Note set #3
Stat402B (Spring 2016)
Last update: January 20, 2016
Stat 402B (Spring 2016): Note set 3
Comparing Several Treatments
Example
•
Data. Let yij represent the j-thP
observation taken under treatment i for
a
i=1,...,a and j=1,...,ni. Let N= i=1 ni
Treatment 1 Treatment 2 . . .
y11
y21
...
y12
y22
...
..
..
y1n1
y2n2
...
Sample Means:
ȳ1.
ȳ2.
...
where ȳi. =
yij
j=1 ni ,
Pni
Treatment a
ya1
ya2
..
yana
ȳa.
i = 1, . . . , a.
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Stat 402B (Spring 2016): Note set 3
Assume
y11, . . . , y1n1
y21, . . . , y2n2
..
iid
∼
N (µ1, σ 2)
iid
N (µ2, σ 2)
∼
iid
ya1, . . . , yana ∼ N (µa, σ 2)
This assumption can be represented by the model (the means model):
iid
yij = µi + eij , where eij ∼ N (0, σ 2)
or equivalently, by the effects model:
yij = µ + τi + eij
where τi are treatment effects expressed as deviations from a fixed value
µ. The null hypothesis of interest is
H0 : µ1 = · · · = µa vs Ha : at least one inequality or, equivalently
H0 : τ1 = · · · = τa
vs Ha : at least one inequality
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Stat 402B (Spring 2016): Note set 3
•
Variation within treatments
Pn1
− ȳ1.)2
, estimates σ 2
=
n1 − 1
Pn2
2
(y
−
ȳ
)
2i
2.
s22 = i=1
, estimates σ 2
n2 − 1
..
Pna
2
(y
−
ȳ
)
ai
a.
, estimates σ 2
s2a = i=1
na − 1
Pool them to obtain one estimator of σ 2:
s21
s2E
i=1 (y1i
=
(n1 −1)s21 +(n2 −1)s22 +···+(na −1)s2a
(n1 −1)+(n2 −1)+···+(na −1)
Pa
=
i=1
Pni
j=1 (yij −ȳi. )
2
N −a
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Stat 402B (Spring 2016): Note set 3
Pa
Pni
j=1 (yij − ȳi. )
2
is called the within treatment sum of squares with
(N-a) d.f. and is denoted by SSE . s2E is called the within treatment
mean square.
i=1
•
Variation Between Treatments
σ2
From the above, we know that ȳi. ∼ N (µi, ni ), i = 1, 2, . . . , a and they
are independent.
Pa
2
2
i=1 ni (ȳi. − ȳ.. )
sT rt =
a−1
Pa Pni
Pa Pni
i=1
j=1 yij
2
.
The
denominator
n
(ȳ
−
ȳ
)
where ȳ.. =
i
i.
..
i=1
j=1
N
is between treatment sum of squares with (a-1) d.f. and is denoted by
SST rt. s2T rt is the between treatment mean squares.
Pa Pni
2
(y
−
ȳ
)
ij
..
i=1
j=1
s2T ot =
N −1
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Stat 402B (Spring 2016): Note set 3
Pa
Pni
− ȳ..) is called the total corrected sum of squares with
(N-1) degrees of freedom and is denoted by SST ot. We have the
partitioning
i=1
j=1 (yij
SST ot = SST rt + SSE
that gives the anova table
Source of Variation d.f.
SS
MS
F
Bet. Trt
a − 1 SST rt M ST rt M ST rt/M SE
Within Trt
N − a SSE
M SE
Total
N − 1 SST rt
We reject H0 at α level of significance if F > Fα,a−1,N −a
Example 3.1: Plasma Etching Experiment
•
An engineer is interested in investigating the relationship between the
RF power setting and the etch rate for a wafer plasma-etching tool. She
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Stat 402B (Spring 2016): Note set 3
is interested in a particular gas (hexafluroethane) and gap (.8 cm) and
wants to test four levels of RF power:160, 180, 200, and 220 W. She
decided to test five wafers at each level of RF power. This is an example
of a single-factor experiments with a = 4 levels and n = 5 replicates.
For this to be a completely randomized design, the 20 tests needs to be
run in a random order i.e., the order in which the testing is done has to
be determined randomly.
•
ANOVA Table (Table 3.4
Source of Variation d.f.
SS
MS
F p − value
RF Power
3 66, 870.55 22, 290.18 66.8
< .0001
Error
16
5339.20
333.70
Total
19 72, 209.75
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Stat 402B (Spring 2016): Note set 3
•
Estimation and Prediction
yij = µi + eij or yij = µ + τi + eij
where eij ∼ N (0, σ 2).
•
Best estimates of µ1, . . . , µa are ȳ1., . . . , ȳ2., . . . , ȳa respectively. i.e.
µ̂i = ȳi, i=1,...,a.
•
Estimate of the difference µp − µq for any p 6= q is ȳp. − ȳq.
•
We cannot estimate τi’s uniquely, but can estimate the difference in a
pair of τi’s. Estimate of τp − τq is ȳp. − ȳq.
•
σ̂ 2 = s2E = M SE
•
Values predicted by the model for yij are: yij = µ̂i = ȳi for each
observation.
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Stat 402B (Spring 2016): Note set 3
•
Residuals rij = yij − ȳi.
•
SSE is also called the residuals sum of squares
•
Residuals are entirely model independent. If the model is adequate,
the residuals should contain no obvious patterns.
•
Data yij
Power(W)
yi.
ni
ȳi
160
575
542
530
539
570
2756
5
551.2
Etch Rate
180
200
565
600
593
651
590
610
579
637
610
629
2937 3127
5
5
587.4 625.4
220
725
700
715
685
710
3535
5
707.0
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Stat 402B (Spring 2016): Note set 3
Power(W)
prediced values ŷij
Power(W)
Residuals yij − ŷij
160
551.20
551.20
551.20
551.20
551.20
160
23.8
−9.2
−21.2
−12.2
18.8
Predicted Values
180
200
587.40 625.40
587.40 625.40
587.40 625.40
587.40 625.40
587.40 625.40
220
707.00
707.00
707.00
707.00
707.00
Residuals
180
200
−22.4 −25.4
5.6
25.6
2.6
−15.4
−8.4
11.6
22.6
3.6
220
18.0
−7.0
8.0
−22.0
3.0
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Stat 402B (Spring 2016): Note set 3
Pairwise Comparison of Means (or Effects)
Pairwise t-tests
H0 : µp = µq vs Ha : µp 6= µq
Reject H0 if kt0k > tα/2,N −a
µp and µq different at α level
P, ai.e., declare
of significance. Where N = i=1 ni, s2E = M SE and tα,N −a is the upper
100( α2 )% point of the t-distribution with (N-a) d.f.
Confidence Intervals (CIs) for Differences
A 100(1 − α)% CI for µp − µq is given by
s
(ȳp. − ȳq.) ± tα/2,N −a · sE
1
1
+
np nq
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Stat 402B (Spring 2016): Note set 3
If this interval does not include zero, we say µp and µq are different at α
level of significance.
Least Significance Difference(LSD)
Declare µp, µq significantly different at α level if
s
|ȳp. − ȳq.| > tα/2,N −a · sE
1
1
+
np nq
The quantity on the right side hand is called the least significance difference
or the LSD. For the balanced case, i.e., n1 = · · · = na = n
r
2
LSDα = tα/2,N −a · sE
n
LSD Procedure (can be used this way only when sample sizes are equal).
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Stat 402B (Spring 2016): Note set 3
Example 3.8: As an illustration of this procedure, compute LSD at α = .05
for the Plasma Etching example
r
r
r
2
2sE
2
2(333.7)
= t.025,16
= 2.12
= 24.49
LSD.05 = t.025,16 · sE
n
n
5
This implies that any pair of means will be found to be significantly different
if the difference in sample means of the pair exceeds 24.49. We may use the
underscoring procedure make comparing every pair simpler. In this case, this
method is unnnecessary as all pairs of means are found to be significantly
different. (See the analysis of Exercise 3.10 below for an illustration of the
underscoring procedure.)
Important Notes:
•
The type I error rate α holds only for testing only a single hypotheis, not
testing all possible differences among the means.
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Stat 402B (Spring 2016): Note set 3
•
When we do all pairwise tests at α level, the actual type I error rate
turns out to be much greater than α.
•
One way to alleviate this problem is to use the Bonferroni adjustment:
Replace α with α/k where k is the number of comparisons being made.
In the case when a means are being compared, k = (a)(a − 1)/2. So for
computing the LSD, one would use a t-value=tα/2k .
•
The problem with this method is it turns-out to be too conservative i.e.:
one may fail to find a few pairs of means that are actually different to
be significant.
•
In any case, make sure that the ANOVA F-statistic is significant at α
level before we use the LSD procedure to ensure that the type I error
rate is somewhat controlled, when making all pairwise comparisons.
•
LSD procedures often leads to conflicting results.
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Stat 402B (Spring 2016): Note set 3
Tukey’s Method
When comparing a means, suppose we wish to state a confidence
interval for µp − µq taking into account the fact that all possible pairwise
differences may be examined. A confidence interval that gives a confidence
of 100(1 − α)% or more when making all pairwise comparisons is given by
Tukey’s Studentized range statistic.
A 100(1 − α)% CI for µp − µq is given by
√
(ȳp. − ȳq.) ∓ (qα,a,f / 2) · sE
s
1
1
+
np nq
where qα,a,f is the upper significant level of the studentized range (see
Table VII). a is the number of means compared and f is the degrees of
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Stat 402B (Spring 2016): Note set 3
freedom associated with s2E = M SE . Calculate
sE
Tukeyα = qα,a,f · √
n
and use this value the same way as the LSD i.e., declare a difference of
means, ȳp. − ȳq. significant if |ȳp. − ȳq.| > Tukeyα If this procedure is used
when making all possible pairwise tests of differences, the total type I error
rate is controlled at α.
Example 3.7: As an illustration of this procedure, compute LSD at .05 α
for the Plasma Etching exampleFor the Etch-rate example,
r
sE
333.7 .
Tukey.05 = q.05,4,16 · √ = 4.05
= 33.09
5
5
Using this value we still find differences of all pairs of means significantly
different.
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Stat 402B (Spring 2016): Note set 3
Exercise 3.10 A product developer is investigating the tensile strength of a
new synthetic fiber that will be used to make cloth for mens shirts. Strength
is usually affected by the percentage of cotton used in the blend of materials
for the fiber. The engineer conducts a completely randomized experiment
with five levels of cotton content and replicated the experiment five times.
The data are:
yi.
ni
ȳi
15
7
7
15
11
9
49
5
9.8
% of Cotton
20
25
30
12
14
19
17
19
25
12
19
22
18
18
19
18
18
23
77
88
108
5
5
5
15.4 17.6 21.6
35
7
10
11
15
11
54
5
10.8
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Stat 402B (Spring 2016): Note set 3
ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Percentage
4 475.76 118.94 14.76
< .0001
Error
20 161.20
8.06
Total
24 636.96
Since the F-value is 14.76 with a corresponding p-value of < 0.0001, reject
the hypothesis that the tensile strength means are all equal. The percentage
of cotton in the fiber appears to have an effect on the tensile strength.
Compute LSD at α = .05:
r
LSD.05 = t.025,20 sE
2
= t.025
n
r
2s2E
n
r
= 2.086
2(8.06)
= 3.75
5
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Stat 402B (Spring 2016): Note set 3
Arrange the sample means in increasing order of magnitude:
ȳ1. ȳ5. ȳ2. ȳ3. ȳ4.
The underscoring procedure gives:
Cotton 15% 35% 20% 25%
Means
9.8 10.8 15.4 17.6
————– ————–
30%
21.6
The following conclusions may be made:
• Mean strength of 30% cotton fiber is significantly different from all other
fiber means and is the strongest.
•
Mean strength of 20% and 25% cotton fiber are not different from each
other but significantly less stronger that the 30% cotton fiber.
•
Mean strength of 15% and 35% cotton fiber are not different from each
other but significantly less stronger that the 20%, 25%, and 30% cotton
fiber.
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