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Chapter 10
The Analysis
Variance
of
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
10.1
Single-Factor
ANOVA
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Analysis of Variance
The analysis of variance (ANOVA),
refers to a collection of experimental
situations and statistical procedures for
the analysis of quantitative responses
from experimental units.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Terminology
The characteristic that differentiates the
treatments or populations from one
another is called the factor under study,
and the different treatments or populations
are referred to as the levels of the factor.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Single-Factor ANOVA
Single-factor ANOVA focuses on a
comparison of more than two
population or treatment means. Let
I = the number of treatments (populations)
being compared.
1  the mean of population 1 or the true average
.
.
response when treatment 1 is applied
 I  the mean of population I or the true average
response when treatment I is applied
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Then the hypotheses of interest are
H 0 : 1  2  ...   I
versus
Ha :
at least two of the  i 's are
different
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Notation
X i, j 
xi , j 
The random variable that denotes the
jth measurement taken from the ith
population, or the measurement taken
on the jth experimental unit that
receives the ith treatment
The observed value of Xi,j when the
experiment is performed
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Assumptions
The I population or treatment distributions
2
are all normal with the same variance  .
Each Xi,j is normally distributed with
E( X i , j )  i V ( X i , j )  
2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Mean Square for Treatments
and Error
Mean square for treatments:
J
2
2
MSTr 
[( X1  X ..)  ...  ( X I   X ..) ]
I 1
J
2

(
X

X
..)

i
I 1 i
Mean square for error:
MSE 
2
S1
2
 S2
2
 ...  S I
I
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Test Statistic
The test statistic for single-factor
ANOVA is F = MSTr/MSE.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Expected Value
When H0 is true,
E(MSTr)  E(MSE)  
2
When H0 is false,
E(MSTr)  E(MSE)  
2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
F Distributions and Test
Let F = MSTr/MSE be the statistic in a
single-factor ANOVA problem involving I
populations or treatments with a random
sample of J observations from each one.
When H0 is true (basic assumptions true) ,
F has an F distribution with v1= I – 1 and
v2= I(J – 1). The rejection region
f  F ,I 1,I ( J 1) specifies a test with
significance level  .
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Formulas for ANOVA
Total sum of squares (SST)
I
J
SST  
i 1 j 1
2
xij
1 2
 x..
IJ
Treatment sum of squares (SSTr)
i
SSTr 
J
I

i 1
xi2.
1 2
 x..
IJ
Error sum of squares (SSE)
I
J

SSE   xij  xi.
i 1 j 1

2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Fundamental Indentity
SST = SSTr + SSE
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Mean Squares
SSTr
MSTr =
I 1
SSE
MSE =
I  J  1
MSTr
F=
MSE
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
ANOVA Table
Source of
Variation
df
Sum of
squares
Mean
Square
f
Treatments
I–1
SSTr
MSTr
MSTr/MSE
Error
I(J – 1)
SSE
MSE
Total
IJ – 1
SST
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
10.2
Multiple
Comparisons in
ANOVA
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Studentized Range Distribution
and Pairwise Differences
With probability 1   ,
X i.  X j.  Q , I , I ( J 1) MSE / J  i   j
 X i.  X j.  Q , I , I ( J 1) MSE / J
for every i and j with i  j.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The T Method for Identifying
Significantly Different i ' s
1. Select  extract Q ,I ,I ( J 1) .
2. Calculate w  Q , I , I ( J 1)  MSE / J
3. List the sample means in increasing
order, underline those that differ by
more than w. Any pair not underscored
by the same line corresponds to a pair
that are significantly different.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Confidence Intervals for Other
Parametric Functions
Let    ci i . Xij’s are normally distributed.

 2
V ˆ  V   ci X i  

 J
 i



2
ci
i
Estimating  2 by MSE and forming ˆ ˆ

ˆ
results in a t variable (   )ˆˆ leads to
 ci xi  t / 2,I ( J 1)
MSE
2
ci
J
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
10.3
More on
Single-Factor
ANOVA
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
ANOVA Model
The assumptions of a single-factor ANOVA
can be modeled by
X ij  i   ij
 ij represents a random deviation from
the population or true treatment mean i .
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
MSTr
J
2
E (MSTr)   
i

I 1
2
Note that when H0 is true

2
i
 0.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
 for the F Test
Consider a set of parameter values
1,...,  n for which H0 is not true.
The probability of a type II error,  , is
the probability that H0 is not rejected
when that set is the set of true values.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Single-Factor ANOVA When
Sample Sizes are Unequal
I
Ji
SST  
i 1 j 1
2
X ij
1 2
 X ..
n
I
df  n  1
1 2 1 2
SSTr   X i.  X ..
n
i 1 J i
df  I  1
SSE = SST – SSTr
df  n  I
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Single-Factor ANOVA When
Sample Sizes are Unequal
Test statistic value:
MSTr
SSTr
f 
where MSTr =
MSE
I 1
SSE
MSE =
nI
Rejection region:
f  F , I 1,n I
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Multiple Comparisons (Unequal
Sample Sizes)
MSE  1
1 
 

Let wij  Q , I ,n1 
2  J i J j 
Then the probability is approximately
1   that
X i.  X j.  wij  i   j  X i.  X j.  wij
for every i and j with i  j.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Data Transformation
If V ( X ij )  g (i ), a known function of i ,
then a transformation h(Xij) that
“stabilizes the variance” so that V[h(Xij)]
is approximately the same for each i is
1/ 2
given by h( x)    g ( x)
dx.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
A Random Effects Model
X ij    Ai  ij with E( Ai )  E(ij )  0
V ( ij )  
2
2
V ( Ai )   A
All Ai’s and  ij 's are normally distributed
and independent of one another.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.