y21, . . . , y2n2
..
y11, . . . , y1n1
N (μ2, σ 2)
iid
iid
∼
N (μ1, σ 2)
∼
iid
Stat 402B (Spring 2016): Note set 3
H 0 : τ 1 = · · · = τa
vs Ha : at least one inequality
2
H0 : μ1 = · · · = μa vs Ha : at least one inequality or, equivalently
yij = μ + τi + eij
where τi are treatment effects expressed as deviations from a fixed value
μ. The null hypothesis of interest is
iid
yij = μi + eij , where eij ∼ N (0, σ 2)
or equivalently, by the effects model:
ya1, . . . , yana ∼ N (μa, σ 2)
This assumption can be represented by the model (the means model):
Assume
Last update: January 20, 2016
Stat402B (Spring 2016)
Note set #3
•
•
yij
j=1 ni ,
ni
i = 1, . . . , a.
s2E
=
=
=
i=1
a
j=1 (yij −ȳi. )
N −a
ni
2
(n1 −1)s21 +(n2 −1)s22 +···+(na −1)s2a
(n1 −1)+(n2 −1)+···+(na −1)
i=1 (y1i
n1
Treatment a
ya1
ya2
..
yana
ȳa.
1
3
Stat 402B (Spring 2016): Note set 3
− ȳ1.)2
, estimates σ 2
n1 − 1
n2
(y2i − ȳ2.)2
2
s2 = i=1
, estimates σ 2
n2 − 1
..
na
(yai − ȳa.)2
s2a = i=1
, estimates σ 2
na − 1
Pool them to obtain one estimator of σ 2:
s21
Variation within treatments
where ȳi. =
Treatment 1 Treatment 2 . . .
y11
y21
...
y12
y22
...
..
..
y1n1
y2n2
...
Sample Means:
ȳ1.
ȳ2.
...
Data. Let yij represent the j-th
observation taken under treatment i for
a
i=1,...,a and j=1,...,ni. Let N= i=1 ni
Example
Stat 402B (Spring 2016): Note set 3
Comparing Several Treatments
•
•
ni
Stat 402B (Spring 2016): Note set 3
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
ANOVA Table (Table 3.4
6
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.
Stat 402B (Spring 2016): Note set 3
4
Variation Between Treatments
2
From the above, we know that ȳi. ∼ N (μi, σni ), i = 1, 2, . . . , a and they
are independent.
a
ni(ȳi. − ȳ..)2
2
sT rt = i=1
a−1
a ni
a ni
y
ij
2
where ȳ.. = i=1 N j=1 . The denominator
j=1 ni (ȳi. − ȳ.. )
i=1
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.
a ni
2
i=1
j=1 (yij − ȳ.. )
2
sT ot =
N −1
2
i=1
j=1 (yij − ȳi. ) 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.
a
•
•
Stat 402B (Spring 2016): Note set 3
that gives the anova table
Stat 402B (Spring 2016): Note set 3
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.
•
•
•
7
Estimate of the difference μp − μq for any p = q is ȳp. − ȳq.
Best estimates of μ1, . . . , μa are ȳ1., . . . , ȳ2., . . . , ȳa respectively. i.e.
μ̂i = ȳi, i=1,...,a.
where eij ∼ N (0, σ 2).
yij = μi + eij or yij = μ + τi + eij
Estimation and Prediction
5
An engineer is interested in investigating the relationship between the
RF power setting and the etch rate for a wafer plasma-etching tool. She
Example 3.1: Plasma Etching Experiment
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
SST ot = SST rt + SSE
•
•
ni
i=1
j=1 (yij − 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
a
H0 : μp = μq vs Ha : μp = μq
(ȳp. − ȳq.) ± tα/2,N −a · sE
A 100(1 − α)% CI for μp − μq is given by
Confidence Intervals (CIs) for Differences
1
1
+
np nq
10
Reject H0 if t0 > tα/2,N −a
, ai.e., declare μp and μq different at α level
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.
Pairwise t-tests
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
1
1
+
np nq
11
LSD Procedure (can be used this way only when sample sizes are equal).
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
2
LSDα = tα/2,N −a · sE
n
|ȳp. − ȳq.| > tα/2,N −a · sE
Declare μp, μq significantly different at α level if
Least Significance Difference(LSD)
If this interval does not include zero, we say μp and μq are different at α
level of significance.
Stat 402B (Spring 2016): Note set 3
160
23.8
−9.2
−21.2
−12.2
18.8
220
707.00
707.00
707.00
707.00
707.00
Stat 402B (Spring 2016): Note set 3
Power(W)
160
551.20
551.20
551.20
551.20
551.20
9
Residuals yij − ŷij
prediced values ŷij
Power(W)
Stat 402B (Spring 2016): Note set 3
Predicted Values
180
200
587.40 625.40
587.40 625.40
587.40 625.40
587.40 625.40
587.40 625.40
8
Pairwise Comparison of Means (or Effects)
yi.
ni
ȳi
220
725
700
715
685
710
3535
5
707.0
Data yij
•
Etch Rate
180
200
565
600
593
651
590
610
579
637
610
629
2937 3127
5
5
587.4 625.4
Residuals are entirely model independent. If the model is adequate,
the residuals should contain no obvious patterns.
•
160
575
542
530
539
570
2756
5
551.2
SSE is also called the residuals sum of squares
•
Power(W)
Residuals rij = yij − ȳi.
•
Stat 402B (Spring 2016): Note set 3
Stat 402B (Spring 2016): Note set 3
Tukey’s Method
15
14
1
1
+
np nq
Using this value we still find differences of all pairs of means significantly
different.
sE
Tukeyα = qα,a,f · √
n
= M SE . Calculate
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,
sE
333.7 .
Tukey.05 = q.05,4,16 · √ = 4.05
= 33.09
5
5
freedom associated with
Stat 402B (Spring 2016): Note set 3
Stat 402B (Spring 2016): Note set 3
s2E
13
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
√
(ȳp. − ȳq.) ∓ (qα,a,f / 2) · sE
A 100(1 − α)% CI for μp − μq is given by
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.
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.
•
•
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 .
•
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.
•
12
The type I error rate α holds only for testing only a single hypotheis, not
testing all possible differences among the means.
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.
•
Important Notes:
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.)
Example 3.8: As an illustration of this procedure, compute LSD at α = .05
for the Plasma Etching example
2s2E
2
2(333.7)
LSD.05 = t.025,16 · sE
= t.025,16
= 2.12
= 24.49
n
n
5
Stat 402B (Spring 2016): Note set 3
%
20
12
17
12
18
18
77
5
15.4
of Cotton
25
30
14
19
19
25
19
22
18
19
18
23
88
108
5
5
17.6 21.6
30%
21.6
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.
•
18
Mean strength of 20% and 25% cotton fiber are not different from each
other but significantly less stronger that the 30% cotton fiber.
•
The following conclusions may be made:
• Mean strength of 30% cotton fiber is significantly different from all other
fiber means and is the strongest.
The underscoring procedure gives:
Cotton 15% 35% 20% 25%
Means
9.8 10.8 15.4 17.6
————– ————–
16
Stat 402B (Spring 2016): Note set 3
35
7
10
11
15
11
54
5
10.8
Arrange the sample means in increasing order of magnitude:
ȳ1. ȳ5. ȳ2. ȳ3. ȳ4.
yi.
ni
ȳi
15
7
7
15
11
9
49
5
9.8
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:
Stat 402B (Spring 2016): Note set 3
LSD.05 = t.025,20 sE
Compute LSD at α = .05:
2
= t.025
n
2s2E
2(8.06)
= 2.086
= 3.75
n
5
17
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.
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