 1) i =1 an b 2 2 E (MS B ) = σ + ∑ β j b 1 − ( ) j =1
(
2
(
)= σ
(
) = σ
E MS AB E MS E
(
2
)
2 a b n τβ τβ + − E ( ) ( ) ∑∑ ij i. (a − 1)(b − 1) i =1 j =1
(
)
2
)
( ) E ( MS AB ). Consider E ( MS A )
(
)
Note that E MS B and E MS E are the results given in Table 8-3. We need to simplify E MS A and
a a E (τ i )2 + E (τβ )i2. + (crossproducts = 0 ) a − 1 i =1 i =1 (a − 1) a bn 2 τ i 2 +a a σ τβ E ( MS A ) = σ 2 + a − 1 i =1 b
(
E MS A
) = σ 2 +
∑
bn
∑
∑
(
E MS A
) = σ 2 + nσ τβ 2 +
a
∑ τ a −1 bn
2
i
i =1
( )ij is NID 0 , a − 1 σ τβ 2 .
since τβ
a
(
E MS AB
(
Consider E MS AB
)= σ
2
+
)
(( ) − (τβ ) ) (a − 1)(b − 1) ∑∑ = = a
n
b
2
E τβ
i 1 j 1
(
)= σ 2 +
(
)= σ
E MS AB
b
i =1 j =1
E MS AB
2
+ nσ
2 τβ
13-13
i.
− 1 a − 1 2 σ τβ a
b ∑∑ (a − 1)(b − 1) b a
n
ij
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(
)
(
)
Thus E MS A and E MS AB agree with Equation 13.9.
13.14. Consider the three-factor factorial design in Example 13.5. Propose appropriate test statistics for all main effects and interactions. Repeat for the case where A and B are fixed and C is random.
If all three factors are random there are no exact tests on main effects. We could use the following: A : F = B : F = C : F =
MS A + MS ABC MS AB + MS AC MS B + MS ABC MS AB + MS BC MS C + MS ABC MS AC + MS BC
If A and B are fixed and C is random, the expected mean squares are (assuming the restricted form of the model):
Factor
F a i
F b j
R c k
R n l
E( MS )
τ i
0
b
c
n
σ 2
2 + bnσ τγ + bcn
β j
a
0
c
n
σ 2
2 + anσ βγ + acn
γ k
a
b
1
n
σ2
+ abn σ γ 2
+ nσ
2 i
∑ (aτ − 1) β 2j
∑ (b − 1)
(τβ ) + cn∑∑ (a − 1)(b − 1) 2
(τβ )ij (τγ )ik ( βγ ) jk (τβγ )ijk ε (ijk )l
0
0
c
n
σ
2
2 τβγ
0
b
1
n
σ
2
2 + bnσ τγ
a
0
1
n
σ2
2 + anσ βγ
0
0
1
n
σ
2
1
1
1
1
σ
2
ij
2 + nσ τβγ
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: Dr op vers us Temp, Op erato r, Gaug e
Fact or Type Level Temp r andom Oper at or r andom Gauge r andom
s Val ues 3 60 4 1 3 1
75 2 2
90 3 3
4
Anal ysi s of Vari ance f or Dr op Sour ce Temp
DF 2
SS 1023. 36
MS 511. 68
13-14
F 2. 30
P 0. 171 x
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Oper at or Gauge Temp*Oper at or Temp*Gauge Oper at or *Gauge Temp*Oper at or *Gauge Err or Tot al
3 2 6 4 6 12 36 71
423. 82 7. 19 1211. 97 137. 89 209. 47 166. 11 770. 50 3950. 32
141. 27 3. 60 202. 00 34. 47 34. 91 13. 84 21. 40
0. 63 0. 06 14. 59 2. 49 2. 52 0. 65
0. 616 x 0. 938 x 0. 000 0. 099 0. 081 0. 788
x Not an exact F- t est. Sour ce 1 2 3 4 5 6 7 8
Var i ance Err or Expect ed Mean Squar e f or component t erm ( usi ng r est r i ct ed model ) Temp 12. 044 * ( 8) + 2( 7) + 8( 5) + 6( 4) Operat or - 4. 544 * (8) + 2( 7) + 6( 6) + 6( 4) Gauge - 2. 164 * ( 8) + 2( 7) + 6( 6) + 8( 5) Temp*Operat or 31. 359 7 ( 8) + 2( 7) + 6( 4) Temp*Gauge 2. 579 7 ( 8) + 2( 7) + 8(5) Oper ator *Gauge 3. 512 7 (8) + 2( 7) + 6( 6) Temp*Oper at or* Gauge - 3. 780 8 ( 8) + 2( 7) Er r or 21. 403 ( 8)
Each Term + 24( 1) + 18( 2) + 24( 3)
* Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 Temp 2 Oper ator 3 Gauge
Err or DF 6. 97 7. 09 5. 98
Err or MS 222. 63 223. 06 55. 54
Synt hesi s of ( 4) + ( 5) ( 4) + ( 6) ( 5) + ( 6) -
Err or MS ( 7) ( 7) ( 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.
Factor
F a i
R b j
R c k
R n l
E( MS )
τ i
0
b
c
n
σ 2
2 2 2 + nσ τβγ + bnσ τγ + cnσ τβ + bcn
β j
a
1
c
n
σ2
2 + anσ βγ + acnσ β 2
γ k
a
b
1
n
σ2
2 + anσ βγ + abnσ γ 2
0
1
c
n
σ2
2 2 + nσ τβγ + cnσ τβ
0
b
1
n
σ
2 2
(τβ )ij (τγ )ik ( βγ ) jk (τβγ )ijk ε ijkl
a
1
1
n
σ
0
1
1
n
σ2
1
1
1
1
σ 2
2 + nσ τβγ + bnσ τγ2
+
τ i2
∑ (a − 1)
2 anσ βγ
2 + nσ τβγ
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:
13-15
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
yijklh
= µ + τ i + β j + γ k + δ l + (τβ ) ij + (τγ )ik + (τδ )il + ( βγ ) jk + ( βδ ) jl + ( γδ ) kl + (τβγ )ijk + (τβδ )ijl + ( βγδ ) jkl + ( τγδ )ikl + (τβγδ ) ijkl + εijklh
To simplify the expected mean square derivations, let capital Latin letters represent the factor effects or variance components. For example, A =
∑ τ
2 i
bcdn
a −1
2 , or B = acdnσ β .
(a) A, B, C , and D are fixed factors. F a i 0 a a a 0
F b j b 0 b b 0
F c k c c 0 c c
F d l d d d 0 d
R n h n n n n n
b b
0 c
d 0
n n
( βγ ) jk
0 0 a
0
0
d
n
( βδ ) jl
a
0
c
0
n
σ
(γδ ) kl
a
b
0
n
σ 2 + CD
(τβγ ) ijk
0
0
0
0 d
n
σ
(τβδ ) ijl
0
0
c
0
n
σ
( βγδ ) jkl
a
0
0
0
n
σ
(τγδ ) ikl
0
b
0
0
n
(τβγδ ) ijkl
0
0
0
0
n
σ
ε (ijkl) h
1
1
1
1
1
σ
Factor τ i
β j γ k δ l
(τβ ) ij (τγ ) ik (τδ ) il
E( MS ) 2
+ A
2
+ B
2
+ C
σ σ σ
σ 2 + D 2
σ
+ AB
σ 2 + AC 2
+ AD
2
+ BC
2
+ BD
σ σ
2
+ ABC
2
+ ABD
2
+ BCD
2
+ ACD
2
+ ABCD
σ
2
There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level s Val ues f i xed 2 H f i xed 2 H f i xed 2 H f i xed 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce A B C D A*B A*C A*D B*C B*D C*D A*B*C A*B*D
DF 1 1 1 1 1 1 1 1 1 1 1 1
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13
F 0. 49 0. 01 0. 09 0. 01 0. 25 0. 25 0. 25 0. 25 0. 25 0. 25 0. 25 2. 27
P 0. 492 0. 921 0. 767 0. 921 0. 622 0. 622 0. 622 0. 622 0. 622 0. 622 0. 622 0. 151
13-16
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
A*C*D B*C*D A*B*C*D Err or Tot al
1 1 1 16 31
3. 13 3. 13 3. 13 198. 00 264. 88
3. 13 3. 13 3. 13 12. 38
0. 25 0. 25 0. 25
0. 622 0. 622 0. 622
Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed 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 Er r or 12. 38 ( 16)
(b) A, B, C , and D are random factors. R c k c c 1 c c
(τδ ) il
1 1
R b j b 1 b b 1 b b
( βγ ) jk
a
1
Factor τ i
β j γ k δ l
(τβ ) ij (τγ ) ik
R a i 1 a a a 1
1 c
R d l d d d 1 d d 1
R n h n n n n n n n
1
d
n
E( MS ) σ 2 2
σ
+ ABCD + ACD + ABD +
ABC + AD + AC + AB + A
+ ABCD + BCD + ABD + ABC + BD + BC + AB + B
2
+ ABCD + ACD + BCD + ABC + AC + BC + CD + C
2
+ ABCD + ACD + BCD + ABD + BD + AD + CD + D
σ σ
σ 2 + ABCD + ABC + ABD + AB σ 2 + ABCD + ABC + ACD + AC 2
+ ABCD + ABD + ACD + AD
2
+ ABCD + ABC + BCD + BC
2
+ ABCD + ABD + BCD + BD
2
+ ABCD + ACD + BCD + CD
σ σ
( βδ ) jl
a
1
c
1
n
(γδ ) kl
a
b
1
1
n
σ
(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
2
+ ABCD + ACD
(τβγδ ) ijkl
1
1
1
1
n
2
σ
+ ABCD
ε (ijkl) h
1
1
1
1
1
σ
σ
σ
2
No exact tests exist on main effects or two-factor interactions. For main effects use statistics such as: A: F =
+ MS ABC + MS ABD + MS ACD MS AB + MS AC + MS AD + MS ABCD MS A
For testing two-factor interactions use statistics such as: AB: F =
The results can also be generated in Minitab as follows:
13-17
+ MS ABCD MS ABC + MS ABD MS AB
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level r andom r andom r andom r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F ** ** 0. 36 ** 0. 11 1. 00 0. 11 1. 00 0. 11 1. 00 1. 00 9. 00 1. 00 1. 00 0. 25
P 0. 843 x 0. 796 0. 667 0. 796 0. 667 0. 796 0. 667 0. 500 0. 205 0. 500 0. 500 0. 622
x x x x x x
x Not an exact F- t est. ** Denomi nator of F- t est i s zero. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) 1 A 1. 7500 * ( 16) + 2( 15) + 4( 13) + 4( 12) + 4( 11) + 8( 5) + 16( 1) 2 B 1. 3750 * ( 16) + 2( 15) + 4( 14) + 4( 12) + 4( 11) + 8( 5) + 16( 2) 3 C - 0. 1250 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 11) + 8( 6) + 16( 3) 4 D 1. 3750 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 12) + 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 Err or 12. 3750 ( 16)
+ 8( 7) + 8( 6) + 8( 9) + 8( 8) + 8( 10) + 8( 8) + 8( 10) + 8( 9)
* Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 A 2 B 3 C 4 D 5 A*B 6 A*C 7 A*D 8 B*C
Err or DF 0. 56 0. 56 0. 14 0. 56 0. 98 0. 33 0. 98 0. 33
Err or MS * * 3. 13 * 28. 13 3. 13 28. 13 3. 13
Synt hesi s of Err or MS ( 5) + ( 6 ) + ( 7 ) - ( 1 1) - ( 1 2) - ( 13) + ( 1 5) ( 5) + ( 8 ) + ( 9 ) - ( 1 1) - ( 1 2) - ( 14) + ( 1 5) ( 6) + ( 8) + ( 10) - ( 11) - ( 13) - ( 14) + ( 15) ( 7) + ( 9) + ( 10) - ( 12) - ( 13) - ( 14) + ( 15) ( 11) + ( 12) - ( 15) ( 11) + ( 13) - ( 15) ( 12) + ( 13) - ( 15) ( 11) + ( 14) - ( 15)
13-18
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
9 B*D 10 C*D
0. 98 0. 33
28. 13 3. 13
( 12) + ( 14) - ( 15) ( 13) + ( 14) - ( 15)
(c) A is fixed and B, C, and D are random. F a i 0 a a a 0
R b j b 1 b b 1
R c k c c 1 c c
R d l d d d 1 d
R n h n n n n n
b b
1 c
d 1
n n
σ
(τδ ) il
0 0
( βγ ) jk
a
1
1
d
n
σ 2 + BCD + BC
( βδ ) jl
a
1
c
1
n
σ
(γδ ) kl
a
b
1
1
n
(τβγ ) ijk
0
1
1
d
n
(τβδ ) ijl
0
1
c
1
n
σ
( βγδ ) jkl
a
1
1
1
n
σ 2 + BCD
(τγδ ) ikl
0
b
1
1
n
σ
(τβγδ ) ijkl
0
1
1
1
n
σ
ε (ijkl ) h
1
1
1
1
1
σ
Factor τ i
β j γ k δ l
(τβ ) ij (τγ ) ik
E( MS ) 2
+ ABCD + ACD + ABD + ABC +
2
+ BCD + BD + BC + B
2
+ BCD + BC + CD + C + BCD + BD + CD + D
σ
σ σ
2
σ
2
+ ABCD + ABC + ABD + AB
2
+ ABCD + ABC + ACD + AC
2
+ ABCD + ABD + ACD + AD
σ
σ
2
+ BCD + BD
2
+ BCD + CD
2
+ ABCD + ABC
2
+ ABCD + ABD
σ σ
2
+ ABCD + ACD
2
+ ABCD
AD + AC + AB + A
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: F =
+ MS ABC + MS ABD + MS ACD MS AB + MS AC + MS AD + MS ABCD MS A
Random main effects could be tested by, for example: D : F =
For testing two-factor interactions involving A use: AB: F =
The results can also be generated in Minitab as follows:
13-19
+ MS BCD MS BD + MS CD
MS D
+ MS ABCD MS ABC + MS ABD MS AB
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level f i xed r andom r andom r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F ** 0. 04 0. 36 0. 04 0. 11 1. 00 0. 11 1. 00 1. 00 1. 00 1. 00 9. 00 1. 00 0. 25 0. 25
P 0. 907 0. 761 0. 907 0. 796 0. 667 0. 796 0. 500 0. 500 0. 500 0. 500 0. 205 0. 500 0. 622 0. 622
x x x x x x
x Not an exact F- t est. ** Denomi nator of F- t est i s zero. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed 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 Err or 12. 3750 ( 16) * Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 A 2 B 3 C 4 D 5 A*B 6 A*C 7 A*D
Err or DF 0. 56 0. 33 0. 33 0. 33 0. 98 0. 33 0. 98
Err or MS * 3. 13 3. 13 3. 13 28. 13 3. 13 28. 13
Synt hesi s of Err or MS ( 5) + ( 6 ) + ( 7 ) - ( 1 1) - ( 1 2) - ( 13) + ( 1 5) ( 8) + ( 9) - ( 14) ( 8) + ( 10) - ( 14) ( 9) + ( 10) - ( 14) ( 11) + ( 12) - ( 15) ( 11) + ( 13) - ( 15) ( 12) + ( 13) - ( 15)
13-20
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(d) A and B are fixed and C and D are random. F a i 0 a a a 0
F b j b 0 b b 0
R c k c c 1 c c
R d l d d d 1 d
R n h n n n n n
b b
1 c
d 1
n n
σ
(τδ ) il
0 0
( βγ ) jk
a
0
1
d
n
σ
( βδ ) jl
a
0
c
1
n
(γδ ) kl
a
b
1
1
n
(τβγ ) ijk
0
0
1
d
n
σ
(τβδ ) ijl
0
0
c
1
n
σ
( βγδ ) jkl
a
0
1
1
n
σ 2 + BCD
(τγδ ) ikl
0
b
1
1
n
σ
Factor τ i
β j γ k δ l
(τβ ) ij (τγ ) ik
E( MS ) σ 2 2
σ
+ ACD + AD + AC +
A
+ BCD + BC + BD + B
2
+ CD + C σ + CD + D σ
2
σ 2 + ABCD + ABC + ABD + AB 2 2
σ
+ ACD + AC + ACD + AD
2
+ BCD + BC
2
+ BCD + BD
2
+ CD
2
+ ABCD + ABC
2
+ ABCD + ABD
σ σ
2
+ ACD
2
+ ABCD
(τβγδ ) ijkl
0
0
1
1
n
σ
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:
+ MS ACD MS AC + MS AD MS B + MS BCD B: F = MS BC + MS BD MS AB + MS ABCD AB: F = MS ABC + MS ABD A: F =
MS A
The results can also be generated in Minitab as follows: Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level f i xed f i xed r andom r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce A B C D A*B A*C A*D B*C B*D
DF 1 1 1 1 1 1 1 1 1
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13
F 1. 96 0. 04 0. 36 0. 04 0. 11 1. 00 1. 00 1. 00 1. 00
P 0. 604 x 0. 907 x 0. 656 0. 874 0. 796 x 0. 500 0. 500 0. 500 0. 500
13-21
Solutions from Montgomery, D. C. (2012) 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 Err or Tot al
1 1 1 1 1 1 16 31
3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
0. 25 1. 00 9. 00 0. 25 0. 25 0. 25
0. 622 0. 500 0. 205 0. 622 0. 622 0. 622
x Not an exact F- t est. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed 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 Err or 12. 3750 ( 16) * Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 A 2 B 5 A*B
Err or DF 0. 33 0. 33 0. 98
Err or MS 3. 13 3. 13 28. 13
Synt hesi s of ( 6) + ( 7) ( 8) + ( 9) ( 11) + ( 12)
Err or MS ( 13) ( 14) - ( 15)
(e) A, B and C are fixed and D is random. F a i 0 a a a 0
F b j b 0 b b 0
F c k c c 0 c c
R d l d d d 1 d
R n h n n n n n
b b
0 c
d 1
n n
σ 2
(τδ ) il
0 0
( βγ ) jk
a
0
0
d
n
σ
( βδ ) jl
a
0
c
1
n
σ
(γδ ) kl
a
b
0
1
n
(τβγ ) ijk
0
0
0
d
n
σ
(τβδ ) ijl
0
0
c
1
n
σ
( 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
Factor τ i
β j γ k δ l
(τβ ) ij (τγ ) ik
13-22
E( MS ) σ 2 + AD + A 2
σ
+ BD + B
2
+ CD + C σ + D σ
2
σ 2 + ABD + AB
2
σ
+ ACD + AC + AD
2
+ BCD + BC
2
+ BD
2
+ CD
2
+ ABCD + ABC
2
+ ABD
σ
2
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
There are exact tests for all effects. The results can also be generated in Minitab as follows: Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level f i xed f i xed f i xed r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F 1. 96 0. 04 0. 36 0. 01 0. 11 1. 00 0. 25 1. 00 0. 25 0. 25 1. 00 2. 27 0. 25 0. 25 0. 25
P 0. 395 0. 874 0. 656 0. 921 0. 795 0. 500 0. 622 0. 500 0. 622 0. 622 0. 500 0. 151 0. 622 0. 622 0. 622
Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed 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 Err or 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 vers us A, B, C, D
Fact or A B C D
Type Level f i xed r andom r andom r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
13-23
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F ** ** 0. 36 ** 0. 11 1. 00 0. 11 1. 00 0. 11 1. 00 1. 00 9. 00 1. 00 1. 00 0. 25
P 0. 843 x 0. 796 0. 667 0. 796 0. 667 0. 796 0. 667 0. 500 0. 205 0. 500 0. 500 0. 622
x x x x x x
x Not an exact F- t est. ** Denomi nator of F- t est i s zero. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng unrest r i ct ed model ) 1 A * ( 16) + 2( 15) + 4( 13) + 4( 12) + 4( 11) + 8( 5) + Q[ 1 ] 2 B 1. 3750 * ( 16) + 2( 15) + 4( 14) + 4( 12) + 4( 11) + 8( 5) + 16( 2) 3 C - 0. 1250 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 11) + 8( 6) + 16( 3) 4 D 1. 3750 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 12) + 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 Err or 12. 3750 ( 16)
+ 8( 7) + 8( 6) + 8( 9) + 8( 8) + 8( 10) + 8( 8) + 8( 10) + 8( 9)
* Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 A 2 B 3 C 4 D 5 A*B 6 A*C 7 A*D 8 B*C 9 B*D 10 C*D
Err or DF 0. 56 0. 56 0. 14 0. 56 0. 98 0. 33 0. 98 0. 33 0. 98 0. 33
Err or MS * * 3. 13 * 28. 13 3. 13 28. 13 3. 13 28. 13 3. 13
Synt hesi s of Err or MS ( 5) + ( 6 ) + ( 7 ) - ( 1 1) - ( 1 2) - ( 13) + ( 1 5) ( 5) + ( 8 ) + ( 9 ) - ( 1 1) - ( 1 2) - ( 14) + ( 1 5) ( 6) + ( 8) + ( 10) - ( 11) - ( 13) - ( 14) + ( 15) ( 7) + ( 9) + ( 10) - ( 12) - ( 13) - ( 14) + ( 15) ( 11) + ( 12) - ( 15) ( 11) + ( 13) - ( 15) ( 12) + ( 13) - ( 15) ( 11) + ( 14) - ( 15) ( 12) + ( 14) - ( 15) ( 13) + ( 14) - ( 15)
A and B are fixed and C and D are random.
13-24
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level f i xed f i xed r andom r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F 1. 96 0. 04 0. 36 ** 0. 11 1. 00 0. 11 1. 00 0. 11 1. 00 1. 00 9. 00 1. 00 1. 00 0. 25
P 0. 604 x 0. 907 x 0. 843 x 0. 796 0. 667 0. 796 0. 667 0. 796 0. 667 0. 500 0. 205 0. 500 0. 500 0. 622
x x x x x x
x Not an exact F- t est. ** Denomi nator of F- t est i s zero. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng unrest r i ct ed model ) 1 A * ( 16) + 2( 15) + 4( 13) + 4( 12) + 4( 11) + Q[ 1, 5] 2 B * ( 16) + 2( 15) + 4( 14) + 4( 12) + 4( 11) + Q[ 2, 5] 3 C - 0. 1250 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 11) + 8( 6) + 16( 3) 4 D 1. 3750 * ( 16) + 2( 15) + 4( 14) + 4( 13) + 4( 12) + 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 Err or 12. 3750 ( 16)
+ 8( 7) + 8( 6) + 8( 9) + 8( 8) + 8( 10) + 8( 8) + 8( 10) + 8( 9)
* Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 1 A 2 B 3 C 4 D 5 A*B 6 A*C 7 A*D 8 B*C 9 B*D 10 C*D
Err or DF 0. 33 0. 33 0. 14 0. 56 0. 98 0. 33 0. 98 0. 33 0. 98 0. 33
Err or MS 3. 13 3. 13 3. 13 * 28. 13 3. 13 28. 13 3. 13 28. 13 3. 13
Synt hesi s of Err or MS ( 6) + ( 7) - ( 13) ( 8) + ( 9) - ( 14) ( 6) + ( 8) + ( 10) - ( 11) - ( 13) - ( 14) + ( 15) ( 7) + ( 9) + ( 10) - ( 12) - ( 13) - ( 14) + ( 15) ( 11) + ( 12) - ( 15) ( 11) + ( 13) - ( 15) ( 12) + ( 13) - ( 15) ( 11) + ( 14) - ( 15) ( 12) + ( 14) - ( 15) ( 13) + ( 14) - ( 15)
13-25
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(e) A, B and C are fixed and D is random. Minitab Output ANOVA: y vers us A, B, C, D
Fact or A B C D
Type Level f i xed f i xed f i xed r andom
s Val ues 2 H 2 H 2 H 2 H
L L L L
Anal ysi s of Var i ance f or y Sour ce 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 Err or Tot al
DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 31
SS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 198. 00 264. 88
MS 6. 13 0. 13 1. 13 0. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 3. 13 28. 13 3. 13 3. 13 3. 13 12. 38
F 1. 96 0. 04 0. 36 ** 0. 11 1. 00 0. 11 1. 00 0. 11 1. 00 1. 00 9. 00 1. 00 1. 00 0. 25
P 0. 395 0. 874 0. 656 0. 795 0. 500 0. 796 x 0. 500 0. 796 x 0. 667 x 0. 500 0. 205 0. 500 0. 500 0. 622
x Not an exact F- t est. ** Denomi nator of F- t est i s zero. Sour ce
Var i ance Err or Expect ed Mean Squar e f or Each Term component t erm ( usi ng unrest r i ct ed 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 Err or 12. 3750 ( 16) * Synt hesi zed Test. Er r or Terms f or Synt hesi zed Test s Source 4 D 7 A*D 9 B*D 10 C*D
Err or DF 0. 56 0. 98 0. 98 0. 33
Err or MS * 28. 13 28. 13 3. 13
Synt hesi s of Err or MS ( 7) + ( 9) + ( 10) - ( 12) - ( 13) - ( 14) + ( 15) ( 12) + ( 13) - ( 15) ( 12) + ( 14) - ( 15) ( 13) + ( 14) - ( 15)
13-26
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
13.19. In Problem 5.19, assume that the three operators were selected at random. Analyze the data under these conditions and draw conclusions. Estimate the variance components. Minitab Output ANOVA: Sc or e vers us C ycl e Tim e, Operat or , Temper atur e
Fact or Type Level Cycl e Ti f i xed Oper at or r andom Temper at f i xed
s Val ues 3 40 3 1 2 300
50 2 350
60 3
Anal ysi s of Var i ance f or Scor e Sour ce Cycl e Ti Oper at or Temper at Cycl e Ti *Oper at or Cycl e Ti *Temper at Oper at or *Temper at Cycl e Ti *Oper at or *Temper at Err or Tot al
DF 2 2 1 4 2 2 4 36 53
SS 436. 000 261. 333 50. 074 355. 667 78. 815 11. 259 46. 185 118. 000 1357. 333
MS 218. 000 130. 667 50. 074 88. 917 39. 407 5. 630 11. 546 3. 278
F 2. 45 39. 86 8. 89 27. 13 3. 41 1. 72 3. 52
P 0. 202 0. 000 0. 096 0. 000 0. 137 0. 194 0. 016
Sour ce 1 2 3 4 5 6 7 8
Var i ance Er r or Expect ed Mean Squar e f or Each Ter m component t erm ( usi ng r est r i ct ed model ) Cycl e Ti 4 ( 8) + 6( 4) + 18Q[ 1] Oper ator 7. 0772 8 ( 8) + 18( 2) Temperat 6 ( 8) + 9( 6) + 27Q[ 3] Cycl e Ti *Operat or 14. 2731 8 ( 8) + 6( 4) Cycl e Ti *Temper at 7 ( 8) + 3( 7) + 9Q[ 5] Oper at or* Temper at 0. 2613 8 ( 8) + 9(6) Cycl e Ti *Operat or* Temperat 2. 7562 8 ( 8) + 3( 7) Er r or 3. 2778 ( 8)
The following calculations agree with the Minitab results:
ˆ2 σ 2 ˆτβγ σ
=
2 ˆτβ σ
2 ˆ βγ σ
n
=
=
ˆ β 2 σ
− MS E
MS ABC MS AB
− MS E
cn
− MS E
MS BC
an
=
MS B
− MS E
acn
= MS E
ˆ 2 = 3.27778 σ
2 σ ˆτβγ
=
11.546296 − 3.277778 = 2.7562 3
=
88.91667 − 3.277778 = 14.27315 2 ( 3)
ˆτγ 2 σ
=
5.629630 − 3.277778 3 (3)
ˆ γ 2 σ
=
2 ˆτβ σ
= 0.26132
130.66667 − 3.277778 = 7.07716 3 ( 2) ( 3)
13.20. Consider the three-factor factorial model
yijk
= µ + τ i + β j + γ k + (τβ )ij + ( βγ ) jk + ε 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.
13-27
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Source
DF
E(MS)
A
a-1
σ2
+ cσ τβ2 + bcσ τ 2
B
b-1
σ2
2 + cσ τβ + a σ βγ2 + acσ β2
C
c-1
σ2
2 + aσ βγ + abσ γ 2
AB
(a-1)(b-1)
σ2
2 + cσ τβ
BC
(b-1)(c-1)
σ2
2 + aσ βγ
Error ( AC + ABC ) Total
b(a-1)(c-1) abc-1
σ 2
MS B There are exact tests for all effects except B. To test B, use the statistic F = MS AB
+ MS E + MS BC
13.21. The three-factor factorial model for a single replicate is
yijk = µ + τ i
+ β j + γ k + (τβ )ij + ( βγ ) jk + (τγ )ik + (τβγ )ijk + ε ijk
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 twofactor interactions and approximate F tests can be conducted for the main effects. For example, to test the main effect of A, use F =
+ MS ABC MS AB + MS AC MS A
If (τβγ ) ijk and (τβ ) ij can be eliminated, the model becomes
yijk = µ
+ τ i + β j + γ k + ( βγ ) jk + (τγ )ik + ε ijk
For this model, the analysis of variance is Source
DF
E(MS)
A
a-1
σ2
+ bσ τγ2 + bcσ τ 2
B
b-1
σ2
2 + aσ βγ + acσ β 2
C
c-1
σ2
2 + aσ βγ + bσ τγ2 + abσ γ 2
AC
(a-1)(c-1)
σ2
2 + bσ τγ
BC
(b-1)(c-1)
σ2
2 + aσ βγ
Error ( AB + ABC ) Total
c(a-1)(b-1) abc-1
σ 2
There are exact tests for all effect except C . To test the main effect of C , use the statistic:
13-28
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
MS C F = MS BC
+ MS E + MS AC
13.22. In Problem 5.8, assume that both machines and operators were chosen randomly. Determine the 2 power of the test for detecting a machine effect such that σ β = σ 2 , where σ β 2 is the variance component
for the machine factor. Are two replicates sufficient?
λ = 1 + 2 If σ β
anσ β 2
σ2
2 + nσ τβ
2 = 7.44 , from the analysis of = σ 2 , then an estimate of σ 2 = σ β 2 = 3.79 , and an estimate of σ τβ
variance table. Then
λ = 1 +
and the other OC curve parameters are υ 1
(3)(2 )(3.79 ) 3.79 + 2(7.44)
= 2.22 = 1.49 This results in β ≈ 0.75 approximately, with
= 3 and υ 2 = 6 .
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 Cov (τβ ) , (τβ ) = − (1 a )2 for ij i' j τβσ
i≠i' .
a
Since
a
∑ (τβ )ij = 0 (constant) we have V ∑ (τβ )ij = 0 , which implies that i =1
i =1 a
∑ i =1
V (τβ )ij
a + 2 Cov (τβ )ij , (τβ )i ' j = 0 2
a! a − 1 σ 2 + ( 2 ) Cov (τβ )ij , (τβ )i ' j = 0 τβ 2!( a − 2 ) ! a ( a − 1) σ τβ 2 + a ( a − 1) Cov (τβ )ij ,τ ( β )i ' j = 0 1 2 Cov τ ( β )ij , (τβ )i ' j = − σ τβ a a
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 2 fixed effects. Let 2 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.
13-29
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
g = As
2
⇒ sˆ 2 = A -1g
Since A -1 always exists then,
( )
( )
(
E s 2 = E A -1 g = A-1 E ( g ) = A -1 As 2 Thus 2 is an unbiased estimator of discussed by Searle (1971a).
2
)=s
2
. This and other properties of the analysis of variance method are
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, 2 write a statement giving the probability that σ τ < 0 in a one-factor analysis of variance. Comment on the usefulness of this probability statement. 2 Suppose σ =
MS1
c
öτ 2 probability that σ
ˆ2 P {σ
− MS 2
, where MS i for i=1,2 are two mean squares and c is a constant. The
< 0 (negative) is
MS < 0} = P {MS1 − MS2 < 0} = P 1 MS2
MS 1 E ( MS1 ) E ( MS1 ) = < E ( MS 1 ) < 1 = P < P Fu ,v E ( MS2 ) E ( MS 2 ) MS 2 E ( MS 2 )
where u is the number of degrees of freedom for MS 1 and v is the number of degrees of freedom for MS 2 . For the one-way model, this equation reduces to
σ 2 1 F ˆ < 0} = P Fa −1, N − a < 2 = < P {σ P a − N − a 1, σ + nσ τ 2 1 + nk 2
where k =
σ τ 2 σ 2
. 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
ˆτ 2 σ
<0.
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: M easur ement vers us Par t, Oper ator
Fact or Part
Type Level s Val ues r andom 10 1 8 Oper at or f i xed 2 1
2 9 2
3 10
4
5
F 7. 33
P 0. 000
Anal ysi s of Vari ance f or Measurem Sour ce Par t
DF 9
SS 99. 017
MS 11. 002
13-30
6
7
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Oper at or Par t *Oper at or Err or Tot al
1 9 40 59
0. 417 5. 417 60. 000 164. 850
0. 417 0. 602 1. 500
0. 69 0. 40
0. 427 0. 927
Sour ce 1 2 3 4
Var i ance Er r or Expect ed Mean Squar e f or Each Term component t erm ( usi ng r est r i ct ed model ) Par t 1. 5836 4 ( 4) + 6( 1) Oper ator 3 ( 4) + 3( 3) + 30Q[ 2] Par t *Operator - 0. 2994 4 ( 4) + 3( 3) Er r or 1. 5000 ( 4)
The second approach is the unrestricted mixed model. Minitab Output ANOVA: M easur ement vers us Par t, Oper ator
Fact or Part
Type Level s Val ues r andom 10 1 8 Oper at or f i xed 2 1
2 9 2
3 10
4
5
6
7
Anal ysi s of Vari ance f or Measurem Sour ce Part Oper at or Par t *Oper at or Err or Tot al
DF 9 1 9 40 59
SS 99. 017 0. 417 5. 417 60. 000 164. 850
MS 11. 002 0. 417 0. 602 1. 500
F 18. 28 0. 69 0. 40
P 0. 000 0. 427 0. 927
Sour ce 1 2 3 4
Var i ance Er r or Expect ed Mean Squar e f or Each Term component t erm ( usi ng unrest r i ct ed model ) Part 1. 7333 3 ( 4) + 3( 3) + 6( 1) Oper ator 3 ( 4) + 3( 3) + Q[ 2] Par t *Operator - 0. 2994 4 ( 4) + 3( 3) Er r or 1. 5000 ( 4)
Source
A
Sum of Squares
DF
Mean Square
99.016667
a-1=9
11.00185
E(MS)
σ2
F -test
F =
+ nσ τβ2 + bnσ τ 2
MS A MS AB
F
18.28
b
∑ β
2 i
σ2
+ nσ τβ 2 + an
B
0.416667
b-1=1
0.416667
AB
5.416667
(a-1)(b-1)=9
0.60185
2 σ 2 + nσ τβ
Error
60.000000
40
1.50000
σ 2
Total
164.85000
nabc-1=59
i =1
b −1
F =
F =
MS B MS AB MS AB MS E
0.692
0.401
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 MS AB will generally be larger than MS E. 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. 12
A) is [ MS AB / bn]
.
13-31
Solutions from Montgomery, D. C. (2012) 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 m..
)
where rows are fixed and columns are random. Now, assuming all model parameters to be independent, we have the following: b
b
( yi .. − y m.. )= τ i − τ m + 1 ∑ (τβ )ij − 1 ∑ (τβ )mj + b j =1
b j =1
b
n
b
∑∑
n
∑∑
1 1 ε ijk − ε mjk bn j =1 k =1 bn j =1 k =1
and 2
2
1 1 1 2 2 V ( yi .. − ym .. ) = bσ τβ + + bσ τβ b b bn Since MS AB estimates σ
2
2
1 bnσ + bn 2
2
bnσ
2
=
(
2 2 σ 2 + nσ τβ
)
bn
2 + nσ τβ , we would use
2 MS AB bn
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.
f E MS E
χ
2 α 2, f E
( 40 )(1.5)
≤ σ 2 ≤ ≤ σ 2 ≤
59.34 1.011 ≤ σ 2
f E MS E
χ 12−α 2, f E
( 40)(1.5)
24.43 ≤ 2.456
(b) Find approximate 95 percent confidence intervals on the other variance components using the Satterthwaite method. 2 2 2 σ τβ and σ τ are negative, and the Satterthwaithe method does not apply. The confidence interval on σ β
is
ˆ β 2 σ
r =
=
MS B
− MS AB
ˆ β 2 σ
an
( MS B − MS AB ) MS B2
+
2
2 MS AB
( b − 1) ( a − 1)( b − 1)
=
=
11.001852 − 0.6018519 2 ( 3)
(11.001852 − 0.6018519) 11.001852 2 (9)
13-32
= 1.7333
2
0.60185192 + (1)( 9)
= 8.01826
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
ˆ O2 r σ χα2 2,r
≤ σ ≤ 2 β
(8.01826 )(1.7333)
≤ σ β 2 ≤
17.55752
ˆ β 2 r σ χ 12−α 2, r
( 8.01826)(1.7333) 2.18950
0.79157 ≤ σ β 2 ≤ 6.34759 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 σ 2. Construct approximate 95 percent confidence interval on the other variance components using the Satterthwaite method.
ˆ2 σ
= MS E
f E MS E
ˆ 2 = 3.79167 σ
≤ σ 2 ≤
χα2 2, f E
(12 )( 3.79167 ) 23.34
≤ σ 2 ≤
1.9494 ≤ σ 2
f E MS E
χ 12−α 2, f E
(12 )(3.79167 ) 4.40
≤ 10.3409
Satterthwaite Method: 2 ˆτβ σ =
r =
MS AB
− MS E n
( MS AB − MS E ) 2 MS AB
( a − 1)( b − 1)
+
=
2 ˆτβ σ
2
MS E 2
=
df E
rσˆ β2
χα2 2,r
7.44444 − 3.79167 = 1.82639 2
( 7.44444 − 3.79167) 7.444442 ( 2 )( 3)
≤ σ ≤ 2 β
+
2
3.79167 2 (12)
= 1.27869
ˆ β 2 r σ χ 12−α 2, r
χ α 2 2, r and χ 12−α 2, r 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.
(1.27869 )(1.82639 ) 5.67991
≤ σ β 2 ≤
(1.27869 )(1.82639 ) 0.014820
0.41117 ≤ σ β 2 ≤ 158.58172
13-33
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
ˆ β 2 < 0 , this variance component does not have a confidence interval using Satterthwaite’s Method. σ
ˆτ 2 σ
r =
=
MS A − MS AB bn
( MS A − MS AB ) MS A2
=
ˆτ 2 σ 2
80.166672 ( 2)
+ ( a − 1) ( a − 1)(b − 1) 2 rσˆτ
χα2 2,r
≤ σ ≤
(1.64108)(9.09028) 6.53293
= 9.09028
( 80.16667 − 7.44444)
=
2 MS AB
80.16667 − 7.44444 4 ( 2)
2 τ
≤ σ τ 2 ≤
2
7.444442 + ( 2 )( 3)
= 1.64108
ˆτ 2 r σ χ 12−α 2, r (1.64108)(9.09028) 0.03281
χ α 2 2, r and χ 12−α 2, r were estimated by extrapolating 1.64108 between degrees of freedom of one and two in Microsoft Excel. More precise methods can be used as well.
2.28349 ≤ σ τ 2
≤ 454.61891
13.30. Consider the three-factor experiment in Problem 5.19 and assume that operators were selected at random. Find an approximate 95 percent confidence interval on the operator variance component.
ˆ β 2 σ
=
r =
− MS E
MS B
ˆ β 2 σ
acn
( MS B − MS E ) 2
MS B
( b − 1)
+
2
2
MS E df E
=
130.66667 − 3.277778 2 ( 3) ( 3 )
=
(130.66667 − 3.27778) 130.66667 2 ( 2)
rσˆ β2
χα2 2,r
≤ σ ≤ 2 β
= 7.07716
2
3.277782 + ( 36 )
= 1.90085
ˆ β 2 r σ χ 12−α 2,r
χ α 2 2, r and χ 12−α 2, r were estimated by extrapolating 1.90085 between degrees of freedom of one and two in Microsoft Excel. More precise methods can be used as well.
(1.90085)( 7.07716) 7.14439
≤ σ β 2 ≤
1.88296 ≤ σ β 2
(1.90085)( 7.07716) 0.04571
≤ 294.28720
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.
13-34
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
σˆ O2
= σ ˆ β 2 =
MS B
G1
=1−
1
H 1
=
− MS AB
ˆ O2 σ
an
=1−
F 0.05,9,∞
1
1 2 χ .95,9
−1 =
F .95,9i , ∞
1 1.88
=
11.001852 − 0.6018519 = 1.7333 2 ( 3)
= 0.46809
−1 =
1 − 1 = 1.7027 0.370
9 Gij
2
(F =
V L
α , fi , f j
− 1) − G12 Fα , f , f − H12 i
j
F α , fi , f j
( 3.18 − 1)2
=
2
− ( 0.46809) ( 3.18) − 1.70272 3.18
= 0.36366
2 = G12 c12 MS B2 + H12 c22 MS AB + G11c1c2 MS B MS AB 2
2
2 2 1 2 1 1 1 V L = ( 0.46809 ) (11.00185) + (1.7027 ) ( 0.60185) + ( 0.36366) (11.00185)( 0.60185) 6 6 6 6 V L = 0.83275 2
ˆ β 2 − V L L = σ
= 1.7333− 0.83275 = 0.82075
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.
MSC
ˆ γ 2 σ
=
G1
= 1−
H 1
=
− MS E
ˆγ 2 σ
abn
1
=1−
F 0.05,3,∞
1 F .95,36,∞
−1 =
130.66667 − 3.277778 = 7.07716 2 ( 3) ( 3 )
=
1 2.60 1
2 χ .95,36
= 0.61538
−1 = .
1 0.64728
−1 = 0.54493
36 Gij
V L
F ( =
α , f i , f j
2
− 1) − G12 Fα , f , f − H12 i
F α , fi , f j
j
( 2.88 − 1)2
=
2
− ( 0.61538 ) ( 2.88 ) − 0.544932 2.88
= 0.74542
2 = G12c12 MS B2 + H12 c22 MS AB + G11c1c2 MSB MS AB 2
2
2 2 1 2 1 1 1 V L = ( 0.61538 ) (130.66667 ) + ( 0.54493) ( 3.27778) + ( 0.74542) ( 130.66667)( 3.27 18 18 18 18 V L = 20.95112 2
ˆ γ 2 − V L L = σ
= 7.07716 − 20.95112 = 2.49992
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.
13-35
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
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 RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Parameter Estimates Term Estimate Intercept 112.29167
0.742044 0.742044 1.94722 112.2917 24
Std Error 1.789728
DFDen 1.831
REML Variance Component Estimates Random Effect Var Ratio Var Component Operator 2.3974359 9.0902778 Machine -0.144689 -0.548611 Operator*Machine 0.481685 1.8263889 Residual 3.7916667 Total 14.159722
t Ratio 62.74
Std Error 10.035225 0.9124188 2.2841505 1.5479414
Covariance Matrix of Variance Component Estimates Random Effect Operator Machine Operator 100.70575 0.3848594 Machine 0.3848594 0.8325081 Operator*Machine -1.154578 -1.539438 Residual 2.151e-12 -1.17e-13
Prob>|t| 0.0005*
95% Lower -10.5784 -2.336919 -2.650464 1.9497217
95% Upper 28.758958 1.239697 6.3032416 10.332013
Operator*Machine -1.154578 -1.539438 5.2173434 -1.198061
Pct of Total 64.198 -3.874 12.898 26.778 100.000
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 RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Parameter Estimates Term Estimate Intercept 49.95
0.388009 0.388009 1.224745 49.95 60
Std Error 0.424591
DFDen 8.563
REML Variance Component Estimates Random Effect Var Ratio Var Component Part Number 1.1555556 1.7333333 Operator -0.004115 -0.006173 Part Number*Operator -0.199588 -0.299383 Residual 1.5 Total 2.9277778
t Ratio 117.64
Std Error 0.8656795 0.0218 0.1464372 0.3354102
13-36
Prob>|t| <.0001*
95% Lower 0.0366326 -0.0489 -0.586394 1.0110933
95% Upper 3.430034 0.0365544 -0.012371 2.4556912
Pct of Total 59.203 -0.211 -10.226 51.233 100.000
