Slide set 8
Stat402 (Spring 2016)
Last update: February 29, 2016
Stat 402 (Spring 2016): Slide Set 8
The Two-way Factorial in a Completely Randomized
Design
•
Example 5.1: Measure maximum output voltage of storage batteries.
All combinations of 3 plate materials and 3 temperatures are used in a
completely randomized design with 4 replications per each combination.
Nine treatment combinations allocated to 36 runs completely at random.
The treatment arrangement is called a 3 × 3 factorial.
9 treatments combinations:
3 materials × 3 temperatures
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Stat 402 (Spring 2016): Slide Set 8
•
Analysis The analysis is best considered in two stages. The first part
of the analysis ignores the treatment structure, i.e., that it is a factorial.
Analyze the data as one-way classification with 9 treatments (i.e., 1
factor with 9 levels) and equal sample sizes. Data will look like:
Treatment 1 Treatment 2 Treatment 3 . . .
y111
y121
...
...
y112
y122
...
...
y113
y123
...
...
y114
y124
...
...
•
Treatment 9
y331
y332
y333
y334
Model:
yijk = µij + ijk , i = 1, . . . , 3 (Material); j = 1, . . . , 3 (Temperature)
k = 1, . . . , 4 (Replications)
i.i.d
and ijk ∼ N (0, σ 2)
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Stat 402 (Spring 2016): Slide Set 8
The Two-way Factorial in a CRD (Cont’d)
•
ANOVA
SV
d.f.
Treatment
8
Error
27
Total
35
Now, consider the modeling of the treatment structure. We consider the
non-additive model
µij = µ + τi + βj + (τ β)ij
for the treatment mean µij where
τi = incremental effect due to plate material i
βj = incremental effect due to plate temperature j
(τ β)ij =incremental effect due to combination of material i and temp. j
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Stat 402 (Spring 2016): Slide Set 8
This partitioning partitions the d.f. due to treatment above as follows:
SV
Trt
Material
Temp
Material × Temp
Error
Total
Look
at
the
ANOVA
table
on
d.f.
8
2
2
4
27
35
p.194
of
Montgomery
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Stat 402 (Spring 2016): Slide Set 8
Two-way Factorial in a Completely Randomized Design
Data:
Factor A
1
2
..
a
1
y111, . . . , y11n
y211, . . . , y21n
..
ya11, . . . , ya1n
ȳ.1.
Factor B
2
...
y121, . . . , y12n . . .
...
..
...
ȳ.2.
...
b
y1ba, . . . , y1bn
yab1, . . . , yabn
ȳ.b.
ȳ1
ȳ1..
ȳ2..
..
ȳa..
ȳ...
where
Pb Pn
ȳi.. = ( j=1 k=1 yijk )/bn (Factor A Means)
Pa Pn
ȳ.j. = ( i=1 k=1 yijk )/an (Factor B Means)
Pn
ȳij. = k=1 yijk /n (Cell Means)
P P P
ȳ... = ( i j k yijk )/abn (Grand Mean)
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Stat 402 (Spring 2016): Slide Set 8
Model:
yijk
=
τi =
βj =
(τ β)ij
=
i = 1, . . . , a
µ + τi + βj + (τ β)ij + ijk
j = 1, . . . , b
k = 1, . . . , n
effect of i-th level of A
a=# levels of Factor A
effect of j-th level of B
b=# levels of Factor B
interaction effect of i-th level n=# of replications of each
of A and j-th level of B
treatment combination
Estimates
µ̂
τ̂i
β̂j
(τˆβ)ij
=
=
=
=




ȳ...
ȳi.. − ȳ...
or µ̂ij = ȳij.
ȳ.j. − ȳ...



ȳij. − ȳi.. − ȳ.j. + ȳ...
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Stat 402 (Spring 2016): Slide Set 8
ANOVA Table
SV
Treatment
A
B
AB
Error
Total
d.f.
ab − 1
a−1
b−1
(a − 1)(b − 1)
ab(n − 1)
abn − 1
SS
SST rt
SSA
SSB
SSAB
SSE
SST
MS
M ST rt
M SA
M SB
M SAB
M SE (= s2E )
F
M ST rt /M SE
1
M SA /M SE 2
M SB /M SE 3
M SAB /M SE Hypotheses Tested
F statistic
What is tested?
1
H0 : τ1 = · · · = τa = 0
vs
Ha : at least one τi 6= 0
≡
[H0 : µ̄1. = · · · = µ̄a.
vs
Ha : at least one ineq.]
2
H0 : β1 = · · · = βb = 0
vs
Ha : at least one βj 6= 0
≡
[H0 : µ̄.1 = · · · = µ̄.b
vs
Ha : at least one ineq.]
3
H0 : (τ β)ij = 0 for all i, j
vs
Ha : at least one 6= 0
≡
[H0 : no interaction
vs
Ha : interaction]
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Stat 402 (Spring 2016): Slide Set 8
Hypotheses Tested (Cont’d)
Predicted Values:
ŷijk
= µ̂ + τ̂i + β̂j + (τˆβ)ij
= ȳ... + (ȳi.. − ȳ...) + (ȳ.j. − ȳ...) + (ȳij. − ȳi.. − ȳ.j. + ȳ...)
Residuals
rijk
= yijk − ŷijk
= yijk − (ȳ... + (ȳi.. − ȳ...) + (ȳ.j. − ȳ...) + (ȳij. − ȳi.. − ȳ.j. + ȳ...))
= (yijk − ȳij.)
SSRes:
a X
b X
n
X
i=1 j=1 k=1
rijk =
a X
b X
n
X
(yijk − ȳij.)2
i=1 j=1 k=1
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Stat 402 (Spring 2016): Slide Set 8
Algebraic Decomposition of Total Variation
Total SS = SST =
SST rt = n



Pa
i=1
SSA
SSB

 SS
AB
Pa
i=1
Pb
Pb
j=1 (yij
j=1
Pn
2
(y
−
ȳ
)
ijk
...
k=1
− ȳij.)2 =
Pa
= bn( i=1(ȳi.. − ȳ...)2)
Pb
= an( j=1(ȳ.j. − ȳ...)2)
Pa Pb
= n( i=1 j=1(ȳij. − ȳi.. − ȳ.j. + ȳ...)2)
SSE =
Pa
i=1
Pb
j=1
Pn
k=1 (yijk
− ȳij.)2
SST = SST rt + SSE = SSA + SSB + SSAB + SSE
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Stat 402 (Spring 2016): Slide Set 8
Suggested procedure
•
•
•
•
•
Test the No Interaction Hypothesis first. If no significant interaction
between A and B then:
Go ahead and test for main effects. These tests compare factor means
i.e. means of each factor averaged over the levels of the other factor
If A or/and B are significant, then use methods such as LSD, Tukey,
Prelanned comparisons, Orthogonal Polynominals to compare factor
means.
A 100(1 − α)% C.I. for τ1 − τ2 (or µ̄1. − µ̄2.)
q
2
(ȳ1.. − ȳ2..) ∓ t α2 ,ν · sE bn
A 100(1 − α) C.I. for β1 − β2 (or µ̄.1 − µ̄.2)
q
2
(ȳ.1. − ȳ.2.) ± t α2 ,ν · sE an
where ν = ab(n − 1)
s2E = M SE
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Stat 402 (Spring 2016): Slide Set 8
Interpretation of Results: Battery Life Data
ANOVA table
Interaction Plot
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Stat 402 (Spring 2016): Slide Set 8
Comparison of Means: Battery Life Data
•
•
•
•
Interaction is significant This implies that rather than comparing battery
means or material means, we need to make comparisons among the 9
cell means (observe the interaction plot)
The interaction plot suggests that it makes sense to compare the average
battery life for the material types at each of the temperature.
For example, from the JMP output, we have, that all pairs of material
type means at each of temperatures 15 and 125 are not different from
each other as evidenced by the interaction plot.
From JMP, 95% CI’s for differences in Material means at Temperature
70 are:
µ1 − µ2 : (−100.2, −24.8)∗
µ1 − µ3 : (−126.2, −50.8)∗
µ2 − µ3 : (−63.7, 11.7005)
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