3
Stat 402 (Spring 2016): Slide set 15
A
+
+
B
+
+
Factorial Effect
AB C AC BC
+
+
+ +
+ + +
+
ABC
+
+
+
+
2
We begin by attempting to estimate the effects using the defining
contrasts in the usual way.
•
I
+
+
+
+
What are the effects that can be estimated?
•
Treatment
Combinations
a
b
c
abc
We shall use the 4 treatment combinations a, b, c, and abc in a CRD.
•
One-half fraction of a 2 factorial
Last update: April 17, 2016
Stat402B (Spring 2016)
Slide set 15
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.
•
3
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.
This says that these two effects cannot be estimated separately (i.e.,
they will always have the same value). They are indistinguishable.
= 1/2[ya − yb − yc + yabc]
 = 1/2[ya − yb − yc + yabc] and BC
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,
One-half fraction of a 2 factorial (Cont’d 1)
Stat 402 (Spring 2016): Slide set 15
•
•
•
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.
•
3
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.
•
1
But the numbers of runs (or experimental units) required for these
complete replicates of 2k factorials increases geometrically as k increases.
So far we have discussed experiments involving all treatment
combinations in 2k factorials (blocked or not).
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•
Stat 402 (Spring 2016): Slide set 15
Fractional Factorial Designs
=
=
=
=
ABC
BC
AC
AB
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.
•
•
•
6
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.
•
•
•
•
A
+
+
B
+
+
C(=AB)
+
+
c
a
b
abc
7
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.
(1)
a
b
ab
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
One-half fraction of a 2 factorial (again)
Stat 402 (Spring 2016): Slide set 15
Stat 402 (Spring 2016): Slide set 15
3
5
One-half fraction of a 2 factorial (Cont’d 4)
3
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:
ABC=1/2[abc+a+b+c-ab-ac-bc-(1)]
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:
One-half fraction of a 2 factorial (Cont’d 3)
Stat 402 (Spring 2016): Slide set 15
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.
•
•
•
3
4
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).
I
A
B
C
We write all this information in the form:
One-half fraction of a 2 factorial (Cont’d 2)
Stat 402 (Spring 2016): Slide set 15
•
•
•
•
•
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3
•
•
•
•
•
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2
Fractional Factorial
Stat 402 (Spring 2016): Slide set 15
a
-
b
-
ab
+
c
-
ac
+
bc
+
abc
-
d
-
ad
+
bd
+
abd
-
cd
+
acd
-
: Analysis
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
10
: Analysis (Cont’d)
These facts are summarized in Table 18.4 (p.327 of the text).
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11
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.
•
I=ABCD
A=BCD AB=CD
B=ACD AC=BD
C=ABD BC=AD
D=ABC
The fact that the above are the only factorial effects that need to be
computed is clear from looking at the alias pattern:
2
4−1
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.
Stat 402 (Spring 2016): Slide set 15
•
•
Stat 402 (Spring 2016): Slide set 15
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.
2
4− 1
abcd
+
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:
2
Stat 402 (Spring 2016): Slide set 15
9
bcd
-
•
4− 1
8
(1), ab, ac, bc, ad, bd, cd, and abcd.
Then select those 16 treatment combinations that has + signs
(1)
ABCD=+
In general 21p fraction of a 2k factorial is called a 2k−p fractional
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:
th
k−p
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12
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)
Stat 402 (Spring 2016): Slide set 15