Post on 21-Dec-2015
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Chapter 5 Introduction to Factorial Designs
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5.1 Basic Definitions and Principles
• Study the effects of two or more factors.• Factorial designs• Crossed: factors are arranged in a factorial design• Main effect: the change in response produced by a
change in the level of the factor
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Definition of a factor effect: The change in the mean response when the factor is changed from low to high
40 52 20 3021
2 230 52 20 40
112 2
52 20 30 401
2 2
A A
B B
A y y
B y y
AB
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50 12 20 401
2 240 12 20 50
92 2
12 20 40 5029
2 2
A A
B B
A y y
B y y
AB
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Regression Model & The Associated Response Surface
0 1 1 2 2
12 1 2
1 2
1 2
1 2
The least squares fit is
ˆ 35.5 10.5 5.5
0.5
35.5 10.5 5.5
y x x
x x
y x x
x x
x x
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The Effect of Interaction on the Response Surface
Suppose that we add an interaction term to the model:
1 2
1 2
ˆ 35.5 10.5 5.5
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y x x
x x
Interaction is actually a form of curvature
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• When an interaction is large, the corresponding main effects have little practical meaning.
• A significant interaction will often mask the significance of main effects.
5.2 The Advantage of Factorials
• One-factor-at-a-time desgin • Compute the main effects of factors
A: A+B- - A-B-
B: A-B- - A-B+
Total number of experiments: 6• Interaction effects
A+B-, A-B+ > A-B- => A+B+ is
better???8
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5.3 The Two-Factor Factorial Design
5.3.1 An Example• a levels for factor A, b levels for factor B and n
replicates• Design a battery: the plate materials (3 levels) v.s.
temperatures (3 levels), and n = 4: 32 factorial design• Two questions:
– What effects do material type and temperature have on the life of the battery?
– Is there a choice of material that would give uniformly long life regardless of temperature?
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• The data for the Battery Design:
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• Completely randomized design: a levels of factor A, b levels of factor B, n replicates
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• Statistical (effects) model:
is an overall mean, i is the effect of the ith level
of the row factor A, j is the effect of the jth
column of column factor B and ( )ij is the
interaction between i and j .
• Testing hypotheses:
1,2,...,
( ) 1,2,...,
1, 2,...,ijk i j ij ijk
i a
y j b
k n
0)( oneleast at : v.s., 0)(:
0 oneleast at : v.s.0:
0 oneleast at : v.s.0:
10
110
110
ijij
jb
ia
HjiH
HH
HH
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• 5.3.2 Statistical Analysis of the Fixed Effects Model
a
i
b
j
n
kijk
ijij
n
kijkij
ja
ij
n
kijkj
ib
ji
n
kijki
abn
yyyy
n
yyyy
an
yyyy
bn
yyyy
1 1
......
1...
..
1.
..
1..
1..
..
1..
1..
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2 2 2... .. ... . . ...
1 1 1 1 1
2 2. .. . . ... .
1 1 1 1 1
( ) ( ) ( )
( ) ( )
a b n a b
ijk i ji j k i j
a b a b n
ij i j ijk iji j i j k
y y bn y y an y y
n y y y y y y
breakdown:
1 1 1 ( 1)( 1) ( 1)
T A B AB ESS SS SS SS SS
df
abn a b a b ab n
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• Mean squares
2
1 1
2
2
1
2
2
1
2
2
))1(
()(
)1)(1(
)(
))1)(1(
()(
1))1/(()(
1))1/(()(
nab
SSEMSE
ba
n
ba
SSEMSE
b
an
bSSEMSE
a
bnaSSEMSE
EE
a
i
b
jij
ABAB
b
jj
BB
a
ii
AA
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• The ANOVA table:
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Response: Life ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]
Sum of Mean FSource Squares DF Square Value Prob > FModel 59416.22 8 7427.03 11.00 < 0.0001A 10683.72 2 5341.86 7.91 0.0020B 39118.72 2 19559.36 28.97 < 0.0001AB 9613.78 4 2403.44 3.56 0.0186Pure E 18230.75 27 675.21C Total 77646.97 35
Std. Dev. 25.98 R-Squared 0.7652Mean 105.53 Adj R-Squared 0.6956C.V. 24.62 Pred R-Squared 0.5826
PRESS 32410.22 Adeq Precision 8.178
Example 5.1
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DESIGN-EXPERT Plot
Life
X = B: TemperatureY = A: Material
A1 A1A2 A2A3 A3
A: MaterialInteraction Graph
Life
B: Temperature
15 70 125
20
62
104
146
188
2
2
22
2
2
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• Multiple Comparisons:– Use the methods in Chapter 3.– Since the interaction is significant, fix the
factor B at a specific level and apply Turkey’s test to the means of factor A at this level.
– See Page 174– Compare all ab cells means to determine which
one differ significantly
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5.3.3 Model Adequacy Checking• Residual analysis: ijijkijkijkijk yyyye ˆ
DESIGN-EXPERT PlotLife
Residual
No
rma
l % p
rob
ab
ility
Normal plot of residuals
-60.75 -34.25 -7.75 18.75 45.25
1
5
10
20
30
50
70
80
90
95
99
DESIGN-EXPERT PlotLife
Predicted
Re
sid
ua
ls
Residuals vs. Predicted
-60.75
-34.25
-7.75
18.75
45.25
49.50 76.06 102.62 129.19 155.75
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DESIGN-EXPERT PlotLife
Run Number
Re
sid
ua
ls
Residuals vs. Run
-60.75
-34.25
-7.75
18.75
45.25
1 6 11 16 21 26 31 36
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DESIGN-EXPERT PlotLife
Material
Re
sid
ua
lsResiduals vs. Material
-60.75
-34.25
-7.75
18.75
45.25
1 2 3
DESIGN-EXPERT PlotLife
Temperature
Re
sid
ua
ls
Residuals vs. Temperature
-60.75
-34.25
-7.75
18.75
45.25
1 2 3
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5.3.4 Estimating the Model Parameters• The model is
• The normal equations:
• Constraints:
ijkijjiijky )(
ijijjiij
j
a
iijj
a
iij
i
b
jij
b
jjii
a
i
b
jij
b
jj
a
ii
ynnnn
ynannan
ynnbnbn
ynanbnabn
)(:)(
)(:
)(:
)(:
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11
1 111
0,0,01111
b
jij
a
iij
b
jj
a
ii
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• Estimations:
• The fitted value:
• Choice of sample size: Use OC curves to choose the proper sample size.
yyyy
yy
yy
y
jiijij
jj
ii
ˆ
ˆ
ˆ
ijijjiijk yy ˆˆˆˆ
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• Consider a two-factor model without interaction:– Table 5.8– The fitted values: yyyy jiijkˆ
• One observation per cell: – The error variance is not estimable because the
two-factor interaction and the error can not be separated.
– Assume no interaction. (Table 5.9)
– Tukey (1949): assume ()ij = rij (Page 183)
– Example 5.2
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5.4 The General Factorial Design
• More than two factors: a levels of factor A, b levels of factor B, c levels of factor C, …, and n replicates.
• Total abc … n observations.• For a fixed effects model, test statistics for each
main effect and interaction may be constructed by dividing the corresponding mean square for effect or interaction by the mean square error.
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• Degree of freedom:– Main effect: # of levels – 1 – Interaction: the product of the # of degrees of
freedom associated with the individual components of the interaction.
• The three factor analysis of variance model:–
– The ANOVA table (see Table 5.12)– Computing formulas for the sums of squares
(see Page 186)– Example 5.3
ijklijkjkik
ijkjiijkly
)()()(
)(
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• Example 5.3: Three factors: the percent carbonation (A), the operating pressure (B); the line speed (C)
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5.5 Fitting Response Curves and Surfaces• An equation relates the response (y) to the factor
(x).• Useful for interpolation.• Linear regression methods• Example 5.4
– Study how temperatures affects the battery life– Hierarchy principle
– Involve both quantitative and qualitative factors
– This can be accounted for in the analysis to produce regression models for the quantitative factors at each level (or combination of levels) of the qualitative factors
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A = Material type
B = Linear effect of Temperature
B2 = Quadratic effect of Temperature
AB = Material type – TempLinear
AB2 = Material type - TempQuad
B3 = Cubic effect of Temperature (Aliased)
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5.6 Blocking in a Factorial Design
• A nuisance factor: blocking• A single replicate of a complete factorial
experiment is run within each block.• Model:
– No interaction between blocks and treatments• ANOVA table (Table 5.20)
ijkkijjiijky )(
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• Example 5.6: – Two factors: ground clutter and filter type– Nuisance factor: operator
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• Two randomization restrictions: Latin square design
• An example in Page 200.• Model:
• Tables 5.23 and 5.24
ijklljkkjiijkly )(