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Stat 402 (Spring 2016): Slide Set 8
Treatment 9
y331
y332
y333
y334
k = 1, . . . , 4 (Replications)
and ijk ∼ N (0, σ 2)
i.i.d
yijk = μij + ijk , i = 1, . . . , 3 (Material); j = 1, . . . , 3 (Temperature)
Model:
Treatment 1 Treatment 2 Treatment 3 . . .
y111
y121
...
...
y112
y122
...
...
y113
y123
...
...
y114
y124
...
...
2
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:
Last update: February 29, 2016
Stat402 (Spring 2016)
Slide set 8
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Stat 402 (Spring 2016): Slide Set 8
3
(τ β)ij =incremental effect due to combination of material i and temp. j
βj = incremental effect due to plate temperature j
τi = incremental effect due to plate material i
for the treatment mean μij where
μij = μ + τi + βj + (τ β)ij
SV
d.f.
Treatment
8
Error
27
Total
35
Now, consider the modeling of the treatment structure. We consider the
non-additive model
ANOVA
The Two-way Factorial in a CRD (Cont’d)
Stat 402 (Spring 2016): Slide Set 8
1
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
The Two-way Factorial in a Completely Randomized
Design
Stat 402 (Spring 2016): Slide Set 8
=
=
Estimates
(τ β)ij
ANOVA
table
on
p.194
d.f.
8
2
2
4
27
35
=
=
=
=
ȳ...
ȳi.. − ȳ...
ȳ.j. − ȳ...
ȳij. − ȳi.. − ȳ.j. + ȳ...
⎪
⎪
⎭
⎫
⎪
⎪
⎬
or μ̂ij = ȳij.
6
[H0 : μ̄1. = · · · = μ̄a.
B Means)
[H0 : μ̄.1 = · · · = μ̄.b
H0 : (τ β)ij = 0 for all i, j
[H0 : no interaction
≡
≡
vs
vs
vs
vs
vs
vs
ȳ1
ȳ1..
ȳ2..
..
ȳa..
ȳ...
F
M ST rt /M SE
1
M SA /M SE 2
M SB /M SE 3
M SAB /M SE Ha : interaction]
Ha : at least one = 0
Ha : at least one ineq.]
Ha : at least one βj = 0
Ha : at least one ineq.]
Ha : at least one τi = 0
MS
M ST rt
M SA
M SB
M SAB
M SE (= s2E )
(Grand Mean)
3
H0 : β1 = · · · = βb = 0
2
≡
What is tested?
H0 : τ1 = · · · = τa = 0
1
F statistic
Hypotheses Tested
SV
Treatment
A
B
AB
Error
Total
SS
SST rt
SSA
SSB
SSAB
SSE
SST
k yijk )/abn
d.f.
ab − 1
a−1
b−1
(a − 1)(b − 1)
ab(n − 1)
abn − 1
j
yab1, . . . , yabn
ȳ.b.
b
y1ba, . . . , y1bn
A Means)
...
...
...
...
...
7
Stat 402 (Spring 2016): Slide Set 8
i
ANOVA Table
ȳ... = (
ȳij. =
ȳ.j. =
b
n
j=1
k=1 yijk )/bn (Factor
a n
( i=1 k=1 yijk )/an (Factor
n
k=1 yijk /n (Cell Means)
ȳi.. = (
Factor B
1
2
y111, . . . , y11n y121, . . . , y12n
y211, . . . , y21n
..
..
ya11, . . . , ya1n
ȳ.1.
ȳ.2.
Stat 402 (Spring 2016): Slide Set 8
where
1
2
..
a
Factor A
Data:
5
Montgomery
Stat 402 (Spring 2016): Slide Set 8
Two-way Factorial in a Completely Randomized Design
4
of
i = 1, . . . , a
j = 1, . . . , b
μ + τi + βj + (τ β)ij + ijk
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
the
μ̂
τ̂i
β̂j
(τˆβ)ij
at
τi =
βj =
yijk
Model:
Look
Material
Temp
Material × Temp
Error
Total
SV
Trt
This partitioning partitions the d.f. due to treatment above as follows:
rijk =
i=1 j=1 k=1
b n
a s2E = M SE
10
A 100(1 − α) C.I. for β1 − β2 (or μ̄.1 − μ̄.2)
2
(ȳ.1. − ȳ.2.) ± t α2 ,ν · sE an
where ν = ab(n − 1)
− ȳij.)2 =
k=1 (yijk
n
− ȳ...)2
i=1
a
j=1
b
k=1 (yijk
n
− ȳij.)2
SST = SST rt + SSE = SSA + SSB + SSAB + SSE
SSE =
Interaction Plot
ANOVA table
Interpretation of Results: Battery Life Data
11
Stat 402 (Spring 2016): Slide Set 8
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j=1 (yij
b
j=1
b
Stat 402 (Spring 2016): Slide Set 8
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.)
2
(ȳ1.. − ȳ2..) ∓ t α2 ,ν · sE bn
•
i=1
a
i=1
a
a
SSA = bn( i=1(ȳi.. − ȳ...)2)
b
SSB = an( j=1(ȳ.j. − ȳ...)2)
a b
⎪
⎩
SSAB = n( i=1 j=1(ȳij. − ȳi.. − ȳ.j. + ȳ...)2)
⎧
⎪
⎨
SST rt = n
Total SS = SST =
9
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Stat 402 (Spring 2016): Slide Set 8
Algebraic Decomposition of Total Variation
8
(yijk − ȳij.)2
Suggested procedure
i=1 j=1 k=1
b n
a = yijk − ŷijk
= yijk − (ȳ... + (ȳi.. − ȳ...) + (ȳ.j. − ȳ...) + (ȳij. − ȳi.. − ȳ.j. + ȳ...))
= (yijk − ȳij.)
SSRes:
rijk
Residuals
= μ̂ + τ̂i + β̂j + (τˆβ)ij
= ȳ... + (ȳi.. − ȳ...) + (ȳ.j. − ȳ...) + (ȳij. − ȳi.. − ȳ.j. + ȳ...)
Predicted Values:
ŷijk
Stat 402 (Spring 2016): Slide Set 8
Hypotheses Tested (Cont’d)
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Stat 402 (Spring 2016): Slide Set 8
μ1 − μ2 : (−100.2, −24.8)∗
μ1 − μ3 : (−126.2, −50.8)∗
μ2 − μ3 : (−63.7, 11.7005)
12
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:
Comparison of Means: Battery Life Data