Slide set 15
Stat402B (Spring 2016)
Last update: April 17, 2016
Stat 402 (Spring 2016): Slide set 15
Fractional Factorial Designs
•
So far we have discussed experiments involving all treatment
combinations in 2k factorials (blocked or not).
•
But the numbers of runs (or experimental units) required for these
complete replicates of 2k factorials increases geometrically as k increases.
•
However, in practice when k is not small, the desired information can
often be obtained by performing only a selected fraction of the 2k
experimental runs needed for a single replicate of a full factorial.
•
For simplicity of discussion, consider a 23 factorial in a 1/2 replication,
i.e., we will run only 1/2 of the 8 treatment combinations.
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial
•
We shall use the 4 treatment combinations a, b, c, and abc in a CRD.
•
What are the effects that can be estimated?
•
We begin by attempting to estimate the effects using the defining
contrasts in the usual way.
Treatment
Combinations
a
b
c
abc
I
+
+
+
+
A
+
+
B
+
+
Factorial Effect
AB C AC BC
+
+
+ +
+ + +
+
ABC
+
+
+
+
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial (Cont’d 1)
•
If we examine the contrasts of the observations that estimate the effects
carefully, it can be observed that the effects A and BC are estimated by
the same contrast. That is,
d = 1/2[ya − yb − yc + yabc]
 = 1/2[ya − yb − yc + yabc] and BC
•
This says that these two effects cannot be estimated separately (i.e.,
they will always have the same value). They are indistinguishable.
•
We say that the effect A is aliased with effect BC. In other words, this
contrast of observations is identified as estimating both A and BC effects.
•
In addition, it is seen that the effects B and AC, the effects C and AB,
and effects I and ABC are all aliased in pairs.
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial (Cont’d 2)
•
We write all this information in the form:
I
A
B
C
•
•
•
•
=
=
=
=
ABC
BC
AC
AB
The above is called the alias pattern. The alias pattern is the most
important idea in fractional factorials.
The alias pattern of a design tells us which of the effects can be estimated.
To be able to use a fractional factorial the experimenter needs to know
in advance whether certain effects can be assumed to be negligible.
For example, in the above, if we knew that the two factors B and C
do not interact from prior knowledge about these two factors, then the
value we estimated by the contrast 1/2(ya − yb − yc + yabc) above should
be an estimate of the effect A (instead of BC).
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial (Cont’d 3)
•
In the above example, we selected the treatment combinations to be run
in the following way. Take the effect aliased with the mean effect I (in
this case ABC), and write the defining contrast for that effect:
ABC=1/2[abc+a+b+c-ab-ac-bc-(1)]
•
•
We used the 4 treatment combinations with the + sign in the experiment.
This is just one method of obtaining the treatment combinations (or the
1/2 fraction) to be run. We will call this Method I.
The relation I=ABC is called the defining relation, and we use it to
obtain the alias pattern in the following way:
1. Take the defining relation I=ABC
2. Multiply both sides by A to give A(I) = A2BC which simplifies to A=BC
3. Similarly multiplying by B and C, respectively, gives B=AC and C=AB.
• Thus, the defining relation is the key to obtaining the alias pattern.
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial (Cont’d 4)
•
Most applications involve
− aliasing main effects with 3-factor or higher order interactions,
and
− aliasing 2-factor interactions with 3-factor or higher order
interactions, etc.
•
Then, we assume that the 3- and higher order interactions do not exist,
in order to estimate main effects and 2 factor interactions.
•
As discussed above, a method for obtaining the treatment combinations
to be run involves writing out the defining contrast for the effect in the
defining relation and choosing the treatment combinations with the +
signs as the desired fraction. (Method I)
•
However, this method becomes too unwieldy for large experiments. The
following method (Method II) is used in those cases.
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Stat 402 (Spring 2016): Slide set 15
One-half fraction of a 23 factorial (again)
•
•
Suppose we want to obtain the treatment combinations to be run for a
1/2 fraction of a 23 factorial, as in the previous example.
Write the defining contrasts for A and B effects as for a 22 factorial. The
generate a contrast for C using C=AB. Thus
(1)
a
b
ab
•
•
A
+
+
B
+
+
C(=AB)
+
+
c
a
b
abc
Now the treatment combinations to be run are obtained by interpreting
the columns as indicating levels of factors A, B, and C to be used.
For example, (-,-,+) is c; (+,-,-) is a, etc. Thus the treatment
combinations to be run are c, a, b, and abc, as also was obtained usin
Method I.
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Stat 402 (Spring 2016): Slide set 15
•
•
•
•
1 th
2p
fraction of a 2 factorial is called a 2k−p fractional
In general
factorial.
For example a 1/2 fraction of a 24 factorial is a 24−1 fractional factorial
and has 8 runs.
An example, of a 24−1 experiment is given in Example 8.1 (p.326 text).
To determine the 8 runs that are chosen to be run in this example,
consider the defining relation I=ABCD and use Method I. Write the
contrast for ABCD:
(1)
ABCD=+
•
2k−p Fractional
Factorial
k
a
-
b
-
ab
+
c
-
ac
+
bc
+
abc
-
d
-
ad
+
bd
+
abd
-
cd
+
acd
-
bcd
-
abcd
+
Then select those 16 treatment combinations that has + signs
(1), ab, ac, bc, ad, bd, cd, and abcd.
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Stat 402 (Spring 2016): Slide set 15
4−1
2
with defining relation I=ABCD
• Alternatively Method II could be used. Write the defining contrast for A,
B, and C as for a 23 factorial and then generate a contrast for D using
D=ABC:
•
Again the treatment combinations to be run are obtained by interpreting
the columns as indicating levels of factors A, B, C, and D, as shown in
the above table.
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Stat 402 (Spring 2016): Slide set 15
•
24−1: Analysis
Computations in the 1/2 fraction of the Filtration Data Example are
shown below. The estimate of effects are obtained by first computing
the contrast and then dividing by 2k−p−1.
Treatment
Combination
(1)
ad
bd
ab
cd
ac
bc
abcd
Contrast
Divisor
Estimate
45
100
45
65
75
60
80
96
I
+
+
+
+
+
+
+
+
566
8
70.75
A
+
+
+
+
76
4
19
B
+
+
+
+
6
4
1.5
Factorial Effects
AB
C
AC
+
+
+
+
+
+
+
+
+
+
+
+
-4
56
-74
4
4
4
-1
14 -18.5
BC
+
+
+
+
76
4
19
(D=ABC)
+
+
+
+
66
4
16.5
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Stat 402 (Spring 2016): Slide set 15
24−1: Analysis (Cont’d)
•
The fact that the above are the only factorial effects that need to be
computed is clear from looking at the alias pattern:
I=ABCD
A=BCD AB=CD
B=ACD AC=BD
C=ABD BC=AD
D=ABC
•
The analysis is based on assuming that the 3 factor and higher order
interactions do not occur. From this table it is not unreasonable to
conclude that main effects A, C, and D are large and that AC(=BD) and
AD(=BC) are also significant.
•
These facts are summarized in Table 18.4 (p.327 of the text).
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Stat 402 (Spring 2016): Slide set 15
•
Also discussed in the text is how the complimentary half fraction given by
I=−ABCD is combined with the above fraction to make more satisfactory
conclusions. (See Example 8.3)
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