Download Analysis of the Fixed Effects Model

Design and Analysis of
Experiments
Dr. Tai-Yue Wang
Department of Industrial and Information Management
National Cheng Kung University
Tainan, TAIWAN, ROC
1/33
Analysis of Variance
Dr. Tai-Yue Wang
Department of Industrial and Information Management
National Cheng Kung University
Tainan, TAIWAN, ROC
2/33
Outline(1/2)

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



Example
The ANOVA
Analysis of Fixed effects Model
Model adequacy Checking
Practical Interpretation of results
Determining Sample Size
3/33
Outline (2/2)
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
Discovering Dispersion Effects
Nonparametric Methods in the ANOVA
4/33
What If There Are More Than
Two Factor Levels?





The t-test does not directly apply
There are lots of practical situations where there are either
more than two levels of interest, or there are several factors
of simultaneous interest
The ANalysis Of VAriance (ANOVA) is the appropriate
analysis “engine” for these types of experiments
The ANOVA was developed by Fisher in the early 1920s,
and initially applied to agricultural experiments
Used extensively today for industrial experiments
5
An Example(1/6)
6
An Example(2/6)


An engineer is interested in investigating the
relationship between the RF power setting
and the etch rate for this tool. The objective of
an experiment like this is to model the
relationship between etch rate and RF power,
and to specify the power setting that will give
a desired target etch rate.
The response variable is etch rate.
7
An Example(3/6)



She is interested in a particular gas (C2F6)
and gap (0.80 cm), and wants to test four
levels of RF power: 160W, 180W, 200W, and
220W. She decided to test five wafers at each
level of RF power.
The experimenter chooses 4 levels of RF
power 160W, 180W, 200W, and 220W
The experiment is replicated 5 times – runs
made in random order
8
An Example --Data
9
An Example – Data Plot
10
An Example--Questions
Does changing the power change the
mean etch rate?
Is there an optimum level for power?
We would like to have an objective way to
answer these questions
The t-test really doesn’t apply here – more
than two factor levels
11
The Analysis of Variance




In general, there will be a levels of the
factor, or a treatments, and n replicates of
the experiment, run in random order…a
completely randomized design (CRD)
N = an total runs
We consider the fixed effects case…the
random effects case will be discussed later
Objective is to test hypotheses about the
equality of the a treatment means
12
The Analysis of Variance
13
The Analysis of Variance

The name “analysis of variance” stems from
a partitioning of the total variability in the
response variable into components that are
consistent with a model for the experiment
14
The Analysis of Variance

The basic single-factor ANOVA model is
 i  1, 2,..., a
yij     i   ij , 
 j  1, 2,..., n
  an overall mean,  i  ith treatment effect,
 ij  experimental error, NID(0,  2 )
15
Models for the Data


There are several ways to write a model for
the data
Mean model
yij  i   ij

Also known as one-way or single-factor
ANOVA
16
Models for the Data


Fixed or random factor?
The a treatments could have been specifically
chosen by the experimenter. In this case, the
results may apply only to the levels
considered in the analysis. fixed effect
models
17
Models for the Data

The a treatments could be a random sample
from a larger population of treatments. In this
case, we should be able to extend the
conclusion to all treatments in the population.
 random effect models
18
Analysis of the Fixed Effects
Model

Recall the single-factor ANOVA for the fixed
effect model
yij     i   ij
yij  i   ij

Define
n
yi .   yij and yi .  yi . / n
i  1,2,..., a
j 1
a
n
y..   yij and y..  y.. / N
i 1 j 1
N  an
19
Analysis of the Fixed Effects
Model

Hypothesis
H 0 : 1  2    a
H1 : i   j for at least one pair(i, j)
a


i 1
i
a

a

i 1
i

0
20
Analysis of the Fixed Effects
Model

Thus, the equivalent Hypothesis
H 0 : 1   2     a  0
H1 :  i  0 for at least one i
21
Analysis of the Fixed Effects
Model-Decomposition

Total variability is measured by the total sum of
squares:
a
n
SST   ( yij  y.. )2
i 1 j 1

The basic ANOVA partitioning is:
a
n
a
n
2
(
y

y
)

[(
y

y
)

(
y

y
)]
 ij ..  i. .. ij i.
2
i 1 j 1
i 1 j 1
a
a
n
 n ( yi.  y.. ) 2   ( yij  yi. ) 2
i 1
SST  SSTreatments  SS E
i 1 j 1
22
Analysis of the Fixed Effects
Model-Decomposition

In detail
 y
a
n
i 1 j 1
ij
 y..
  y
2
a
n
i 1 j 1
i.
 y..  yij  yi .

a
 n  yi .  y..
i 1
a
n

 
2

   y
2
a
n
i 1 j 1

ij
 2 yi .  y.. yij  yi .
 yi .


i 1 j 1
(=0)
23
2
Analysis of the Fixed Effects
Model-Decomposition

Thus
 y
a
n
i 1 j 1
ij
 y..
  y
2
a
n
i 1 j 1
a
SST
i.

 y..  yij  yi .
 n  yi .  y..
i 1
SSTreatments

 
2
   y
2
a
n
i 1 j 1
ij
 yi .

SSE
24
2
Analysis of the Fixed Effects
Model-Decomposition
SST  SSTreatments  SS E



A large value of SSTreatments reflects large
differences in treatment means
A small value of SSTreatments likely indicates
no differences in treatment means
Formal statistical hypotheses are:
H 0 : 1  2 
 a
H1 : At least one mean is different
25
Analysis of the Fixed Effects
Model-Decomposition

For SSE
a

n
SSE   yij  yi .
i 1 j 1

Recall
 y
n
Si2 
n
j 1
ij
 yi .
 n
2
   yij  yi . 
i 1  j 1

a



2
n 1

  yij  yi .
j 1

2

2
 ( n  1) Si2
26
Analysis of the Fixed Effects
Model-Decomposition

Combine a sample variances
a

SS E   ( n  1) S
i 1
2
i

( n  1) S12  ( n  1) S22    ( n  1) Sa2

1
 ( n  1) S12  ( n  1) S22    ( n  1) Sa2 

N  a 

 

1
(
n

1
)

(
n

1
)



(
n

1
)



SS E
( n  1) S12  ( n  1) S22    ( n  1) Sa2


N  a 
( n  1)  ( n  1)    ( n  1)

The above formula is a pooled estimate of
the common variance with each a
treatment.
27
Analysis of the Fixed Effects
Model-Mean Squares

Define
SSTreatments
MSTreatments 
a  1
df
and
SS E
MS E 
N  a 
df
28
Analysis of the Fixed Effects
Model-Mean Squares

By mathematics,
E ( MS E )   2
a
E ( MSTreatments)   2 


n  i2
i 1
a 1
That is, MSE estimates σ2.
If there are no differences in treatments
means, MSTreatments also estimates σ2.
29
Analysis of the Fixed Effects
Model-Statistical Analysis
Cochran’s Theorem
Let Zi be NID(0,1) for i=1,2,…,v and

v
Z
i 1
i
 Q1  Q2    Qs
where s  v,
and Qi has vi degrees of freedom (i  1,2,, s )
then Q1, q2,…,Qs are independent chisquare random variables withv1, v2, …,vs
degrees of freedom, respectively, If and
only if v  v1  v2    vs
30
Analysis of the Fixed Effects
Model-Statistical Analysis

Cochran’s Theorem implies that
SSTreatments /  2 and SS E /  2
are independently distributed chi-square
random variables
 Thus, if the null hypothesis is true, the ratio
F0 
SSTreatments /( a  1)
SS E /( N  a )
is distributed as F with a-1 and N-a degrees of
freedom.
31
Analysis of the Fixed Effects
Model-Statistical Analysis

Cochran’s Theorem implies that
SSTreatments /  2 and SS E /  2
Are independently distributed chi-square
random variables
 Thus, if the null hypothesis is true, the ratio
F0 
SSTreatments /( a  1)
SS E /( N  a )
is distributed as F with a-1 and N-a degrees of
freedom.
32
Analysis of the Fixed Effects
Models-- Summary Table


The reference distribution for F0 is the Fa-1,
a(n-1) distribution
Reject the null hypothesis (equal treatment
means) if F0  F ,a1,a ( n1)
33
Analysis of the Fixed Effects
Models-- Example

Recall the example of etch rate,

Hypothesis
H 0 : 1  2  3  4
H1 : some means are different
34
Analysis of the Fixed Effects
Models-- Example
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ANOVA table
35
Analysis of the Fixed Effects
Models-- Example
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Rejection region
36
Analysis of the Fixed Effects
Models-- Example
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P-value
P-value
37
Analysis of the Fixed Effects
Models-- Example

Minitab
One-way ANOVA: Etch Rate versus Power
Source DF SS MS
F
P
Power 3 66871 22290 66.80 0.000
Error 16 5339 334
Total 19 72210
S = 18.27 R-Sq = 92.61% R-Sq(adj) = 91.22%
Individual 95% CIs For Mean Based on Pooled StDev
Level N Mean StDev ---+---------+---------+---------+-----160 5 551.20 20.02 (--*---)
180 5 587.40 16.74
(--*---)
200 5 625.40 20.53
(--*---)
220 5 707.00 15.25
(--*---)
---+---------+---------+---------+-----550
600
650
700
Pooled StDev = 18.27
38
Analysis of the Fixed Effects
Model-Statistical Analysis


Coding the observations will not change the
results
Without the assumption of randomization,
the ANOVA F test can be viewed as
approximation to the randomization test.
39
Analysis of the Fixed Effects
Model- Estimation of the model parameters

Reasonable estimates of the overall mean
and the treatment effects for the singlefactor model
yij     i   ij
are given by

  y..

 i  yi .  y..
40
Analysis of the Fixed Effects
Model- Estimation of the model parameters

Confidence interval for μi
yi.  t / 2,N a

MS E
MS E
 i  yi.  t / 2,N a
n
n
Confidence interval for μi - μj
yi.  y j.  t / 2,N a
MS E
MS E
 i  yi.  y j.  t / 2,N a
n
n
41
Analysis of the Fixed Effects
Model- Unbalanced data

For the unbalanced data
a
2
ni
y..
2
SST   y ij 
N
i 1 j 1
SSTreatments 
a

i 1
y i2.
2
y..

ni
N
42
A little (very little) humor…
43
Model Adequacy Checking

Assumptions on the model
yij     i   ij


Errors are normally distributed and
independently distributed with mean zero and
constant but unknown variances σ2
Define residual

eij  yij  yij

where yij is an estimate of yij
44
Model Adequacy Checking
--Normality

Normal probability plot
45
Model Adequacy Checking
--Normality

Four-in-one
46
Model Adequacy Checking
--Plot of residuals in time sequence

Residuals vs run order
47
Model Adequacy Checking
--Residuals vs fitted

Residuals vs fitted
48
Model Adequacy Checking
--Residuals vs fitted

Defects



Horn shape
Moon type
Test for equal variances

Bartlett’s test
H 0 :  12   22     a2
H1 : above not true for at least one  i2
49
Model Adequacy Checking
--Residuals vs fitted

Test for equal variances

Bartlett’s test
Test for Equal Variances: Etch Rate versus Power
95% Bonferroni confidence intervals for standard deviations
Power N Lower StDev Upper
160 5 10.5675 20.0175 83.0477
180 5 8.8384 16.7422 69.4591
200 5 10.8357 20.5256 85.1557
220 5 8.0496 15.2480 63.2600
Bartlett's Test (Normal Distribution)
Test statistic = 0.43, p-value = 0.933
Levene's Test (Any Continuous Distribution)
Test statistic = 0.20, p-value = 0.898
50
Model Adequacy Checking
--Residuals vs fitted

Test for equal variances

Bartlett’s test
51
Model Adequacy Checking
--Residuals vs fitted

Variance-stabilizing transformation


Deal with nonconstant variance
If observations follows Poisson distribution
 square root transformation
yij* 

yij or yij* 
1  yij
If observations follows Lognormal distribution
 Logarithmic transformation
yij*  log yij
52
Model Adequacy Checking
--Residuals vs fitted

Variance-stabilizing transformation

If observations are binominal data
Arcsine transformation
yij*  arcsin

yij
Other transformation  check the relationship
among observations and mean.
53
Practical Interpretation of
Results – Regression Model

The one-way ANOVA model
yij     i   ij
is a regression model and is similar to
yij  0  1 xi   ij
54
Practical Interpretation of
Results – Regression Model

Computer Results
Regression Analysis: Etch Rate versus Power
The regression equation is
Etch Rate = 138 + 2.53 Power
Predictor Coef SE Coef
T
P
Constant 137.62 41.21 3.34 0.004
Power
2.5270 0.2154 11.73 0.000
S = 21.5413 R-Sq = 88.4% R-Sq(adj) = 87.8%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
18
19
SS
63857
8352
72210
MS
F
63857 137.62
464
P
0.000
Unusual Observations
Obs Power Etch Rate Fit SE Fit Residual St Resid
11 200 600.00 643.02 5.28 -43.02 -2.06R
55
R denotes an observation with a large standardized residual.
The Regression Model
56
Practical Interpretation of
Results – Comparison of Means




The analysis of variance tests the hypothesis
of equal treatment means
Assume that residual analysis is satisfactory
If that hypothesis is rejected, we don’t know
which specific means are different
Determining which specific means differ
following an ANOVA is called the multiple
comparisons problem
57
Practical Interpretation of
Results – Comparison of Means


There are lots of ways to do this
We will use pairwised t-tests on
means…sometimes called Fisher’s Least
Significant Difference (or Fisher’s LSD)
Method
58
Practical Interpretation of
Results – Comparison of Means
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons among Levels of Power
Simultaneous confidence level = 81.11%
Power = 160 subtracted from:
Power Lower
180
11.71
200
49.71
220 131.31
Center Upper
36.20
60.69
74.20
98.69
155.80 180.29
----+---------+---------+---------+----(--*-)
(-*--)
(--*-)
----+---------+---------+---------+-----100
0
100
200
59
Practical Interpretation of
Results – Comparison of Means
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons among Levels of Power
Simultaneous confidence level = 81.11%
Power = 180 subtracted from:
Power Lower Center Upper ----+---------+---------+---------+----200
13.51 38.00 62.49
(--*-)
220
95.11 119.60 144.09
(-*-)
----+---------+---------+---------+-----100
0
100
200
Power = 200 subtracted from:
Power Lower Center Upper ----+---------+---------+---------+----220
57.11 81.60 106.09
(-*--)
----+---------+---------+---------+-----100
0
100
200
60
Practical Interpretation of
Results – Graphical Comparison of Means
61
Practical Interpretation of
Results – Contrasts

A linear combination of parameters
a
a
i 1
i 1
   ci i and  ci  0

So the hypothesis becomes
a
H 0 :  ci i  0
i 1
a
H1 :  ci i  0
i 1
62
Practical Interpretation of
Results – Contrasts

Examples
H 0 : 3  4  0
H1 : 3  4  0
H 0 : 1  2  3  4  0
H1 : 1  2  3  4  0
H 0 : 31  2  3  4  0
H1 : 31  2  3  4  0
63
Practical Interpretation of
Results – Contrasts

Testing

t-test

Contrast average
C
a
c
i 1

Contrast Variance
yi .
i

V C  
2
n
a

Test statistic
t0 
c
i 1
MS E
n
i
a
c
i 1
i
yi .
a
c
i 1
i
64
Practical Interpretation of
Results – Contrasts
2


  ci yi . 
MS C
SSC / 1
2
i 1


F0  t 0 


MS E a 2
MS E
MS E
c

n i 1 i
a

Testing

F-test

Test statistic
1

MS E
 a

  ci yi . 
 i 1

1 a 2
ci

n i 1
2


  ci yi . 

where SSC   i 1 a
1
2
c

n i 1 i
a
2
65
Practical Interpretation of
Results – Contrasts

Testing

F-test

Reject hypothesis if
F0  F ,1, N a
66
Practical Interpretation of
Results – Contrasts

Confidence interval
a
 ci yi.  t / 2,N a
i 1

a
2
c
 i
i 1
a
c 
i 1

MS E
n
a
i
i
 ci yi.  t / 2,N a
i 1
MS E
n
a
2
c
 i
i 1
67
Practical Interpretation of
Results – Contrasts

Standardized contrast


Standardize the contrasts when more than one
contrast is of interest
Standardized contrast
a
*
c
 i yi .
i 1
where
c 
*
i
ci
1 a 2
ci

n i 1
68
Practical Interpretation of
Results – Contrasts

Unequal sample size

Contrast
a
   ci i
i 1
a
n c
i 1

i i
0
t-statistic
a
t0 
c y
i 1
i
i.
a
c i2
i 1
ni
MS E 
69
Practical Interpretation of
Results – Contrasts

Unequal sample size

Contrast sum of squares


  ci yi . 
i 1


SSC 
a c2
i
a
2
n
i 1
i
70
Practical Interpretation of
Results – Orthogonal Contrasts

Define two contrasts with coefficients {ci}
and {di} are orthogonal contrasts if
a
c d
i 1

i
i
0
Unbalanced design
a
n c d
i 1
i i
i
0
71
Practical Interpretation of
Results – Orthogonal Contrasts

Why use orthogonal contrasts ?
 For a treatments, the set of a-1 orthogonal
contrasts partition the sum of squares due to
treatments into a-1 independent singledegree-of freedom
tests performed on orthogonal contrasts are
independent.
72
Practical Interpretation of
Results – Orthogonal Contrasts

example
73
Practical Interpretation of
Results – Orthogonal Contrasts

Example for contrast
74
Practical Interpretation of
Results – Scheffe’s method for



comparing all contrasts
Comparing any and all possible contrasts
between treatment means
Suppose that a set of m contrasts in the
treatment means
u  c1u 1  c2u 2    cau a u  1,2,..., m
of interest have been determined.
The corresponding contrast in the treatment
averages yi. is
Cu  c1u. y1.  c2u y2.    cau ya. u  1,2,..., m
75
Practical Interpretation of
Results – Scheffe’s method for

comparing all contrasts
The standard error of this contrast is
SCu  MS E  ciu2 / ni 
a
i 1

The critical value against which Cu should be
compared is
S ,u  SCu

a-1Fα,a 1,N a
If |Cu|>Sα,u , the hypothesis that contrast Γu
equals zero is rejected.
76
Practical Interpretation of
Results – Scheffe’s method for

comparing all contrasts
The simultaneous confidence intervals with
type I error α
Cu  S ,u  u  Cu  S ,u
77
Practical Interpretation of
Results – example for Scheffe’s method

Contrast of interest
1  1  2  3  4
2  1  4

The numerical values of these contrasts are
C1  y1.  y2.  y3.  y4.
 551.2  587.4  625.4  707.0  193.8
C2  y1.  y4.
 551.2  707.0  155.8
78
Practical Interpretation of
Results – example for Scheffe’s method

Standard error
SC1  16.34
SC2  11.55

One percent critical values are
S0.01,1  65.09
S0.01,2  45.97

|C1|>S0.01,1 and |C1|>S0.01,1 , both contrast
hypotheses should be rejected.
79
Practical Interpretation of
Results –comparing pairs of treatment



means
Tukey’s test
Fisher’s Least significant Difference (LSD)
method
Hsu’s Methods
80
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
One-way ANOVA: Etch Rate versus Power
Source DF SS
MS
F
P
Power 3 66871 22290 66.80 0.000
Error 16 5339
334
Total 19 72210
S = 18.27 R-Sq = 92.61% R-Sq(adj) = 91.22%
Individual 95% CIs For Mean Based on Pooled StDev
Level
160
180
200
220
N
5
5
5
5
Mean
551.20
587.40
625.40
707.00
StDev
20.02
16.74
20.53
15.25
---+---------+---------+---------+-----(--*---)
(--*---)
(--*---)
(--*---)
---+---------+---------+---------+-----550
600
650
700
Pooled StDev = 18.27
81
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
Grouping Information Using Tukey Method
Power N Mean Grouping
220 5 707.00 A
200 5 625.40
B
180 5 587.40
C
160 5 551.20
D
Means that do not share a letter are significantly different.
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Power
Individual confidence level = 98.87%
Power = 160 subtracted from:
Power Lower Center Upper -----+---------+---------+---------+---180
3.11 36.20 69.29
(---*--)
200
41.11 74.20 107.29
(--*---)
220 122.71 155.80 188.89
(---*--)
-----+---------+---------+---------+----100
0
100
200
82
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
Power = 180 subtracted from:
Power Lower Center Upper -----+---------+---------+---------+---200 4.91
38.00 71.09
(---*--)
220 86.51 119.60 152.69
(--*--)
-----+---------+---------+---------+----100
0
100
200
Power = 200 subtracted from:
Power Lower Center Upper -----+---------+---------+---------+---220
48.51 81.60 114.69
(--*--)
-----+---------+---------+---------+----100
0
100
200
83
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
Hsu's MCB (Multiple Comparisons with the Best)
Family error rate = 0.05
Critical value = 2.23
Intervals for level mean minus largest of other level means
Level Lower Center Upper ---+---------+---------+---------+-----160 -181.53 -155.80
0.00 (---*------------------)
180 -145.33 -119.60
0.00
(--*--------------)
200 -107.33 -81.60
0.00
(--*---------)
220
0.00
81.60 107.33
(---------*--)
---+---------+---------+---------+------160
-80
0
80
84
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
Grouping Information Using Fisher Method
Power
220
200
180
160
N
5
5
5
5
Mean Grouping
707.00 A
625.40 B
587.40
C
551.20
D
Means that do not share a letter are significantly different.
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons among Levels of Power
Simultaneous confidence level = 81.11%
Power = 160 subtracted from:
Power Lower Center
180
11.71 36.20
200
49.71 74.20
220 131.31 155.80
Upper ----+---------+---------+---------+----60.69
(--*-)
98.69
(-*--)
180.29
(--*-)
----+---------+---------+---------+-----100
0
100
200
85
Practical Interpretation of
Results –comparing pairs of treatment
means—Computer output
Power = 180 subtracted from:
Power Lower Center Upper ----+---------+---------+---------+----200
13.51 38.00 62.49
(--*-)
220
95.11 119.60 144.09
(-*-)
----+---------+---------+---------+-----100
0
100
200
Power = 200 subtracted from:
Power Lower Center Upper ----+---------+---------+---------+----220
57.11 81.60 106.09
(-*--)
----+---------+---------+---------+-----100
0
100
200
86
Practical Interpretation of
Results –comparing treatment means
with a control



Donntt’s method
Control– the one to be compared
Totally a-1 comparisons
Dunnett's comparisons with a control
Family error rate = 0.05
Individual error rate = 0.0196
Control = level (160) of Power
Critical value = 2.59
Intervals for treatment mean minus control mean
Level Lower Center Upper ---------+---------+---------+---------+
180
6.25 36.20 66.15 (-----*-----)
200
44.25 74.20 104.15
(-----*-----)
220 125.85 155.80 185.75
(-----*-----)
---------+---------+---------+---------+
50
100
150
200
87
Determining Sample size
-- Minitab






StatPower and sample size One-way ANOVA
One-way ANOVA
Alpha = 0.01 Assumed standard deviation = 25
Number of level ->4
Factors: 1 Number of levels: 4
Sample size ->
Maximum Sample Target
Difference Size
Power Actual Power
Max. difference  75
75
6
0.9
0.915384
Power value  0.9
The sample size is for each level.
SD  25
88
Dispersion Effects



ANOVA for location effects
Different factor level affects variability
dispersion effects
Example

Average and standard deviation are measured for a
response variable.
89
Dispersion Effects


ANOVA found no location effects
Transform the standard deviation to
y   ln (s)
Ratio
Obser.
Control
Algorithm
1
2
3
4
5
6
1
-2.99573
-3.21888
-2.99573
-2.81341
-3.50656
-2.99573
2
-3.21888
-3.91202
-3.50656
-2.99573
-3.50656
-2.40795
3
-2.40795
-2.04022
-2.20727
-1.89712
-2.52573
-2.12026
4
-3.50656
-3.21888
-2.99573
-2.99573
-3.50656
-3.91202
90
Dispersion Effects

ANOVA found dispersion effects
One-way ANOVA: y=ln(s) versus Algorithm
Source DF
SS
MS
F
P
Algorithm 3 6.1661 2.0554 21.96 0.000
Error
20 1.8716 0.0936
Total
23 8.0377
S = 0.3059 R-Sq = 76.71% R-Sq(adj) = 73.22%
91
Nonparametric methods
in the ANOVA


When normality is invalid
Use Kruskal-Wallis test




Rank observation yij in ascending order
Replace each observation by its rank, Rij
In case tie, assign average rank to them
test statistic
1  a Ri2. N ( N  1)2 
H  2  

S  i 1 ni
4

92
Regression and ANOVA
Regression Analysis: Etch Rate versus Power
The regression equation is
Etch Rate = 138 + 2.53 Power
Predictor Coef
SE Coef
T
P
Constant 137.62
41.21
3.34 0.004
Power
2.5270 0.2154 11.73 0.000
S = 21.5413 R-Sq = 88.4% R-Sq(adj) = 87.8%
Analysis of Variance
Source
DF SS MS
F
P
Regression
1 63857 63857 137.62 0.000
Residual Error 18 8352
464
Total
19 72210
Unusual Observations
Obs Power Etch Rate Fit
SE Fit Residual St Resid
11 200 600.00
643.02 5.28 -43.02
-2.06R
R denotes an observation with a large standardized residual.
93
Nonparametric methods
in the ANOVA

Kruskal-Wallis test

test statistic
1  a Ri2. N ( N  1)2 
H  2  

S  i 1 ni
4

where
2
a nj


1
N
(
N

1
)
2
2
S 
 Rij 

N  1  i 1 j 1
4

94
Nonparametric methods
in the ANOVA

Kruskal-Wallis test

If
H  2,a1
the null hypothesis is rejected.
95