Chapter 5 Introduction to Factorial Designs
Topics: Factorial Designs, main e↵ects, interactions
5.3 The Two-Factor Factorial Design
The battery life experiment
• Two factors: Material type (qualitative) and Temperature (quantitative)
• The engineer can control the temperature during the experiment.
• However, when the battery is manufactured and shipped to the field, theengineer has no control over the temperature.
• Response: life (in hours) of the battery.
Material Temperature (oF)type 15 70 1251 130,155,74,180 34,40,80,75 20,70,82,582 150,188,159,126 136,122,106,115 25,70,58,453 138,110,168,160 174,120,150,139 96,104,82,60
Questions:
• What e↵ects do material type and temperature have on life?
• Is there a choice of material that would give long life regardless of tem-perature (a robust product)?
This is a two-factor factorial design (or two-way layout).
• Factor A has a levels and factor B has b levels.
• There are total ab combinations (treatments), each replicated n times.
• The order in which the abn observations are taken is selected at randomso that this design is a completely randomized design.
21
Comparison with randomized block design (RBD)
• RBD has one blocking factor and one experimental factor
• A 2-factor factorial studies two experimental factors
• In a 2-factor factorial, randomization is applied to all ab units.
• In a RBD, randomization is applied within each block.
• A block represents a restriction on randomization.
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tempM
ean
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pons
e
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material312
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Interaction Plot
material
Mea
n R
espo
nse
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temp
7015125
An interaction plot shows the mean responses of the treatments. It is auseful tool to see the relationship between two factors.
• If there is no interaction e↵ect, the lines would be nearly parallel.
22
Linear model (or e↵ects model) for the two-factor factorial design is
yijk
= µ+ ↵i
+ �j
+ (↵�)ij
+ ✏ijk
, i = 1, . . . , a; j = 1, . . . , b; k = 1, . . . , n
where yijk
is the observation for the kth replicate of the ith level of factor Aand the jth level of factor B, ↵
i
is the ith main e↵ect for A, �j
is the jthmain e↵ect for B, (↵�)
ij
is the (i, j)th interaction e↵ect between A and Band ✏
ijk
are NID(0, �2) errors.
The Zero-Sum Constraints
aX
i=1
↵i
=bX
j=1
�j
= 0,
aX
i=1
(↵�)ij
= 0 for j = 1, . . . , b
bX
j=1
(↵�)ij
= 0 for i = 1, . . . , a
The ANOVA table
Sources DF SS MSA a� 1 SS
A
MSA
= SSA
/(a� 1)B b� 1 SS
B
MSB
= SSB
/(b� 1)A⇥ B (a� 1)(b� 1) SS
AB
MSAB
= SSAB
/((a� 1)(b� 1))residual ab(n� 1) SS
E
MSE
= SSE
/(ab(n� 1))total abn� 1 SS
total
The ANOVA decomposition:
aX
i=1
bX
j=1
nX
k=1
(yijk
� y···
)2 =aX
i=1
nb(yi··
� y···
)2 +bX
j=1
na(y·j·
� y···
)2
+aX
i=1
bX
j=1
n(yij·
� yi··
� y·j·
+ y···
)2 +aX
i=1
bX
j=1
nX
k=1
(yijk
� yij·
)2
SStotal
= SSA
+ SSB
+ SSAB
+ SSE
.
23
For the null hypothesis H0 : ↵1 = · · · = ↵a
(i.e., no di↵erence between thefactor A main e↵ects), reject H0 at level ↵ if
F =MS
A
MSE
=SS
A
/(a� 1)
SSE
/ab(n� 1)> F
a�1,ab(n�1),↵,
Test of the factor B main e↵ects is similar.For the null hypothesis H0: all (↵�)ij are equal (i.e., no interaction e↵ect
between A and B), reject H0 at level ↵ if
F =MS
AB
MSE
=SS
AB
/((a� 1)(b� 1))
SSE
/ab(n� 1)> F(a�1)(b�1),ab(n�1),↵,
For the battery experiment, the ANOVA table is
Df Sum Sq Mean Sq F value Pr(>F)
material 2 10684 5342 7.9114 0.001976 **
temperature 2 39119 19559 28.9677 1.909e-07 ***
material:temperature 4 9614 2403 3.5595 0.018611 *
Residuals 27 18231 675
Both main e↵ects are very significant. There are some moderate interac-tion between the material type and temperature. The interaction e↵ect issignificant at 5% level but not significant at 1% level.
After rejecting the nulls, it is usually of interest to make comparisons be-tween the individual factor levels to discover the specific di↵erences. BecauseAB interaction is significant, one can compare levels of A (or B) for fixedlevel of B (or A); one can also compare all ab treatments. See the text fordetails on multiple comparisons.
Conclusion:
• There are some moderate interaction between material type and tem-perature. The lower the temperature (within the experimental range),the longer the battery life.
• Batteries made from material type 3 tend to have a uniform long liferegardless of temperature.
24
Residual analysis for the battery experiment
60 80 100 140
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20
Fitted values
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Theoretical Quantiles
Stan
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resi
dual
s
Normal Q−Q plot
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material
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idua
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temp
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idua
ls
The normal probability plot shows one potential low outlier.The residual plots show some mild inequality of variance, with the treat-
ment combination of 15oF and material type 1 possibly having larger variancethan the others.
Overall, the problem is not severe enough to have a dramatic impact onthe analysis and conclusions.
25
5.4 The General Factorial Design
Consider a three-factor factorial design
• A has a levels, B has b levels and C has c levels.
• There are total abc treatments, each replicated n times.
• For each replicate, abc units are randomly assigned to the abc treatments.
The linear model for the three-factor factorial design is:
yijkl
= µ+ ↵i
+ �j
+ �k
+ (↵�)ij
+ (↵�)ik
+ (��)jk
+ (↵��)ijk
+ ✏ijkl
,
i = 1, . . . , a; j = 1, . . . , b; k = 1, . . . , c, l = 1, . . . , n, (19)
where ↵i
, �j
and �k
are the A, B and C main e↵ects; (↵�)ij
, (↵�)ik
and (��)jk
are the A ⇥ B, A ⇥ C and B ⇥ C two-factor interaction e↵ects; (↵��)ijk
isthe A⇥B⇥C three-factor interaction e↵ect, and ✏
ijkl
are NID(0, �2) errors.Let
yijkl
= µ+ ↵i
+ �j
+ �k
+ [(↵�)ij
+ d(↵�)ik
+ d(��)jk
+ \(↵��)ijk
+ rijkl
where
µ = y····
,
↵i
= yi···
� y····
,
�j
= y·j··
� y····
,
�k
= y··k·
� y····
,[(↵�)
ij
= yij··
� yi···
� y·j··
+ y····
,d(↵�)
ik
= yi·k·
� yi···
� y··k·
+ y····
,d(��)
jk
= y·jk·
� y·j··
� y··k·
+ y····
,\(↵��)
ijk
= yijk·
� yij··
� yi·k·
� y·jk·
+ yi···
+ y·j··
+ y··k·
� y····
,
rijkl
= yijkl
� yijk·
are the estimates of the parameters under the zero-sum constraints.
26
The ANOVA table is
Degrees of Sum ofSource Freedom Squares
A a� 1P
a
i=1 nbc(↵i
)2
B b� 1P
b
j=1 nac(�j)2
C c� 1P
c
k=1 nab(�k)2
A⇥ B (a� 1)(b� 1)P
a
i=1
Pb
j=1 nc([(↵�)
ij
)2
A⇥ C (a� 1)(c� 1)P
a
i=1
Pc
k=1 nb(d(↵�)
ik
)2
B ⇥ C (b� 1)(c� 1)P
b
j=1
Pc
k=1 na(d(��)
jk
)2
A⇥ B ⇥ C (a� 1)(b� 1)(c� 1)P
a
i=1
Pb
j=1
Pc
k=1 n(\(↵��)
ijk
)2
residual abc(n� 1)P
a
i=1
Pb
j=1
Pc
k=1
Pn
l=1 (yijkl � yijk·
)2
total abcn� 1P
a
i=1
Pb
j=1
Pc
k=1
Pn
l=1 (yijkl � y···
)2
SStotal
= SSA
+ SSB
+ SSC
+ SSAB
+ SSAC
+ SSBC
+ SSABC
+ SSE
The F statistic for testing H0 : ↵1 = . . . = ↵a
(i.e., no di↵erence amongfactor A main e↵ects)
F =MS
A
MSE
=SS
A
/(a� 1)
SSE
/abc(n� 1),
which has a� 1 and abc(n� 1) DF.The F statistics for other hypotheses are similar.
27
Chapter 6 The 2k Factorial Design
Topics: main e↵ects, interactions, half-normal quantile plot.
• A special and important case of the general k-factor factorial design.
• Each factor has two levels (low/high, �/+, 0/1, etc.)
6.2 The 22 Design
The chemical process experiment
• To investigate the e↵ect of reactant concentration (A) and the catalystamount (B) on the conversion (yield y) in a chemical process.
• A: low (15%) and high (25%); B: low (1 lb) and high (2 lbs).
• The experiment is replicated 3 times.
• The order in which the runs are made is random.
ReplicateA B 1 2 3 Total� � 28 25 27 80+ � 36 32 32 100� + 18 19 23 60+ + 31 30 29 90
Notation
• The 4 treatment combinations are labeled as (1), a, b, ab.
• (1), a, b, ab also represent the total of all n replicates taken at thetreatment combinations.
28
Main e↵ects
A = y(A = +)� y(A = �) = [ab+ a� b� (1)]/(2n).
B = y(B = +)� y(B = �) = [ab+ b� a� (1)]/(2n).
Interaction e↵ect
AB = y(AB = +)� y(AB = �) = [ab+ (1)� a� b]/(2n).
The estimates are A = 8.33, B = �5, AB = 1.67.
Linear model is
y = �0 + �1xA + �2xB + �3xAxB + ✏,
where xA
= ±1, xB
= ±1 and ✏ is NID(0, �2) error.
The Model Matrix for a single replicate is
Treatment Factorial E↵ectCombination I A B AB
(1) +1 �1 �1 +1a +1 +1 �1 �1b +1 �1 +1 �1ab +1 +1 +1 +1
• The model matrix X is orthogonal, i.e., XTX = (4n)I4.
• Each column (except for I) represents a contrast.
• All the contrasts are orthogonal to each other.
The least square estimates are
�0 = y···
= [ab+ a+ b+ (1)]/(4n)
�1 = [ab+ a� b� (1)]/(4n) = A/2
�2 = [ab+ b� a� (1)]/(4n) = B/2
�3 = [ab+ (1)� a� b]/(4n) = AB/2
• Factorial e↵ects are twice least squares estimates (under the zero-sum constraints)
29
The ANOVA table for the chemical process experiment is
Df Sum Sq Mean Sq F value Pr(>F)
A 1 208.333 208.333 53.1915 8.444e-05 ***
B 1 75.000 75.000 19.1489 0.002362 **
A:B 1 8.333 8.333 2.1277 0.182776
Residuals 8 31.333 3.917
The regression summary is
Estimate Std. Error t value Pr(>|t|)
(Intercept) 27.5000 0.5713 48.135 3.84e-11 ***
A 4.1667 0.5713 7.293 8.44e-05 ***
B -2.5000 0.5713 -4.376 0.00236 **
A:B 0.8333 0.5713 1.459 0.18278
Residual standard error: 1.979 on 8 degrees of freedom
Multiple R-Squared: 0.903,Adjusted R-squared: 0.8666
F-statistic: 24.82 on 3 and 8 DF, p-value: 0.0002093
• Both main e↵ects are very significant and the interaction e↵ect is notsignificant.
• Notice the connections between two outputs?
We shall drop the interaction term. The final model is
Estimate Std. Error t value Pr(>|t|)
(Intercept) 27.500 0.606 45.377 6.13e-12 ***
A 4.167 0.606 6.875 7.27e-05 ***
B -2.500 0.606 -4.125 0.00258 **
Residual standard error: 2.099 on 9 degrees of freedom
Multiple R-Squared: 0.8772,Adjusted R-squared: 0.8499
F-statistic: 32.14 on 2 and 9 DF, p-value: 7.971e-05
• Note that the estimates are NOT changed.
The final model in terms of coded values is
y = 27.5 + 4.167xA
� 2.5xB
The final model in terms of actual factor values is
y = 27.5 + 4.167(Conc� 20)/5� 2.5(Catalyst� 1.5)/0.5
30
Residual analysis for the final model: Are the assumptions reasonable?
Interaction and residual plots for the chemical process experiment
2022
2426
2830
32
AB Interaction
A
mea
n of
y
−1 1
B
−11
2022
2426
2830
32
BA Interaction
B
mea
n of
y
−1 1
A
−11
22 24 26 28 30 32 34
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12
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−1.5
−0.5
0.5
1.0
1.5
Theoretical Quantiles
Stan
dard
ized
resi
dual
s
Normal Q−Q plot
7
1
5
31
6.5 A Single Replicate of the 2k Design
• A single replicate of the 2k design is an unreplicated 2k factorial design.
• These designs are very widely used in practice for run size economy.
• Risks: modeling noise?
• Lack of replication causes potential problems in statistical testing
– Replication admits an estimate of “pure error”
– With no replication, fitting the full model results in zero degrees offreedom for error
• Potential solutions to this problem
– Pooling high-order interactions to estimate error
– Normal or half-normal quantile plotting of e↵ects
– Other methods; see text, pp. 234
The Filtration Rate Experiment
• A 24 factorial was used to investigate the e↵ects of four factors on thefiltration rate of a chemical product.
• The factors are A = temperature, B = pressure, C = concentration offormaldehyde, D= stirring rate
• The 16 runs are made in random order.
• The goal is to maximize the filtration rate.
32
Factor FiltrationRun A B C D Rate (y)
1 �1 �1 �1 �1 452 1 �1 �1 �1 713 �1 1 �1 �1 484 1 1 �1 �1 655 �1 �1 1 �1 686 1 �1 1 �1 607 �1 1 1 �1 808 1 1 1 �1 659 �1 �1 �1 1 4310 1 �1 �1 1 10011 �1 1 �1 1 4512 1 1 �1 1 10413 �1 �1 1 1 7514 1 �1 1 1 8615 �1 1 1 1 7016 1 1 1 1 96
The treatment contrast table (model matrix for the full model) is
Run I A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD1 1 �1 �1 �1 �1 1 1 1 1 1 1 �1 �1 �1 �1 12 1 1 �1 �1 �1 �1 �1 �1 1 1 1 1 1 1 �1 �13 1 �1 1 �1 �1 �1 1 1 �1 �1 1 1 1 �1 1 �14 1 1 1 �1 �1 1 �1 �1 �1 �1 1 �1 �1 1 1 15 1 �1 �1 1 �1 1 �1 1 �1 1 �1 1 �1 1 1 �16 1 1 �1 1 �1 �1 1 �1 �1 1 �1 �1 1 �1 1 17 1 �1 1 1 �1 �1 �1 1 1 �1 �1 �1 1 1 �1 18 1 1 1 1 �1 1 1 �1 1 �1 �1 1 �1 �1 �1 �19 1 �1 �1 �1 1 1 1 �1 1 �1 �1 �1 1 1 1 �110 1 1 �1 �1 1 �1 �1 1 1 �1 �1 1 �1 �1 1 111 1 �1 1 �1 1 �1 1 �1 �1 1 �1 1 �1 1 �1 112 1 1 1 �1 1 1 �1 1 �1 1 �1 �1 1 �1 �1 �113 1 �1 �1 1 1 1 �1 �1 �1 �1 1 1 1 �1 �1 114 1 1 �1 1 1 �1 1 1 �1 �1 1 �1 �1 1 �1 �115 1 �1 1 1 1 �1 �1 �1 1 1 1 �1 �1 �1 1 �116 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Factorial e↵ect✓ = y(✓ = +1)� y(✓ = �1)
33
Estimated factorial e↵ects are
A B C D AB AC AD BC
21.625 3.125 9.875 14.625 0.125 -18.125 16.625 2.375
BD CD ABC ABD ACD BCD ABCD
-0.375 -1.125 1.875 4.125 -1.625 -2.625 1.375
Q: Which factorial e↵ects are significant?
Normal Quantile Plot of Factorial E↵ects:
• Plot ordered factorial e↵ects against standard normal quantiles. Let✓(1) · · · ✓(I) be ordered e↵ects.
– plot ✓(i) versus ��1([i� 0.5]/I) for i = 1, . . . , I
• Large e↵ects far from the line are important.
Half-Normal Quantile Plot of Factorial E↵ects:
• Plot ordered absolute factorial e↵ects against standard half normal quan-tiles
– plot |✓|(i) versus ��1(0.5 + 0.5[i� 0.5]/I) for i = 1, . . . , I.
• Large e↵ects above the line are important.
• Half-normal quantile plot is preferred.
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00
1020
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normal quantiles
effe
cts C
DA:D
A:C
A
●●●●●
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●
●
●
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●
0.0 1.0 2.0
05
1015
20
Half−Normal Quantile Plot
half−normal quantiles
abso
lute
effe
cts
C
DA:DA:C
A
34
• A, AC, AD, D, C are important.
• Factor B does not appear in any significant e↵ects.
Design Projection
• Projection onto factors A, C and D is a replicated 23 factorial design(hidden replication).
• By projecting, we have an estimate of error.
The regression summary for the replicated 23 design:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 70.0625 1.1842 59.164 7.40e-12 ***
A 10.8125 1.1842 9.131 1.67e-05 ***
C 4.9375 1.1842 4.169 0.003124 **
D 7.3125 1.1842 6.175 0.000267 ***
A:C -9.0625 1.1842 -7.653 6.00e-05 ***
A:D 8.3125 1.1842 7.019 0.000110 ***
C:D -0.5625 1.1842 -0.475 0.647483
A:C:D -0.8125 1.1842 -0.686 0.512032
Residual standard error: 4.737 on 8 degrees of freedom
Multiple R-Squared: 0.9687,Adjusted R-squared: 0.9413
F-statistic: 35.35 on 7 and 8 DF, p-value: 2.119e-05
The final model is
y = 70.0625+ 10.8125xA
+4.9375xC
+7.3125xD
� 9.0625xA
xC
+8.3125xA
xD
We shall perform residual analysis for the final model. Are the modeland assumptions reasonable?
How to choose optimal factor settings to maximize filtration rate?
• Use the final model : A = , C = , D =
• Use interaction plots: A = , C = , D =
35
Interaction and residual plots for the filtration rate experiment
4050
6070
8090
AC Interaction
A
mea
n of
y
−1 1
C
−11
4050
6070
8090
AD Interaction
A
mea
n of
y
−1 1
D
−11
50 60 70 80 90 100
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02
46
Fitted values
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dard
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resi
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s
Normal Q−Q plot
14 5
7
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Main e↵ects and Interaction plots for the filtration rate experiment
4050
6070
8090
100
Main Effect A
A
mea
n of
y
−1 1
4050
6070
8090
100
Main Effect C
C
mea
n of
y
−1 1
4050
6070
8090
100
Main Effect D
D
mea
n of
y
−1 1
4050
6070
8090
100
Interaction AC
A
mea
n of
y
−1 1
C
−11
4050
6070
8090
100
Interaction AD
A
mea
n of
y
−1 1
D
−11
4050
6070
8090
100
Interaction CD
C
mea
n of
y
−1 1
D
−11
5060
7080
90
Interaction ACD
A
mea
n of
y
−1 1
CD
−1−1−111−111
5060
7080
90
Interaction ACD
D
mea
n of
y
−1 1
AC
−1−1−111−111
5060
7080
90
Interaction ACD
C
mea
n of
y
−1 1
AD
−1−1−111−111
37
Chapter 7 Blocking and Confounding in the 2k Factorial
7.2 Blocking a Replicated 2k Factorial Design
The chemical process experiment (c.f. §6.2) in 3 blocks
• Suppose only 4 experimental trials can be made from a single batch ofraw material.
• Three batches of raw material is required to run all 3 replicates.
• Each raw material is a block.
• Runs within the block are randomized.
• This is a randomized complete block design.
BlockA B 1 2 3 Total� � 28 25 27 80+ � 36 32 32 100� + 18 19 23 60+ + 31 30 29 90Total 113 106 111 330
Linear model is
y = �0 + ↵j
+ �1xA + �2xB + �3xAxB + ✏,
where ↵j
is the jth block e↵ect, xA
= ±1, and xB
= ±1.The ANOVA table is
Df Sum Sq Mean Sq F value Pr(>F)
block 2 6.500 3.250 0.7852 0.4978348
A 1 208.333 208.333 50.3356 0.0003937 ***
B 1 75.000 75.000 18.1208 0.0053397 **
A:B 1 8.333 8.333 2.0134 0.2057101
Residuals 6 24.833 4.139
• Note that factorial estimates are NOT changed.
• The block e↵ect is not significant.
38
7.3-7.4 Confounding in the 2k Factorial Design
The Filtration Rate Experiment (c.f. §6.5) in 2 Blocks
• Suppose only 8 treatment combinations can be run from a single batchof raw material.
• Two batches of raw materials are needed for 16 runs.
• The batches are blocks.
• This is an incomplete block design.
Factor FiltrationRun A B C D Block Rate (y)
1 �1 �1 �1 �1 1 252 1 �1 �1 �1 2 713 �1 1 �1 �1 2 484 1 1 �1 �1 1 455 �1 �1 1 �1 2 686 1 �1 1 �1 1 407 �1 1 1 �1 1 608 1 1 1 �1 2 659 �1 �1 �1 1 2 4310 1 �1 �1 1 1 8011 �1 1 �1 1 1 2512 1 1 �1 1 2 10413 �1 �1 1 1 1 5514 1 �1 1 1 2 8615 �1 1 1 1 2 7016 1 1 1 1 1 76
• The original responses are
45, 71, 48, 65, 68, 60, 80, 65, 43, 100, 45, 104, 75, 86, 70, 96
• All responses in block 1 is 20 units lower than the responses in theoriginal data.
Block e↵ectBlock e↵ect = y(block 1)� y(block 2)
• Cannot estimate all 15 factorial e↵ects and one block e↵ect.
• One factorial e↵ect is fully confounded with the block e↵ect.
39
The treatment contrast table (model matrix) with block e↵ect is
Run I A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD Block1 1 �1 �1 �1 �1 1 1 1 1 1 1 �1 �1 �1 �1 1 12 1 1 �1 �1 �1 �1 �1 �1 1 1 1 1 1 1 �1 �1 �13 1 �1 1 �1 �1 �1 1 1 �1 �1 1 1 1 �1 1 �1 �14 1 1 1 �1 �1 1 �1 �1 �1 �1 1 �1 �1 1 1 1 15 1 �1 �1 1 �1 1 �1 1 �1 1 �1 1 �1 1 1 �1 �16 1 1 �1 1 �1 �1 1 �1 �1 1 �1 �1 1 �1 1 1 17 1 �1 1 1 �1 �1 �1 1 1 �1 �1 �1 1 1 �1 1 18 1 1 1 1 �1 1 1 �1 1 �1 �1 1 �1 �1 �1 �1 �19 1 �1 �1 �1 1 1 1 �1 1 �1 �1 �1 1 1 1 �1 �110 1 1 �1 �1 1 �1 �1 1 1 �1 �1 1 �1 �1 1 1 111 1 �1 1 �1 1 �1 1 �1 �1 1 �1 1 �1 1 �1 1 112 1 1 1 �1 1 1 �1 1 �1 1 �1 �1 1 �1 �1 �1 �113 1 �1 �1 1 1 1 �1 �1 �1 �1 1 1 1 �1 �1 1 114 1 1 �1 1 1 �1 1 1 �1 �1 1 �1 �1 1 �1 �1 �115 1 �1 1 1 1 �1 �1 �1 1 1 1 �1 �1 �1 1 �1 �116 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
• Block e↵ect and ABCD are fully confounded.
Block e↵ect = ABCD
• The estimate of ABCD is lost.
• Is this a good blocking scheme? Why?
Estimated e↵ects are
A B C D AB AC AD BC
21.625 3.125 9.875 14.625 0.125 -18.125 16.625 2.375
BD CD ABC ABD ACD BCD Block+ABCD
-0.375 -1.125 1.875 4.125 -1.625 -2.625 -18.625
• Estimated block e↵ect (+ABCD) is 1.375� 20 = �18.625.
Design question: To arrange a 2k design in 2p blocks of size 2k�p, whiche↵ects should be confounded with block e↵ects? (Stats c225)
40
Chapter 8 Two-Level Fractional Factorial Designs
Topics: Alias, resolution, minimum aberration, supersaturated design
8.1 Introduction
• Motivation for fractional factorials is obvious; as the number of factorsbecomes large, the size of the designs grows very quickly
• Emphasis is on factor screening: to e�ciently identify the factors withlarge e↵ects
• There may bemany variables (often because we don’t know much aboutthe system)
• Almost always run as unreplicated factorials, but often with centerpoints
Why do fractional factorial designs work?
• The sparsity of e↵ects principle
– There may be lots of factors, but few are important
– System is dominated by main e↵ects, low-order interactions
• The projection property
– Every fractional factorial contains full factorials in fewer factors
• Sequential experimentation
– Can add runs to a fractional factorial to resolve di�culties (or am-biguities) in interpretation
41
8.2 The One-Half Fraction of the 2k Design
Notation: because the design has 2k/2 runs, it is referred to as a 2k�1 design
• Consider a really simple case, the 23�1 design defined by I = ABC
• For the principal fraction, notice that the contrast for estimating themain e↵ect A is exactly the same as the contrast used for estimating theBC interaction.
• This phenomena is called aliasing and it occurs in all fractional designs
• Aliases can be found directly from the columns in the table of + and �
signs
42
Aliasing in the 23�1 design
• A main e↵ect is aliased with a two-factor interaction (or me = 2fi)
A = BC,B = AC,C = AB
• Aliases can be found from the defining relation I = ABC:
A = AI = A(ABC) = A2BC = BC
B = BI = B(ABC) = AB2C = AC
C = CI = C(ABC) = ABC2 = AB
• Textbook notation for aliased e↵ects:
[A] �! A+BC, [B] �! B + AC, [C] �! C + AB
The alternate fraction of the 23
• I = �ABC is the defining relation
• Implies slightly di↵erent aliases: A = �BC,B = �AC, and C = �AB
• Both designs belong to the same family, defined by
I = ±ABC
• Suppose that after running the principal fraction, the alternate fractionwas also run
• The two groups of runs can be combined to form a full factorial
– an example of sequential experimentation
Design resolution: the length of the shortest word
• Resolution III Designs: example 23�1III
I = ABC
– me = 2fi
• Resolution IV Designs: example 24�1IV
I = ABCD
– me=3fi and 2fi = 2fi
43
• Resolution V Designs: example 25�1V
I = ABCDE
– me=4fi and 2fi = 3fi
• Resolution R Designs: no p-factor e↵ect is aliased with another e↵ectcontaining < R� p factors
Construction of a One-half Fraction
• write a full 2k�1 design as the basic design
• add the kth column according to the design generator I = ±ABC · · ·K
Projection of Fractional Factorials
• Every fractional factorial contains full factorials in fewer factors
• A one-half fraction 2k�1 will project into a full factorial in any k � 1 ofthe original factors (if resolution R = k)
44
Example 8-1 A half fraction of the filtration rate experiment
Recall: The estimated factorial e↵ects from the full 24 design are
A B C D AB AC AD BC
21.625 3.125 9.875 14.625 0.125 -18.125 16.625 2.375
BD CD ABC ABD ACD BCD ABCD
-0.375 -1.125 1.875 4.125 -1.625 -2.625 1.375
• [A] = A+BCD = 21.625� 2.625 = 19
• Interpretation of results often relies on making some assumptions
– Main e↵ects are more likely to be important than two-factor inter-actions
• Ockhams razor: the simplest interpretation is usually the correct one
• Confirmation experiments can be important
45
• Adding the alternate fraction can resolve the ambiguity due to aliasing
Possible strategies for follow-up experimentation following a frac-tional factorial design
46
8.3 The One-Quarter Fraction of the 2k Design
The injection molding process experiment:
• Parts manufactured in the process are showing excessive shrinkage.
• investigation of six factors: (A) mold temperature, (B) screw speed, (C)holding time, (D) cycle time, (E) gate size, (F) hold pressure
• objective: learn how each factor a↵ects shrinkage and something abouthow the factors interact.
• A 26�2 design with complete defining relation
I = ABCE = BCDF = ADEF
Basic Design ObservedRun A B C D E = ABC F = BCD shrinkage (⇥10)1 � � � � � � 62 + � � � + � 103 � + � � + + 324 + + � � � + 605 � � + � + + 46 + � + � � + 157 � + + � � � 268 + + + � + � 609 � � � + � + 810 + � � + + + 1211 � + � + + � 3412 + + � + � � 6013 � � + + + � 1614 + � + + � � 515 � + + + � + 3716 + + + + + + 52
• generators: E = ABC and F = BCD
• Defining words: ABCE,BCDF,ADEF
• Resolution = shortest word length
47
• This is a resolution IV design, so me=3fi, 2fi=2fi
• Each e↵ect is aliased with other three e↵ects.
• Can estimate 15 factorial e↵ects, one from each alias set
A,B,C,D,E, F,AB = CE,AC = BE,AD = EF,
AE = BC = DF,AF = DE,BD = CF,BF = CD, [ABD], [ABF ]
Estimates of aliased e↵ects
A B C D E F A:B A:C
13.875 35.625 -0.875 1.375 0.375 0.375 11.875 -1.625
A:D A:E A:F B:D B:F A:B:D A:B:F
-5.375 -1.875 0.625 -0.125 -0.125 0.125 -4.875
• [A] �! A+BCE +DEF + ABCDF = 13.875
• [AB] �! AB + CE + ACDF +BDEF = 11.875
48
●●●●●●●●●●
● ●
●●
●
0.0 1.0 2.0
010
2030
Half−Normal Plot
half−normal quantiles
abso
lute
effe
cts
A:BA
B
1030
50
AB Interaction plot
A
mea
n of
shr
inka
ge−1 1
B1−1
• [A], [B] and [AB] are important
• Since neither C nor E is important, it is reasonable to assume CE isnegligible.
• Conclusion: Two factors (A, B) and the AB interaction are important
A final model with A, B and AB, which has R2 = .96.
y = 27.312 + 6.937xA
+ 17.812xB
+ 5.937xA
xB
Regression summary
Estimate Std. Error t value Pr(>|t|)
(Intercept) 27.312 1.138 23.996 1.65e-11 ***
A 6.937 1.138 6.095 5.38e-05 ***
B 17.812 1.138 15.649 2.39e-09 ***
A:B 5.937 1.138 5.216 0.000216 ***
Residual standard error: 4.553 on 12 degrees of freedom
Multiple R-Squared: 0.9626,Adjusted R-squared: 0.9533
• To minimize the response, choose A = and B =
• From the interaction plot, the key is to choose B =
49
Residual analysis
10 20 30 40 50
−50
5
Fitted values
Res
idua
ls
●
● ●
●
●
●
●
●
●
● ● ●
●
●
●
●
Residuals vs Fitted13
7 16
●
●●
●
●
●
●
●
●
●●●
●
●
●
●
−1.0 0.0 0.5 1.0
−6−2
26
Residuals vs holding time (C)
C
resi
d(g)
• there are some dispersion e↵ects
Uses of the alternate fractions
I = ±ABCE, I = ±BCDF
Projection
• Any 4-factor subset of the original six variables that is not a word in thecomplete defining relation will result in a full factorial design
– Consider ABCD (full factorial)
– Consider ABCE (replicated half fraction)
– Consider ABCF (full factorial)
8.4 The General 2k�p Fractional Factorial Design
• 2k�1 = one-half fraction, 2k�2 = one-quarter fraction, 2k�3 = one-eighthfraction, . . ., 2k�p = 1/2p fraction
• Add p columns to the basic design (a full 2k�p design); select p indepen-dent generators
• Important to select generators so as to maximize resolution
50
Example: Two 26�2 designs (Which design is better?)
• Design 1: E = ABC,F = BCD
• Design 2: E = ABCD,F = BCD
Resolution may not be su�cient.
Minimum aberration designs
• minimize the number of words in the defining relation that are of mini-mum length
• see Table 8-14 for k 15 factors and up to n 128 runs
• see Xu (2009, Technometrics) for k up to 40–160 factors and n = 128–4096 runs.
Other topics
• Projection: a design of resolution R contains full factorials in any R�1of the factors
• Blocking
– How to optimally arrange 2k�p designs into 2q blocks?
– see Xu (2006, Annals of Statistics), Xu and Lau (2006, JSPI), Xuand Mee (2010, JSPI) for minimum aberration blocked 2k�p designs
Research problems:
• How to construct minimum aberration designs with � 256 runs?
• How to construct minimum aberration blocked designs with � 256 runs?
51
8.5 Resolution III Designs
• Main e↵ects are aliased with two-factor interactions (me=2fi)
• Often used for screening (5–7 variables in 8 runs, 9–15 variables in 16runs, for example)
• A saturated design of N runs has k = N � 1 variables
A saturated 27�4III
design
• alias structure (ignoring 3fi or higher):
[A] �! A+BD + CE + FG, [B] �! B + AD + CF + EG,
[C] �! C + AE +BF +DG, [D] �! D + AB + CG+ EF,
[E] �! E + AC +BG+DF, [F ] �! F +BC + AG+DE,
[G] �! G+ CD +BE + AF
• can use the fold-over techniques to dealiases all main e↵ects from theirtwo-factor interaction alias chains
Plackett-Burman designs
• These are a di↵erent class of resolution III design
• The number of runs, N , need only be a multiple of four
N = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40,
52
• The designs where N = 12, 20, 24, etc. are called nonregular designs
• See text for comments on construction of Plackett-Burman designs
The Weld Repaired Cast Fatigue Experiment (Wu and Hamada, 2009)
• used a 12-run design to study the e↵ects of seven factors on the fatiguelife of weld repaired castings.
• The response is the logged lifetime of the casting
• The goal of the experiment was to identify the factors that a↵ect thecasting lifetime.
LevelFactor � +
A. initial structure as received � treatB. bead size small largeC. pressure treat none HIPD. heat treat anneal solution treat/ageE. cooling rate slow rapidF. polish chemical mechanicalG. final treat none peen
Factor LoggedRun A B C D E F G 8 9 10 11 Lifetime1 + + � + + + � � � + � 6.0582 + � + + + � � � + � + 4.7333 � + + + � � � + � + + 4.6254 + + + � � � + � + + � 5.8995 + + � � � + � + + � + 7.0006 + � � � + � + + � + + 5.7527 � � � + � + + � + + + 5.6828 � � + � + + � + + + � 6.6079 � + � + + � + + + � � 5.81810 + � + + � + + + � � � 5.91711 � + + � + + + � � � + 5.86312 � � � � � � � � � � � 4.809
53
> g=lm(y ~ A+B+C+D+E+F+G, dat); summary(g)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.73025 0.17114 33.482 4.75e-06 ***
A 0.16292 0.17114 0.952 0.3950
B 0.14692 0.17114 0.858 0.4390
C -0.12292 0.17114 -0.718 0.5123
D -0.25808 0.17114 -1.508 0.2060
E 0.07492 0.17114 0.438 0.6842
F 0.45758 0.17114 2.674 0.0556 .
G 0.09158 0.17114 0.535 0.6209
• Only F is significant at 10%.
• A model with F only has R2 = 0.45 (does not fit well).
• A model with F and D has R2 = 0.59.
Aliasing
• The alias structure is complex in this PB design
• Every main e↵ect is partially aliased with every 2fi not involving itself
[A] �! A+1
3(�BC � BD � BE +BF + · · ·� FG)
• Partial aliasing can greatly complicate analysis and interpretation
• More elaborate analysis shows that F and FG are significant
• A model with F and FG has R2 = 0.89.
y = 5.7303 + 0.4576F � 0.4588FG
Projections onto 3 or 4 factors
• projection onto any 3 factors contains a full 23 deign (projectivity 3)
• regular resolution III fractions has projectivity 2.
54
8.7 Supersaturated Designs
• A design is saturated if the # of variables k = N � 1, where N is the# of runs
• A design is supersaturated if the number of variables k > N � 1
• Supersaturated designs are commonly used in screening experiments
– goal is to identify sparse and dominant active e↵ects with low cost
• potential applications in many areas such as industrial, medical sciences,engineering
55
Why does it work?
• E↵ect sparsity: the number of relatively important e↵ects in a factorialexperiment is small.
Construction
• Basic idea: Choose a design with small correlations among the columns
Analysis
• Very challenging due to the existence of complex aliasing among e↵ects
• Can NOT fit a linear regression model directly
• forward stepwise variable selection
– easy but has large type I and type II errors
• Bayesian variable selection
– flexible but not easy to use
• Dantzig selector
– a good tool and easy to use
Dantzig selector proposed by Candes and Tao (2007, Annals of Statistics)
• chooses the best subset of variables or active factors by solving a verysimple convex program, which can be recast as a convenient linear pro-gram.
• successfully used in biomedical imaging, analog-to-digital data conver-sion and sensor networks, where the goals are to recover some sparsesignals from some massive data.
Consider a general linear regression model
y = X� + ✏,
where X is an n⇥p model matrix. The model is supersaturated when p > n.
56
The Dantzig selector is the solution to the l1-regularization problem
min�
k�kl1 subject to kXT (y �X�)k
l1 � (20)
where for a vector a, kakl1 =
P|a
i
| and kakl1 = max |a
i
|, and � is a tuningparameter.
The Dantzig selector can be recast as a linear program
minX
i
ui
subject to � u � u and � �1 XT (y �X�) �1, (21)
where the optimization variables are u, � 2 Rp and 1 is a p-dimensional vectorof ones. This is equivalent to the standard linear program in matrix form
min cTx subject to Ax � b and x � 0, (22)
where
c =
✓10
◆, A =
0
@XTX �XTX
�XTX XTX2I
p
�Ip
1
A , b =
0
@�XTy � �1XTy � �1
0
1
A , x =
✓u
u+ �
◆.
When X is an orthonormal matrix, the Dantzig selector has a simple closeform
�i
= max(|bi
|� �, 0)sign(bi
),
where bi
is the least squares estimate of �i
. Note that the Dantzig selector isshifted toward the origin, a soft-thresholding phenomena. For an arbitraryX, the method continues to exhibit a soft-thresholding type of behavior andas a result, may underestimate the true value of the nonzero parameters.
A two-stage procedure for bias correction
1. Estimate I = {i : �i
6= 0} with I = {i : |�i
| > �} for some small � � 0with � as the solution to the linear program (21).
2. Construct the least squares estimate �I
= (XT
I
XI
)�1XT
I
y and set theother coordinates to 0.
In other words, we rely on the Dantzig selector to estimate the model I andthen construct a new estimator by regressing y onto the model I.
57
Example: The cast fatigue experiment. Consider two models:
• (a) a main e↵ects model
– X is a 12⇥ 7 matrix and orthogonal
• (b) a main e↵ects plus 2-factor interactions model
– X is a 12⇥ (7 + 21) matrix and supersaturated
• center the response y and obtain the Dantzig selectors � by solving thelinear program (21) with � varying from 0 to 6.
• Make a profile plot by plotting � against �
– (left) main e↵ects, (right) main e↵ects + 2-factor interactions
0 1 2 3 4 5 6
−0.4
0.0
0.2
0.4
delta
beta
F
D
0 1 2 3 4 5 6
−0.4
0.0
0.2
0.4
delta
beta
F
FGAE
• For the main e↵ects model, F is the most significant andD is moderatelysignificant.
• When we entertain the two-factor interactions, F and FG are very sig-nificant and AE is moderately significant.
Reference: Phoa, F. K. H., Pan, Y.-H. and Xu, H. (2009). Analysis ofSupersaturated Designs via the Dantzig Selector. Journal of Statistical Plan-ning and Inference, 139, 2362-2372.
58
