Slide set 13
Stat402 (Spring 2016)
Last update: April 22, 2016
Stat 402 (Spring 2016): slide set 13
A 23 Factorial in Blocks of Size 4
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Now consider a 23 factorial experiment, where we only have blocks of
size 4 available (i.e., need 2 blocks to accomodate the 8 treatment
combination.)
Suppose we are going to take 3 replications, i.e., each treatment
combination will occur in the design 3 times.
We also decide to confound the ABC interaction in each of the
replications.
To do this, first look at the defining contrast of the ABC interaction
effect. Since ABC:(a − 1)(b − 1)(c − 1), ABC effect is defined by
1/2[abc + a + b + c − ab − ac − bc − (1)
Put all the treatment combinations with + sign in one block and the
treatment combinations with - sign in the other.
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Stat 402 (Spring 2016): slide set 13
A 23 Factorial in Blocks of Size 4 (Cont’d 1)
Rep 1
Block I Block II
abc
ab
a
ac
b
bc
c
(1)
•
Rep 2
Block III Block IV
abc
ab
a
ac
b
bc
c
(1)
Rep 3
Block V Block VI
abc
ab
a
ac
b
bc
c
(1)
If we carry out a similar analysis as before (same for each replicate), we
see that we can estimate A, B, C, AB, AC and BC effects free of block
effects and that ABC interaction effect is confounded in each of the reps.
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Stat 402 (Spring 2016): slide set 13
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For example using the appropriate model,
 =
1
4 [µabc + β1 + abc + µab + β2 + ab + µac + β2 + ac
+µa + β1 + a − µb − β1 − b − µc − β1 − c − µbc −
β2 − bc
−µ(1) − β2 − (1)]
=
1
4 [µabc + µab + µac + µa
+ 14 [abc + ab + ac + a
=
1
4 [contrast
\ =
ABC
=
− µb − µc − µbc − µ(1)]
− b − c − bc − (1)]
of means] + 14 [contrast of errors]
1
4 [µabc
+c −
+ β1 + abc + µa + β1 + a + µb + β1 + b + µc + β1
µab − β2 − ab − µac − β2 − ac − µbc − β2 − bc
−µ(1) − β2 − (1)]
1
4 [µabc + µa + µb + µc
+ 14 [abc + a + c + b
− µab − µac − µbc − µ(1)] + 12 (β1 − β)
− ab − ac − bc − (1)]
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Stat 402 (Spring 2016): slide set 13
A 23 Factorial in Blocks of Size 4 (Cont’d 2)
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We say in this case that the ABC interaction is completely confounded
with blocks, meaning that it cannot be estimated from any of the reps
(replications).
SV
Treatments
A
B
C
AB
BC
AC
Blocks
Error
Total
•
DF
6
1
1
1
1
1
1
5
12
23
Notes: No information about ABC is available from the above analysis.
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Stat 402 (Spring 2016): slide set 13
Partial Confounding
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Remarks:
1. We have been able to put 12 of the 8 treatment combinations in 1 block
and the other half in another and come up with a design which gives
all the information provided by a complete block design except that one
effect is confounded with blocks.
2. See Example 7-2 for a computational example of a factorial in an
incomplete block experiment.
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Consider the same 23 factorial used in the example used before, where
we have blocks of size 4 and replicate the experiment 3 times. Instead of
confounding the same effect in each replicate we may decide to confound
a different effect in each of the replicates.
•
For example, in Rep 1 confound ABC with blocks, in Rep 2 confound
AB with blocks, in Rep 3 confound BC with blocks.
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Stat 402 (Spring 2016): slide set 13
A 23 Factorial in Blocks of Size 4
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Consider the same 23 factorial used in the example used before, where
we have blocks of size 4 and replicate the experiment 3 times.
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Instead of confounding the same effect in each replicate we may decide
to confound a different effect in each of the replicates.
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Here we will confound different effects with blocks in each rep:
− in Rep 1 confound ABC with blocks,
− in Rep 2 confound AB with blocks,
− in Rep 3 confound BC with blocks.
Block
abc
a
b
c
Rep 1
I
Block II
(1)
ab
ac
bc
Rep 2
Block III
Block IV
abc
a
c
b
ab
ac
(1)
bc
Rep 3
Block V
Block VI
abc
ab
a
ac
bc
b
(1)
c
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Stat 402 (Spring 2016): slide set 13
Partial Confounding (contd.)
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We can estimate A, B, C, and AC from all 3 reps, thus using 100% of
the available information.
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We can estimate ABC effect from Rep 2 and Rep 3, thus using 66.7% of
the information.
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We can estimate AB effect from Rep 1 and Rep 3 or use 66.7% of the
information.
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We can estimate BC from Rep 1 and Rep 2 or use 66.7% information.
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We say that the effects ABC, AB,and BC are partially confounded with
blocks.
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We have not lost 100% of the information of any one effect as before
but have lost part of the information about AB, BC, ABC.
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Stat 402 (Spring 2016): slide set 13
ANOVA Table
Partial Confounding (Cont’d)
SV
Treatments
A
B
AB
C
AC
BC
ABC
Blocks
Error
Total
•
DF
7
1
1
1
1
1
1
1
5
11
23
(Relative Information)
100%
100%
66.6%
100%
100%
66.7%
66.7%
Remarks Estimate of partially confounded effects have a larger variance
than the unconfounded effects which imply that we have more information
about unconfounded effects as shown above.
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Stat 402 (Spring 2016): slide set 13
Remarks and Definition
Definition
An incomplete block design for a factorial experiment involves Balanced
Confounding if all effects of the same order are confounded with blocks an
equal number of times.
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We use a 23 factorial experiment with blocks of size 4 as before and use
3 replicates. Suppose that we use the following design.
Rep 1: confound the AB effect
Rep 2: confound the BC effect
Rep 3: confound the AC effect.
We see that all 2nd order interactions are confounded an equal number
of times, i.e., once in each rep.
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Stat 402 (Spring 2016): slide set 13
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Another Example:
A 23 factorial with blocks of size 4 and use 5 reps.
Rep
Rep
Rep
Rep
Rep
1:
2:
3:
4:
5:
confound
confound
confound
confound
confound
the
the
the
the
the
AB effect
ABC effect
BC effect
ABC effect
AC effect
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We use 100% information to estimate A, B, and C effects, 80% of
information to estimate AB, AC and BC effects and 60% of information
to estimate the ABC effects. We are using balanced confounding.
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Remarks
− If balanced confounding is used the estimate of all effects of the
same order have the same variance.
− See Example 7.3.
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