ST 5203 Experimental Design Tutorial 8
Question 11 (Page 355)
In this question, we have a 28-2 design with effects A, B, C, D, E, F, G and H. Suppose a
first experiment (which we shall henceforth call MAIN) was conducted and the following
results are either obtained or assumed:
1. All three-factor interactions and higher order interactions are considered to be
zero except possibly those listed in (4) below.
2. All main effects are estimated cleanly.
3. All two-factor interactions are estimated cleanly or are known from previous
experimentation, except for those listed in (4) below.
4. The following alias “pairs” (after dropping other terms from the alias group) have
significant results:
CDF=ABF
CF=CGH
AB=CD
BF=BGH
AF=AGH
DF=DGH
Problem: We wish to design an experiment in order to estimate cleanly each of the
twelve effects listed above, using the minimum number of treatment
combinations.
We first make two important observations before designing the experiment.
1. In each of the alias pairs given above, for example CDF=ABF, if one of the
effects (say, CDF) can be estimated cleanly using the new experiment that we
are going to design, then we can estimate the other (ABF) by substituting the
estimate of CDF into the original estimate of the sum (or difference) of CDF
and ABF (obtained from experiment MAIN) to obtain the clean estimate of
ABF.
2. In view of point 1 above, we try to design our experiment in such a way that
in each of the 6 alias pairs above, at least one of the two effects can be
estimated cleanly. Observe that in the 6 pairs above, only effects A, B, C, D, F
(five in total) appears on the left hand side while effects A, B, C, D, F, G, H
(seven in total) appears on the right hand side. This gives us an indication on
how we can minimize the number of treatment combinations used.
Let us consider a 25-2 fractional factorial design (which we shall denote as SUB) with the
following defining relation:
I=ACD=BDF=ABCF
Consequently, the alias rows are as follows:
1)
2)
3)
4)
5)
6)
7)
I
A
B
C
D
F
AB
AF
=
=
=
=
=
=
=
=
ACD
CD
ABCD
AD
AC
ACDF
BCD
CDF
=
=
=
=
=
=
=
=
BDF
ABDF
DF
BCDF
BF
BD
ADF
ABD
=
=
=
=
=
=
=
=
ABCF
BCF
ACF
ABF
ABCDF
ABC
CF
BC
Recall that the six alias pairs in MAIN are CDF=ABF, CF=CGH, AB=CD, BF=BGH,
AF=AGH and DF=DGH. The effects that appear in the alias rows above are
highlighted.
We make the following observations regarding the alias rows above:
Row 1): CD can be estimated cleanly, since we have the clean estimate of A (from
MAIN) and assumed that ABDF=BCF=0.
Row 2): DF can be estimated cleanly, since we have the clean estimate of B (from
MAIN) and assumed that ABCD=ACF=0.
Row 3): ABF can be estimated cleanly, since we have the clean estimates of C and AD
(from MAIN) and assumed that BCDF=0.
Row 4): BF can be estimated cleanly, since we have the clean estimates of D and AC
(from MAIN) and assumed that ABCDF=0
We now describe how our design SUB can be used to accomplish the task of estimating
the twelve effects that are aliased in MAIN.
1. Effects CD, DF, ABF and BF are can be estimated cleanly as described in the
observations above.
2. Since CDF=ABF in MAIN, the clean estimate of ABF obtained in SUB can be
used to estimate CDF cleanly.
3. Since AB=CD in MAIN, the clean estimate of CD obtained in SUB can be used
to estimate AB cleanly.
4. Since AB=CF in SUB (see Row 6, noting that BCD=ADF=0 by our assumption),
the clean estimate of AB obtained in step 3 above can be used to estimate CF
cleanly.
5. Since CF=CGH in MAIN, the clean estimate of CF obtained in step 4 above can
be used to estimate CGH cleanly.
6. Since BF=BGH in MAIN, the clean estimate of BF obtained in SUB can be used
to estimate BGH cleanly.
7. Since AF=CDF=BC in SUB (see Row 7, noting that ABD=0 by our assumption),
the clean estimate of AF can be obtained using the clean estimate of CDF
obtained in step 2 above together with the clean estimate of BC obtained in
MAIN.
8. Since AF=AGH in MAIN, the clean estimate of AGH can be obtained using the
clean estimate of AF obtained from step 7 above.
9. Since DF=DGH in MAIN, the clean estimate of DF obtained in SUB can be used
to estimate DGH cleanly.
This concludes the discussion on how our design SUB can be used to accomplish the task
of estimating (cleanly) the twelve effects in the six alias pairs in MAIN.
------------------------------------------End of question 11-----------------------------------------Question 12 (Page 356)
In this question, we have 7 factors at two levels each and we wish to design a 27-3
experiment assuming that

all interaction effects, except AB, AC, AD, AE, AF, AG, are zero.
Note that in a 27-3 experiment, we will lose 23-1=7 effects completely. Thus we have to
decide which seven effects are to be lost completely and these seven should not include
main effects or any of the interaction effects that cannot be assumed to be zero. (So we
cannot choose AB, AC, AD, AE, AF, AG).
Note that we also do not choose three-factor interaction effects involving A, like ABC in
the defining equation. If we do so, then there will be a two-factor interaction effect
involving A being in the same alias group as a main effect. For example if I=ABC in the
defining equation, then AB=C in some alias group, and in this case, both AB and C
cannot be estimated cleanly.
Recall that we can only choose three effects in the defining equation since the other four
are automatically determined from the three we choose by using mod 2 multiplication.
Suppose we choose the following 3 three-factor interaction effects in our defining
equation BCD, BEG and CFG. Then the other four effects are CDEG, BDFG, BCEF and
DEF. So the defining equation is
I=BCD=BEG=CFG=DEF=CDEG=BDFG=BCEF
In this case, the alias groups involving main effects and two-factor interaction effects
involving effect A are as follows:
I
A
B
C
D
E
F
G
AB
AC
AD
AE
AF
AG
=
=
=
=
=
=
=
=
=
=
=
=
=
=
BCD
4
CD
BD
BC
4
4
4
3
3
3
5
5
5
=
=
=
=
=
=
=
=
=
=
=
=
=
=
BEG
4
EG
4
4
BG
4
BE
3
5
5
3
5
3
=
=
=
=
=
=
=
=
=
=
=
=
=
=
CFG
4
4
FG
4
4
CG
CF
5
3
5
5
3
3
=
=
=
=
=
=
=
=
=
=
=
=
=
=
DEF
4
4
4
EF
EF
DE
4
5
5
3
3
3
5
=
=
=
=
=
=
=
=
=
=
=
=
=
=
CDEG
5
5
3
3
3
5
3
6
4
4
4
6
4
=
=
=
=
=
=
=
=
=
=
=
=
=
=
BDFG
5
3
5
3
5
3
3
4
6
4
6
4
4
=
=
=
=
=
=
=
=
=
=
=
=
=
=
BCEF
5
3
3
5
3
3
5
4
4
6
4
4
6
In fact, without computing the table above, we can see why all main effects and twofactor interaction effects involving effect A will be estimated cleanly. This is because:
1. Main effect A will be alias with 4 four-factor interaction effects and 3 five-factor
interaction effects since the defining equation contains 4 three-factor interaction
effects and 3 four-factor interaction effects and does not involve any interaction
effects with A. This enables us to estimate A cleanly.
2. Main effects B, C, D, E, F, G will be alias with either

Two-factor interaction effects not involving A (since the three-factor
interaction effects in the defining equation does not involve A); or
 Three-factor or higher order interaction effects.
This enables us to estimate effects B,C, D, E, F and G cleanly.
3. Two-factor interaction effects involving A will be alias with three-factor or higher
order interaction effects since all the effects in the defining equation are three or
four-factor interaction effects that does not involve A. This implies that all the
two-factor interaction effects involving A can also be estimated cleanly.
In conclusion, we see that by performing a 27-3 fractional factorial design with the
defining equation given above will allow us to estimate all main effects and all two-factor
interactions involving A cleanly, assuming that all other interactions equal to zero.
------------------------------------------End of question 12------------------------------------------
Question 15 (Page 356)
In the magazine advertising study, the following 5 factors (each at 2 levels) were studied:
A: Size of advertisement (Eighth of a page vs. Quarter of a page)
B: Color (Black and white vs. Two colors)
C: Location (Top of page vs. Bottom of page)
D: Rest of page (Mostly advertisements vs. Mostly article)
E: Advertisement layout (Cluttered vs. More white)
F: Magazine (Magazine X vs. Magazine Y)
We assume that all interaction effects were zero, except two-factor interactions involving
factor F.A 26-2 fractional factorial design with the following defining equation was used:
I=ABCD=ABEF=CDEF
This defining equation divides the 26=64 treatment combinations into 4 blocks in the
following way: (principal block given in the text, other blocks obtained by performing
mod 2 multiplication on principal block).
Principal Block
1
ab
cd
abcd
ef
abef
cdef
abcdef
ace
bce
ade
bde
acf
bcf
adf
bdf
Block 2
a
b
acd
bcd
aef
bef
acdef
bcdef
ce
abce
de
abde
cf
abcf
df
abdf
Block 3
c
abc
d
abd
cef
abcef
def
abdef
ae
be
acde
bcde
af
bf
acdf
bcdf
Block 4
e
abe
cde
abcde
f
abf
cdf
abcdf
ac
bc
ad
bd
acfe
bcfe
adfe
bdef
The treatment combinations that were run (the principal block) are highlighted in the
table above. Data was given for 16 treatment combinations with 4 replicates each. The
data was entered into MINITAB with the following modification:

We exchange D and E in the defining equation. This is to enable us to input the
generators in the form that is acceptable to MINITAB. With this change, the
defining equation is now I=ABCE=ABDF=CDEF.
The following outlines the steps in which the data is keyed into MINITAB:
1. Stat>>DOE>>Factorial>>Create factorial design.
2. Type of design: 2-level factorial (specific generators)
 Number of effects = 4 = 6-2 (since we are using a 26-2 design)
3. Designs:
 Select Full factorial
 Number of replicates = 4
 Generators: Since I=ABCE=ABDF=CDEF, we have E=ABC and
F=ABD. Thus, generators are E=ABC and F=ABD.
4. Label the factors: Size(A), Color(B), Location(C), Ad Layout(E), Rest of Pg(D),
Mag(F).
5. Sort the worksheet by standard order and then key in the data under a new
“Response” column.
6. Stat>>DOE>>Factorial>>Analyze factorial design.
7. Response: “Response”.
 Terms: Since we assume that all interaction effects except those two-factor
interaction effects involving F are zero, we choose the following terms in
our model: A, B, C, D, E, F, AF, BF, CF, DF, EF.
The following gives the output from MINITAB.
Fractional Factorial Design
Factors:
Runs:
Blocks:
6
64
none
Design Generators:
Base Design:
Replicates:
Center pts (total):
E = ABC
4, 16
4
0
Resolution: IV
Fraction: 1/4
F = ABD
Fractional Factorial Fit: Response versus Size(A), Color(B), ...
Estimated Effects and Coefficients for Response (coded units)
Term
Constant
Size(A)
Color(B)
Location
Ad Layou
Rest of
Mag(F)
Size(A)*Mag(F)
Color(B)*Mag(F)
Location*Mag(F)
Ad Layou*Mag(F)
Rest of*Mag(F)
Effect
-0.719
201.281
0.844
3.844
-0.906
-3.031
-0.906
-2.031
2.156
-1.594
-1.969
Coef
228.578
-0.359
100.641
0.422
1.922
-0.453
-1.516
-0.453
-1.016
1.078
-0.797
-0.984
SE Coef
7.713
7.713
7.713
7.713
7.713
7.713
7.713
7.713
7.713
7.713
7.713
7.713
T
29.63
-0.05
13.05
0.05
0.25
-0.06
-0.20
-0.06
-0.13
0.14
-0.10
-0.13
P
0.000
0.963
0.000
0.957
0.804
0.953
0.845
0.953
0.896
0.889
0.918
0.899
Analysis of Variance for Response (coded units)
Source
Main Effects
2-Way Interactions
Residual Error
Lack of Fit
Pure Error
Total
DF
6
5
52
4
48
63
Seq SS
648642
256
197999
197565
434
846898
Adj SS
648642
256
197999
197565
434
Adj MS
108107
51
3808
49391
9
F
28.39
0.01
P
0.000
1.000
5E+03
0.000
We draw the following conclusions from the output above:
1. From the estimated effects and coefficients for response table, we see that only
effect B (Color) is significant. All the other effects are not significant even at very
high -levels.
2. From the ANOVA table, we see that the sum of all the two-way interactions is not
significant (p-value=1).
Next steps:
1. Remove interaction effects one at a time. We observe that when one interaction
effect is removed, only effect B (color) remains significant. Other interaction
effects and main effects remain insignificant.
2. The reduced model with all interaction effects removed:
Estimated Effects and Coefficients for Response (coded units)
Term
Constant
Size(A)
Color(B)
Location
Ad Layou
Rest of
Mag(F)
Effect
-0.719
201.281
0.844
3.844
-0.906
-3.031
Coef
228.578
-0.359
100.641
0.422
1.922
-0.453
-1.516
SE Coef
7.372
7.372
7.372
7.372
7.372
7.372
7.372
T
31.01
-0.05
13.65
0.06
0.26
-0.06
-0.21
P
0.000
0.961
0.000
0.955
0.795
0.951
0.838
After delete main effects one by one (backwards selection), the final model is
Y=B. Note that you should check assumptions once you get your final model. The
assumption checking is routine and skipped here.
------------------------------------------End of question 15-----------------------------------------Question 16 (Page 357)
Suppose now we are told that the four replications in question 15 are such that
replications 1 and 2 are men, replications 3 and 4 are women. The gender is assumed to
be a blocking factor.
Just revise the column “Blocks” to reflect this change and fit the data as in Question 15.
Here are Effects and Anova table:
Estimated Effects and Coefficients for Response (coded units)
Term
Constant
Block
Size(A)
Color(B)
Location
Ad Layou
Rest of
Mag(F)
Size(A)*Mag(F)
Color(B)*Mag(F)
Location*Mag(F)
Ad Layou*Mag(F)
Rest of*Mag(F)
Effect
-0.719
201.281
0.844
3.844
-0.906
-3.031
-0.906
-2.031
2.156
-1.594
-1.969
Coef
228.578
-0.484
-0.359
100.641
0.422
1.922
-0.453
-1.516
-0.453
-1.016
1.078
-0.797
-0.984
SE Coef
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
7.788
T
29.35
-0.06
-0.05
12.92
0.05
0.25
-0.06
-0.19
-0.06
-0.13
0.14
-0.10
-0.13
P
0.000
0.951
0.963
0.000
0.957
0.806
0.954
0.846
0.954
0.897
0.890
0.919
0.900
Analysis of Variance for Response (coded units)
Source
Blocks
Main Effects
2-Way Interactions
Residual Error
Lack of Fit
Pure Error
Total
DF
1
6
5
51
19
32
63
Seq SS
15
648642
256
197984
197693
291
846898
Adj SS
15
648642
256
197984
197693
291
Adj MS
15
108107
51
3882
10405
9
F
0.00
27.85
0.01
P
0.951
0.000
1.000
1E+03
0.000
As can be seen, there is only one significant effect (B- Color). Do backward model
selection as in Question 15, then we will come out the same final model: Y=B. Note that
to delete the “Blocks” effect from model in Minitab, you must tick off “include blocks in
the model” when you select terms in step 7 as stated in Question 15.
------------------------------------------End of question 16------------------------------------------