IE 361 Module 23
Design and Analysis of Experiments: Part 4
(Fractional Factorial Studies and Analyses With 2-Level Factors)
Reading: Section 7.1, Statistical Quality Assurance Methods for
Engineers
Prof. Steve Vardeman and Prof. Max Morris
Iowa State University
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IE 361 Module 23
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Large p Complete Factorials are Impossible in Practice
In this module we discuss what can be done in a many-factor context
where the number of possible combinations of levels of p factors is so large
that doing a complete factorial experiment is not practically possible. For
p factors, even 2p gets big fast, for example.
for p = 10, 2p = 1024
So in practice, for p of any size, one must make do with information from
only some fraction of a complete factorial in p factors.
We consider rational approaches to design and analysis of fractional
factorial studies, and here limit our discussion to fractions of 2p studies.
We will …rst consider half fractions of these, and then other fractions that
are powers of 1/2.
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IE 361 Module 23
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Large p Complete Factorials are Impossible in Practice
Example 23-1 (From an article by Hendrix in 1979 Chemtech)
A 15-factor chemical experiment had factors and levels as
Factor
Levels
A-Coating Roll Temp
115
B-Solvent
Factor
vs 125
Levels
J-Feed Air to Dryer Preheat
Yes vs No
Recycled vs Re…ned
K-Dibutylfutile in Formula
12% vs 15%
C-Polymer X-12 Preheat
No vs Yes
L-Surfactant in Formula
.5% vs 1%
D-Web Type
LX-14 vs LB-17
M-Dispersant in Formula
.1% vs .2%
E-Coating Roll Tension
30 vs 40
N-Wetting Agent in Formula
1.5% vs 2.5%
F-Number of Chill Roll
1 vs 2
O-Time Lapse
10min vs 30min
G-Drying Roll Temp
75
P-Mixer Agitation Speed
100rpm vs 250rpm
H-Humidity of Air Feed
75% vs 90%
vs 80
and response variable
y = a measure of product cold crack resistance
A full factorial would require at least 215 = 32, 768 runs of the process!
An obvious "solution" is to collect data for only some (a fraction) of all
possible combinations of levels of the factors.
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IE 361 Module 23
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"Obvious" Limitations and Objective
Qualitative points that ought to be “obvious” a priori if a fraction of a
factorial is used as an experimental design are that
there must be some information loss (relative to the full factorial),
some ambiguity must inevitably follow because of the loss, and
careful planning and wise analysis are needed to hold these to a
minimum.
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IE 361 Module 23
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"Obvious" Limitations and Objective
Example 23-2 (Hypothetical/unrealistic but instructive half fraction of a 2X2 Factorial)
The full 22 factorial structure is shown in below. Here, if combination (1)
is used, combination ab must be employed or else one learns nothing about
the action of one of the factors. (If a is used, then b must be employed or
one learns nothing about the action of one of the factors.) On the other
hand, if there is a big di¤erence in response between the two combinations
included in the half fraction, one doesn’t know whether to attribute this to
one or the other or to the interaction e¤ect of the factors A and B.
Figure: A 22 Full Factorial Layout
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IE 361 Module 23
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Potential and Problems
Example 23-3 (Hypothetical)
Suppose that 23 factorial e¤ects are
µ... = 10, α2 = 3, β2 = 1, γ2 = 2, αβ22 = 2,
αγ22 = 0, βγ22 = 0, αβγ222 = 0
The corresponding means are then as pictured below.
Figure: A 23 Factorial With a "Good" Half Fraction Indicated With Circled
Corners
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IE 361 Module 23
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Potential and Problems
Example 23-3 continued
Suppose further that one gets data adequate to essentially reveal the mean
responses for combinations a, b, c and abc (the 4 corners circled on panel
6) but has no data on the other combinations. How then might one try to
assess the 23 factorial e¤ects if presented with information from the circled
corners on panel 6?
Remember that with the 23 as pictured on panel 6,
α2 = "right face average"
"grand average" = 3
A "half-fraction version" of this might be
α2
= "available right face average"
= 13 10 = 3
"available grand average"
Here α2 = α2 !!! Is this something for nothing? Can we learn about the
A main e¤ect using data from only "4 corners"?
Vardeman and Morris (Iowa State University)
IE 361 Module 23
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Potential and Problems
Example 23-3 continued
But note that a similar calculation for the C main e¤ect gives
γ2
= "available back face average"
= 14 10 = 4
and 4 = γ*2 6= γ2 = 2
"available grand average"
???
The general story here is that for this 23
1
fractional factorial
α2 = α2 + βγ22 and γ2 = γ2 + αβ22
We were able to "recover" α2 using only α2 (based on only half of the
corners of the cube) because βγ22 = 0. We were unable to "recover" γ2
using only γ2 (based on only half of the corners of the cube) because
αβ22 = 2 6= 0. We can really only know α2 + βγ22 (and not α2 alone)
and γ2 + αβ22 (and not γ2 alone) based on the half fraction. This is an
example of confounding/aliasing/ambiguity that of necessity comes with
use of only a fractional factorial data collection plan.
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IE 361 Module 23
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An Aside
Example 23-3 continued
As a bit of an aside at this point, notice that the choice of 4 corners on
panel 6 has admirable symmetry. If one collapses the cube in any direction
one is left with a complete factorial arrangement in the two other factors.
That means that if, in fact, one of the 3 factors is "inert" doing nothing to
a¤ect response, one has a full factorial in the 2 factors that matter.
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IE 361 Module 23
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Issues to Address
The issues to be addressed in order to use 2p
how to rationally choose
1
2q
q
fractional factorials are:
out of 2p combinations for study,
how to determine the corresponding aliasing/confounding pattern, and
how to do data analysis.
We proceed to consider these questions …rst in the context of half
fractions (the q = 1 case), then for general 1/2q fractions.
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IE 361 Module 23
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Choice of a Half Fraction
To choose what we will call a "standard" (a "good") half fraction of a 2p
factorial, we will
write out signs for specifying levels for all possible combinations of
levels of the “…rst” p 1 factors, and
then “multiply” these together for a given combination of the “…rst”
factors to arrive at a corresponding level to use for the “last” factor.
Vardeman and Morris (Iowa State University)
IE 361 Module 23
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Choice of a Half Fraction
Example 23-4 (A Hypothetical Half Fraction of a 4-Factor Study)
With 4 two-level factors A, B, C and D, one proceeds as in the following
table. (One multiplies
and + signs as if they were 1’s and the last
column gives the indicated combination of factor levels for the row, using
the special 2p naming convention introduced in Module 22.)
A
B
C
Product (used for D)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Vardeman and Morris (Iowa State University)
+
+
Combination
(1)
ad
bd
ab
cd
ac
bc
abcd
IE 361 Module 23
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Choice of a Half Fraction
Example 23-5
R. Snee in a 1985 ASQC Technical Supplement discussed a 25 1 chemical
process study. The factors and their levels were as in the following table.
Factor
A-Solvent/Reactant
B-Catalyst/Reactant
C-Temperature
D-Reactant Purity
E-pH of Reactant
( )
low
.025
150
92%
8.0
vs
vs
vs
vs
vs
(+)
high
.035
160
96%
8.7
In Snee’s study, the response variable was
y = color index
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IE 361 Module 23
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Choice of a Half Fraction
Example 23-5
Snee’s (unreplicated) data were as below.
Combination
e
a
b
abe
c
ace
bce
abc
y
.63
2.51
2.68
1.66
2.06
1.22
2.09
1.93
Combination
d
ade
bde
abd
cde
acd
bcd
abcde
y
6.79
6.47
3.45
5.68
5.22
9.38
4.30
4.05
These are data from half of all 32 combinations of 2 levels of each of the 5
factors (half of all possible labels of combinations based on the 5 letters
a,b,c,d and e are given above, namely those involving an odd number of
letters). In fact, Snee followed the standard recommendation for choosing
this half fraction.
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IE 361 Module 23
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Aliasing Structure of a Half Fraction
To understand the pattern of ambiguities one is left with upon using a
standard half fraction of a 2p factorial, we will use a method of formal
multiplication, beginning from a so-called “generator” that represents the
way in which the half fraction was chosen. The generator is of the form
name of “last” factor $ product of names of “…rst” factors
The rules of formal multiplication are that
any letter I$the same letter
letter same letter$I
(We will use the words "aliasing" and "confounding" interchangeably to
refer to the ambiguities left by a fractional factorial study.)
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IE 361 Module 23
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Aliasing Structure of a Half Fraction
Example 23-3 continued
In the hypothetical 23
1
example, the generator is
C $ AB
We can multiply through by C to obtain the so called “de…ning relation”
I $ ABC
This …rst says that the ABC 3-factor interaction αβγ222 is aliased with the
grand mean. That is, only
µ
...
+ αβγ222
can be evaluated based on the half-fraction, not αβγ222 alone.
Multiplying through the de…ning relation by any set of letters of interest
produces a statement of what e¤ect(s) are “aliased with” the
corresponding e¤ect.
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IE 361 Module 23
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Aliasing Structure of a Half Fraction
Example 23-3 continued
For example
A $ BC
(read “the A main e¤ect is aliased with the BC 2-factor
interaction). Similarly
C $ AB
as was illustrated earlier. In fact, the whole alias structure is
I $ ABC and A $ BC and B $ AC and C $ AB
and we see that 23 factorial e¤ects are aliased in 4 pairs.
The technical meaning of aliasing is that only sums of e¤ects can be
learned from the half fraction study, not individual e¤ects. (This makes
sense. With only 4 conditions studied, one should be able to resolve only
4 di¤erent quantities!)
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IE 361 Module 23
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Aliasing Structure of a Half Fraction
Example 23-4 continued
The hypothetical 24
1
study had generator
D $ ABC
So the de…ning relation is
I $ ABCD
From this we see, e.g., that the AB 2-factor interaction is aliased with the
CD 2-factor interaction.
In fact, the student is encouraged to write out the entire alias structure
and see that the 24 factorial e¤ects are aliased in 8 pairs.
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IE 361 Module 23
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Aliasing Structure of a Half Fraction
Example 23-5 continued
Snee’s 25
1
study had generator
E $ ABCD
and hence de…ning relation
I $ ABCDE
From this one sees, e.g., that the AB 2-factor interaction is aliased with
the CDE 3-factor interaction. (There are 16 pairs of aliased/
indistinguishable 25 factorial e¤ects. One can hope to learn only about
the sum of a pair, not the individual e¤ects making up a pair.)
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IE 361 Module 23
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Data Analysis for a Half Fraction
To do data analysis for a half fraction of a 2p study, one may
initially temporarily ignore the “last” factor, treat the data as a full
factorial in the “…rst” p 1 factors, and judge the statistical
signi…cance and practical importance of estimates derived from the
Yates algorithm, and
then interpret these estimates in light of the alias structure (as
estimates of appropriate sums of 2p e¤ects).
For judging statistical signi…cance, where there is some replication (not all
2p 1 sample sizes are 1) con…dence intervals can be made for the (sums
of) e¤ects. Lacking any replication, normal plotting of the output of the
Yates algorithm (ignoring the “last” factor) is the only available method.
Vardeman and Morris (Iowa State University)
IE 361 Module 23
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Data Analysis for a Half Fraction
To be completely explicit about the making of con…dence intervals based
on the output of the Yates, algorithm, we use
r
1
1
b tspooled
E
2p 1 ∑ ncomb
where
2
spooled
=
2
∑ (ncombination 1) scombination
∑ (ncombination 1)
and the appropriate degrees of freedom for t are
∑ (ncombination
Vardeman and Morris (Iowa State University)
1) = n
IE 361 Module 23
2p
1
21 / 47
Data Analysis for a Half Fraction
Example 23-6 (Another Hypothetical Half Fraction of a 2X2X2 Study)
Suppose
na = 1, ya = 5, nb = 2, ȳb = 3, sb2 = 1.5, nc = 1, yc = 2.5, and
2
nabc = 3, ȳabc = 5.5, sabc
= 1.8
Then listing the 4 combinations in Yates standard order as regards the
"…rst" 2 factors A and B (i.e. ignoring the "last" factor, C), the
(p 1 = 2 cycle) Yates algorithm is applied to the following table.
Combination
c
a
b
abc
ȳ
2.5
5.0
3.0
3.0
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Data Analysis for a Half Fraction
Example 23-6 continued
Con…dence intervals based on the output of the algorithm would be made
using
0 + (2 1)1.5 + 0 + (3 1)1.8
2
spooled
=
0 + (2 1) + 0 + (3 1)
These have the form
b
E
t3 spooled
Vardeman and Morris (Iowa State University)
1
23 1
r
1 1 1 1
+ + +
1 2 1 3
IE 361 Module 23
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Data Analysis for a Half Fraction
Example 23-5 continued
Snee’s 25 1 study had no replication. Ignoring factor E temporarily, the
(4-cycle) Yates algorithm can be applied to the 16 responses exactly as
listed earlier (they are in Yates order as regards the …rst 4 factors). The
result is the set of estimates below.
Figure: Estimates for Snee’s Chemical Data
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IE 361 Module 23
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Data Analysis for a Half Fraction
Example 23-5 continued
A normal plot of the (last 15) Snee estimates is below and suggests that
at most 4 sums of e¤ects are distinguishable from background variation.
Figure: A Normal Plot of 15 Fitted Sums of E¤ects From Snee’s 25 1 Study
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IE 361 Module 23
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Data Analysis for a Half Fraction
Example 23-5 continued
Tentative engineering conclusions based on Snee’s study were that for
uniform color index, attention must be paid to controlling/reducing
variation in the following (in decreasing order of importance):
Factor D, Reactant Purity
Factor B, Catalyst/Reactant Ratio
Factor E, pH of Reactant
Factor A, Solvent/Reactant Ratio
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IE 361 Module 23
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Issues to Address
The issues to be addressed in order to use 2p
how to rationally choose
1
2q
q
fractional factorials remain:
out of 2p combinations for study,
how to determine the corresponding aliasing/confounding pattern, and
how to do data analysis.
The answers for smaller-than-half fractions are the natural generalizations
of the half fraction answers just discussed.
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IE 361 Module 23
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Choice of a Fractional Factorial
To choose a 1/2q fraction of a 2p factorial, we will
write out signs for specifying levels for all possible combinations of
levels of the “…rst” p q factors, and
pick q di¤erent groups of the …rst p q factors and use the products
of the signs corresponding to members of the groups to specify levels
for the “last” q factors.
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IE 361 Module 23
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Choice of a Fractional Factorial
Example 23-7
Hanson and Best in a presentation at the 1986 annual meeting of the
American Statistical Association reported on an experiment for the
development of a catalyst for producing ethyleneamines by the amination
of monoethanolamine involving p = 5 factors. These factors and their
levels were
Factor
A-Ni/Re Ratio
B-Precipitant
C-Calcining Temp
D-Reduction Temp
E-Support Used
( )
2/1
(NH4 )2 CO
300
300
alpha-alumina
vs
vs
vs
vs
vs
(+)
20/1
none
500
500
silica alumina
The response of interest was
y = % water produced
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IE 361 Module 23
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Choice of a Fractional Factorial
Example 23-7 continued
The investigators decided against a full factorial, choosing instead to use a
25 2 design. That is, they chose to use q = 2 and a 14 fraction of the full
25 study. They ran 25 2 = 8 out of the 25 = 32 possible A, B, C, D, E
combinations.
The (somewhat arbitrary) choice was made to use ABC sign products to
choose levels of D, and BC sign products to choose levels of E. (Other
choices are possible and lead to di¤erent aliasing patterns that might for
some other studies be preferred by the engineer in charge.) The choice
used is summarized in the table on panel 31, whose last column speci…es
the 8 combinations of levels of the 5 factors used in the study in the
special 2p factorial notation.
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IE 361 Module 23
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Choice of a Fractional Factorial
Example 23-7 continued
A
B
C
ABC Product (for D)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Vardeman and Morris (Iowa State University)
BC Product (for E)
+
+
+
+
+
+
IE 361 Module 23
Combination
e
ade
bd
ab
cd
ac
bce
abcde
31 / 47
Choice of a Fractional Factorial
Example 23-7 continued
The data reported by Hanson and Best were as listed and summarized
below.
Combination
e
ade
bd
ab
cd
ac
bce
abcde
y
8.70, 11.60, 9.00
26.80
24.88
33.15
28.90, 30.98
30.20
8.00, 8.69
29.30
Vardeman and Morris (Iowa State University)
y
9.767
26.800
24.880
33.150
29.940
30.200
8.345
29.300
IE 361 Module 23
s2
2.543
2.163
.238
32 / 47
Aliasing Structure of a Smaller-Than-Half Fraction
To understand the pattern of ambiguities one faces with a particular
choice of a 21q fraction of a 2p factorial, we will use the formal
multiplication, beginning from the q generators that represent the way in
which the 21q fraction was chosen. To …nd the de…ning relation (the list
of all products "equivalent to" I) we …rst convert the generators to
statements of products equivalent to I, and then multiply these in pairs,
then in triples, then in sets of four, etc. The letter I will ultimately have
2q 1 equivalent products, i.e. the 2p factorial e¤ects are aliased in 2p q
di¤erent groups of 2q each.
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IE 361 Module 23
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Aliasing Structure of a Smaller-Than-Half Fraction
Example 23-7 continued
In the catalyst example, the generators were
D $ ABC and E $ BC
so
I $ ABCD and I $ BCE
Further, multiplying these two we get
I I $ (ABCD) (BCE) i.e. I $ ADE
So the whole de…ning relation for the catalyst study is
I $ ABCD $ BCE $ ADE
and therefore e¤ects are aliased in 8 groups of 4. For example,
multiplying through the de…ning relation by A gives
A $ BCD $ ABCE $ DE
and we see that the A main e¤ect is aliased with the DE 2-factor
interaction (among other things).
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IE 361 Module 23
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Data Analysis for a Smaller-Than-Half Fraction
To do data analysis for a 2p
q
study, one may
initially temporarily ignore the “last” q factors, treat the data as a full
factorial in the “…rst” p q factors, and judge the statistical
signi…cance and practical importance of estimates derived from the
Yates algorithm, and
then interpret these estimates in light of the alias structure (as
estimates of appropriate sums of 2p e¤ects).
For judging statistical signi…cance, where there is some replication (not all
2p q sample sizes are 1) con…dence intervals can be made for the (sums
of) e¤ects. Lacking any replication, normal plotting of the output of the
Yates algorithm (ignoring the “last” factor) is the only available method.
Vardeman and Morris (Iowa State University)
IE 361 Module 23
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Data Analysis for a Smaller-Than-Half Fraction
To be explicit, the form of con…dence intervals for the (sums of) e¤ects is
r
1
1
b tspooled
E
2p q ∑ ncomb
where
2
spooled
=
2
∑ (ncombination 1) scombination
∑ (ncombination 1)
and the appropriate degrees of freedom for t are
∑ (ncombination
Vardeman and Morris (Iowa State University)
1) = n
IE 361 Module 23
2p
q
36 / 47
Data Analysis for a Smaller-Than-Half Fraction
Example 23-7 continued
In the catalyst example the 8 sample means, ȳ , listed before were in Yates
standard order for factors A, B and C (the “…rst” p q = 3) ignoring D
and E (the “last” q = 2). So the (p q = 3 cycle) Yates algorithm can
be applied to them in the order listed. The following table shows the …rst
two and last columns of the Yates table and then records what the
estimates produced by the algorithm attempt to approximate.
Combination
e
ade
bd
ab
cd
ac
bce
abcde
y
9.767
26.800
24.880
33.150
29.940
30.200
8.345
29.300
Vardeman and Morris (Iowa State University)
Estimate
24.048
5.815
.129
1.492
.399
.511
5.495
3.682
Sum Estimated
grand mean+aliases
A main e¤ect+aliases
B main e¤ect+aliases
AB interaction+aliases
C main e¤ect+aliases
AC interaction+aliases
BC interaction+aliases
ABC interaction+aliases
IE 361 Module 23
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Data Analysis for a Smaller-Than-Half Fraction
Example 23-7 continued
Since the original data had 3 sample sizes larger than 1, statistical
signi…cance/detectability of these can be judged using con…dence limits for
sums of e¤ects. First,
2
spooled
=
(3
1)(2.543) + (2 1)(2.163) + (2 1)(.238)
= 1.872
(3 1) + (2 1) + (2 1)
p
So spooled = 1.872 = 1.368, and this can be used as a measure of
background noise and as a basic ingredient of con…dence intervals for the
sums of e¤ects.
spooled has 4 associated degrees of freedom. So if, e.g., 95% con…dence
intervals for the sums of e¤ects are desired, the “+/ part” of the
con…dence interval formula becomes
r
1
1 1 1 1 1 1 1 1
2.776(1.368) 3
+ + + + + + +
i.e.
1.195
2
3 1 1 1 2 1 2 1
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IE 361 Module 23
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Data Analysis for a Smaller-Than-Half Fraction
Example 23-7 continued
So a margin of error to associate with any one of the values produced by
the Yates algorithm is 1.195. We might therefore judge any estimate
larger in absolute value than 1.195 to represent a sum of e¤ects clearly
large enough to see above the background experimental variation.
Then the “detectable” sums are (in decreasing order of magnitude):
Sum
Estimate
α2 + βγδ222 + αβγe2222 + δe22
5.815
βγ22 + αδ22 + e2 + αβγδe22222
5.495
αβγ222 + δ2 + αe22 + βγδe2222
3.682
αβ22 + γδ22 + αγe222 + βδe222
1.492
Happily, the last of these is smaller in magnitude than the other 3, but
there are at least 4 a priori equally plausible interpretations of the
possibility that each one of these 3 are really driven by a single e¤ect.
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IE 361 Module 23
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Data Analysis for a Smaller-Than-Half Fraction
Example 23-7 continued
From this data analysis alone, it is equally plausible that there are
important
A main e¤ects, E main e¤ects, and D Main e¤ects,
A main e¤ects, E main e¤ects, and AE 2-factor interactions,
A main e¤ects, AD 2-factor interactions, and D main e¤ects, or
DE 2-factor interactions, E main e¤ects, D main e¤ects.
In fact, a follow-up study con…rmed the importance of the D main e¤ect
(and made the …rst of these most attractive).
If the A (Ni/Re ratio) main e¤ect, the E (Support Type) main e¤ect and
the D (Reduction Temp) main e¤ect are indeed the most important
determiners of y , and large y is desirable, the signs of the estimates
indicate the need for “high A” (20/1 Ni/Re ratio), “low E”
(alpha-alumina support) and “high D” (500 reduction temp).
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IE 361 Module 23
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Perspective
By now it should be obvious that the larger is q, the larger the inevitable
ambiguity of interpretation of the fractional factorial results and the more
likely the need for follow-up study. Small fractions are really most useful
as screening studies, to pick a few likely candidates out of many
potentially important factors for subsequent more detailed study.
We end with an extreme example involving a large q, i.e. a small fraction.
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IE 361 Module 23
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Example 23-1 continued
The Hendrix chemical process study involved p = 15 factors A, B, C, D, E,
F, G, H, J, K, L, M, N, O, P (the factor names and levels were given
earlier).
Here p q = 4, i.e., only 24 = 16 combinations were run !!!!! This was a
1
1
= 2048
fraction !!!!
2 15 4
The 11 generators used were:
E $ ABCD,F $ BCD,G $ ACD,H $ ABC,J $ ABD,
K $ CD,L $ BD,M $ AD,N $ BC,O $ AC,P $ AB
These led to the 16 combinations and (ultimately) the data on panel 43.
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IE 361 Module 23
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Example 23-1 continued
Figure: Hendrix 215 11 Data
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Example 23-1 continued
Pretty clearly it isn’t sensible to write out the whole de…ning relation here.
E¤ects are going to be aliased in 16 groups of 211 = 2048 e¤ects. But for
a most tentative interpretation, let’s see what we might glean if the
physical system is so simple that only main e¤ects dominate. (Whether
this is physically reasonable is a question that needs to be answered by a
process expert.)
The 16 observations are listed in Yates order for factors A,B,C and D
(ignoring the rest). We therefore begin by running them through the
Yates algorithm, with the results on panel 45.
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Example 23-1 continued
Figure: Result of the Yates Algorithm for the Hendrix Data
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Example 23-1 continued
There is no replication in this data set, so we’re driven to normal plotting
in order to judge statistical signi…cance of these estimates. A normal plot
of the (last 15) estimates is below.
Figure: Normal Plot of 15 Estimated Sums of E¤ects from the 215 11 Chemical
Process Study
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Example 23-1 continued
The plot shows that there are two …tted sums of e¤ects that are clearly
statistically detectable. As each of these stands for a sum of 2048 e¤ects,
no solid …nal conclusions can be made. But if one assumes that the "big"
sums are primarily driven by the main e¤ects appearing in them, a plausible
tentative interpretation is that the most important factors appear to be
B (Solvent) and F (# of Chill Rolls)
and for large cold crack resistance "high B" (re…ned solvent) and "high F"
(2 chill rolls) appear best.
Note that the analysis does point out what is in retrospect quite obvious,
namely that it is those combinations in the data set with “high B” and
“high F” that have the largest y ’s.
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