Transcript Design of Engineering Experiments Part 7 – The 2k

Chapter 13
Design & Analysis of Experiments
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Design of Engineering Experiments Experiments with Random Factors
• Text reference, Chapter 13
• Previous chapters have focused primarily on fixed factors
– A specific set of factor levels is chosen for the experiment
– Inference confined to those levels
– Often quantitative factors are fixed (why?)
• When factor levels are chosen at random from a larger
population of potential levels, the factor is random
– Inference is about the entire population of levels
– Industrial applications include measurement system studies
– It has been said that failure to identify a random factor as random and
not treat it properly in the analysis is one of the biggest errors
committed in DOX
• The random effect model was introduced in Chapter 3
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Extension to Factorial Treatment Structure
• Two factors, factorial experiment, both factors
random (Section 13.2, pg. 574)
 i  1, 2,..., a

yijk     i   j  ( )ij   ijk  j  1, 2,..., b
k  1, 2,..., n

V ( i )   2 , V (  j )   2 , V [( )ij ]   2 , V ( ijk )   2
V ( yijk )   2   2   2   2
• The model parameters are NID random variables
• Random effects model
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Testing Hypotheses - Random Effects Model
• Once again, the standard ANOVA partition is appropriate
• Relevant hypotheses:
H 0 :  2  0
H 0 :  2  0
H 0 :  2  0
H1 :  2  0
H1 :  2  0
H1 :  2  0
• Form of the test statistics depend on the expected mean
squares: E ( MS )   2  n 2  bn 2  F  MS A

A

0
MS AB
E ( MS B )   2  n 2  an 2  F0 
MS B
MS AB
E ( MS AB )   2  n 2
MS AB
MS E
 F0 
E ( MS E )   2
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Estimating the Variance Components
– Two Factor Random model
• As before, we can use the ANOVA method; equate expected
mean squares to their observed values:
MS A  MS AB
ˆ 
bn
MS B  MS AB
2
ˆ
 
an
MS AB  MS E
2
ˆ 
n
ˆ 2  MS E
• These are moment estimators
• Potential problems with these estimators
2
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Example 13.1
A Measurement Systems Capability Study
• Gauge capability (or R&R) is of interest
• The gauge is used by an operator to measure a critical
dimension on a part
• Repeatability is a measure of the variability due only to
the gauge
• Reproducibility is a measure of the variability due to the
operator
• See experimental layout, Table 13.1. This is a two-factor
factorial (completely randomized) with both factors
(operators, parts) random – a random effects model
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Example 13.1
Minitab Solution – Using Balanced ANOVA
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Example 13.2
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
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Example 13.1
Minitab Solution – Reduced Model
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Example 13.1
Minitab Solution – Reduced Model
• Estimating gauge capability:
2
ˆ gauge
 ˆ 2  ˆ 2
 0.88  0.01
 0.89
• If interaction had been significant?
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Maximum Likelihood Approach
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CI on the Variance Components
The lower and upper bounds on the CI are:
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The Two-Factor Mixed Model
• Two factors, factorial experiment, factor A fixed,
factor B random (Section 13.3, pg. 581)
 i  1, 2,..., a

yijk     i   j  ( )ij   ijk  j  1, 2,..., b
k  1, 2,..., n

V (  j )   2 , V [( )ij ]  [(a  1) / a] 2 , V ( ijk )   2
a

i 1
a
i
 0,  ( )ij  0
i 1
• The model parameters  j and  ijk are NID random
variables, the interaction effects are normal, but
not independent
• This is called the restricted model
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Testing Hypotheses - Mixed Model
• Once again, the standard ANOVA partition is appropriate
• Relevant hypotheses:
H0 : i  0
H 0 :  2  0
H 0 :  2  0
H1 :  i  0
H1 :  2  0
H1 :  2  0
• Test statistics depend on the expected
mean squares:
a
 F0 
MS A
MS AB
E ( MS B )   2  an 2
 F0 
MS B
MS E
E ( MS AB )   2  n 2
 F0 
MS AB
MS E
E ( MS A )   2  n 2 
Chapter 13
bn  i2
E ( MS E )   2
i 1
a 1
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Estimating the Variance Components
– Two Factor Mixed model
• Use the ANOVA method; equate expected mean
squares to their observed values:
MS B  MS E
an
MS AB  MS E

n
ˆ 2  MS E
ˆ 2 
ˆ2
• Estimate the fixed effects (treatment means) as
usual
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•
•
•
•
Example 13.2
The Measurement Systems Capability
Study Revisited
Same experimental setting as in example
13.1
Parts are a random factor, but Operators are
fixed
Assume the restricted form of the mixed
model
Minitab can analyze the mixed model and
estimate the variance components
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Example 13.2
Minitab Solution – Balanced ANOVA
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Example 13.2
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
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The Unrestricted Mixed Model
• Two factors, factorial experiment, factor A fixed,
factor B random (pg. 583)
 i  1, 2,..., a

yijk     i   j  ( )ij   ijk  j  1, 2,..., b
k  1, 2,..., n

V ( j )   2 ,V [( )ij ]   2 , V ( ijk )   2
a

i 1
i
0
• The random model parameters are now all
assumed to be NID
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Testing Hypotheses – Unrestricted Mixed Model
• The standard ANOVA partition is appropriate
• Relevant hypotheses:
H 0 : i  0
H 0 :  2  0
H 0 :  2  0
H1 : i  0
H1 :  2  0
H1 :  2  0
• Expected mean squares determine
the test statistics:
a
 F0 
MS A
MS AB
E ( MS B )   2  n 2  an 2
 F0 
MS B
MS AB
E ( MS AB )   2  n 2
 F0 
MS AB
MS E
E ( MS A )   2  n 2 
Chapter 13
bn  i2
E ( MS E )   2
i 1
a 1
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Estimating the Variance Components
– Unrestricted Mixed Model
• Use the ANOVA method; equate expected mean
squares to their observed values:
MS B  MS AB
ˆ  
an
MS AB  MS E
2
ˆ
  
n
ˆ 2  MS E
2
• The only change compared to the restricted mixed
model is in the estimate of the random effect
variance component
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Example 13.3
Minitab Solution – Unrestricted Model
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Finding Expected Mean Squares
• Obviously important in determining the form of the test
statistic
• In fixed models, it’s easy:
E(MS )   2  f (fixed factor)
• Can always use the “brute force” approach – just apply the
expectation operator
• Straightforward but tedious
• Rules on page 523-524 work for any balanced model
• Rules are consistent with the restricted mixed model – can
be modified to incorporate the unrestricted model
assumptions
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JMP Output for the Mixed Model – using REML
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Approximate F Tests
• Sometimes we find that there are no exact tests for certain
effects (page 525)
• Leads to 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
• 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
• Minitab will analyze these experiments, although their
“synthetic” mean squares are not always the best choice
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Notice that there are no exact tests for main effects
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We chose a different linear
combination
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