1 the two-factor mixed model two factors, factorial experiment, factor a fixed, factor b random...
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1
The Two-Factor Mixed Model• Two factors, factorial experiment, factor A fixed,
factor B random (Section 13-3, pg. 495)
• The model parameters are NID random variables, the interaction effect is normal, but not independent
• This is called the restricted model
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and j ijk
2
Testing Hypotheses - Mixed Model
• Once again, the standard ANOVA partition is appropriate
• Relevant hypotheses:
• Test statistics depend on the expected mean squares:
2 20 0 0
2 21 1 1
: 0 : 0 : 0
: 0 : 0 : 0
i
i
H H H
H H H
2
2 2 10
2 20
2 20
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( ) 1
( )
( )
( )
a
ii A
AAB
BB
E
ABAB
E
E
bnMS
E MS n Fa MS
MSE MS an F
MS
MSE MS n F
MS
E MS
Ho is rejected if
Fo > F,a-1,(a-1)(b-1)
Fo > F,b-1,ab(n-1)
Fo > F,(a-1)(b-1), ab(n-1)
3
Estimating the Variance Components – Two Factor Mixed model
• Use the ANOVA method; equate expected mean squares to their observed values:
• Estimate the fixed effects (treatment means) as usual
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B E
AB E
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MS MS
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4
Example 13-3 (pg. 497) The Measurement Systems Capability
Study Revisited• Same experimental setting as in example 13-2• Parts are a random factor, but Operators are fixed• Assume the restricted form of the mixed model• Minitab can analyze the mixed model• The variance components can also be estimated as
99.0ˆ
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99.071.0ˆ
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99.039.62ˆ
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EOperatorsPartsOperatorsParts
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5
Example 13-3 (pg. 497) Minitab Solution – Balanced ANOVA
Source DF SS MS F P
Part 19 1185.425 62.391 62.92 0.000
Operator 2 2.617 1.308 1.84 0.173
Part*Operator 38 27.050 0.712 0.72 0.861
Error 60 59.500 0.992
Total 119 1274.592
Source Variance Error Expected Mean Square for Each Term
component term (using restricted model)
1 Part 10.2332 4 (4) + 6(1)
2 Operator 3 (4) + 2(3) + 40Q[2]
3 Part*Operator -0.1399 4 (4) + 2(3)
4 Error 0.9917 (4)
6
Example 13-3 Minitab Solution – Balanced ANOVA
• There is a large effect of parts (not unexpected)• Small operator effect• No Part – Operator interaction• Negative estimate of the Part – Operator
interaction variance component• Fit a reduced model with the Part – Operator
interaction deleted• This leads to the same solution that we found
previously for the two-factor random model
7
The Unrestricted Mixed Model• Two factors, factorial experiment, factor A fixed,
factor B random (pg. 498)
• The random model parameters are now all assumed to be NID . is no longer assumed – unrestricted model
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8
Testing Hypotheses – Unrestricted Mixed Model
• The standard ANOVA partition is appropriate
• Relevant hypotheses:
• Expected mean squares determine the test statistics:
2 20 0 0
2 21 1 1
: 0 : 0 : 0
: 0 : 0 : 0
i
i
H H H
H H H
2
2 2 10
2 2 20
2 20
2
( ) 1
( )
( )
( )
a
ii A
AAB
BB
AB
ABAB
E
E
bnMS
E MS n Fa MS
MSE MS n an F
MS
MSE MS n F
MS
E MS
9
Estimating the Variance Components – Unrestricted Mixed Model
• Use the ANOVA method; equate expected mean squares to their observed values:
• The only change compared to the restricted mixed model is in the estimate of the random effect variance component
• Which model to use? o They are fairly close in many caseso The restricted model is slightly more generalo The restricted model is mostly preferred
2
2
2
ˆ
ˆ
ˆ
B AB
AB E
E
MS MS
anMS MS
n
MS
10
Example 13-4 (pg. 499) Minitab Solution – Unrestricted Model
Source DF SS MS F P
Part 19 1185.425 62.391 87.65 0.000
Operator 2 2.617 1.308 1.84 0.173
Part*Operator 38 27.050 0.712 0.72 0.861
Error 60 59.500 0.992
Total 119 1274.592
Source Variance Error Expected Mean Square for Each Term
component term (using unrestricted model)
1 Part 10.2798 3 (4) + 2(3) + 6(1)
2 Operator 3 (4) + 2(3) + Q[2]
3 Part*Operator -0.1399 4 (4) + 2(3)
4 Error 0.9917 (4)
11
Sample Size Determination with Random Effects
• Consider a single-factor random effects model
• Power = 1 – P(Reject HoHo is false)
P(Fo > F,a-1,N-a Ho is false)
• Fo = MSTreatments/MSE (dofs are needed to determine the OC curve)
• The operating characteristic curves (Chart VI, Appendix) can be used
• The curves plot the probability of type II error against the parameter
2
2
1 n
12
Sample Size Determination with Random Effects – Example 13-5
• Five treatments randomly selected (a = 5)• Six observations per treatment (n = 6)• = 0.05, a – 1 = 4 (v1), N – a = 25 (v2)• Assume that • Then
0.20
646.2)1(61
22
13
Sample Size Determination with Random Effects
• Use the percentage increase in the standard deviation of an observation
• If the treatments are homogeneous,
• If the treatments are different,
• P is the fixed percentage increase in the standard deviation
• Then
]1)01.01[(11 22
2
Pnn
22
P01.012
22
15
Finding Expected Mean Squares
• Obviously important in determining the form of the test statistic
• In fixed models, it’s easy:
• Can always use the “brute force” approach – just apply the expectation operator
• Straightforward but tedious• Rules on page 502-503 [due to Cornfield and Tukey
(1956)] work for any balanced model• Rules are consistent with the restricted mixed model
2( ) (fixed factor)E MS f
17
Approximate F Tests• One possibility: assume that certain interactions are
negligible – needs conclusive evidence• If we cannot assume that certain interactions are
negligible, then use an approximate F test (“pseudo” F test)
• Test procedure is due to Satterthwaite (1946), and uses linear combinations of the original mean squares to form the F-ratio
• For example:
MS’ = MSr + …+ MSs
MS’’ = MSu + …+ MSv
• The mean squares are chosen so that E(MS’) – E(MS’’) is a multiple of the effect considered in the null hypothesis
• F is distributed approximately as Fp,q
SM
SMF
18
Approximate F Tests
• The linear combinations of the original mean squares are sometimes called “synthetic” mean squares
• Adjustments are required to the degrees of freedom
• Refer to Example 13-7, page 505• Minitab will analyze these experiments, although
their “synthetic” mean squares are not always the best choice
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