Analysis of Variance
or
ANOVA
In ANOVA, we are interested in comparing
the means of different populations (usually
more than 2 populations).
Since this technique has frequently been
applied to medical research, samples from
the different populations are referred to as
different treatment groups.
Main Idea behind Analysis of Variance
• Measure the variability within the treatment groups
and also the variability between the treatment
groups.
• The “between” variability is due to both differences
in treatment and the random error component.
• The “within” variability is due only to the random
error component.
• So, if there are significant treatment differences, the
“between” variability should be substantially larger
than the “within” variability.
In ANOVA, the null hypothesis H0 is always
essentially “there are no differences in
treatment” and the alternative hypothesis H1
is always essentially “there is a difference in
treatment.”
If the problem says “test whether there are no
treatment effects,” H0 is “there are NO differences
in treatment.”
If the problem says “test whether there are treatment
effects,” H0 is “there are NO differences in
treatment.”
We will begin with one-factor ANOVA.
Suppose we’re looking at the effects of different types of
heart medication on pulse rate.
An individual’s pulse rate is the sum of different effects.
pulse rate
 (effect of being on heart medication in general)
 (effect of this medication relative to heart medication s in general)
 (random error of this individual )
If there is no difference in the effects of the various heart
medications or treatments, then the middle effect above is zero.
That is what we would like to test.
Example 1: A manager is comparing the output of 3 machines. He has
recorded 5 randomly selected minutes of operation for machine 1, 10 minutes
for machine 2, and 6 minutes for machine 3. The output is as shown below.
Machine 1
Y1
(Y1 - Y1 )2
Machine 2
Y2
(Y2 - Y2 )2
Machine 3
Y3
10
6
11
6
7
8
8
9
13
12
4
10
6
6
10
10
12
5
6
8
6
(Y3 - Y3 )2
In this example, we have
n = # of observations = 21
and c = # of groups (or # of machines here) = 3
Machine 1
Y1
(Y1 - Y1 )2
Machine 2
Y2
(Y2 - Y2 )2
Machine 3
Y3
10
6
11
6
7
8
8
9
13
12
4
10
6
6
10
10
12
5
6
8
6
(Y3 - Y3 )2
We compute the mean output per minute for each machine.
The grand mean for all the machines combined is (42+67+64)/21 = 8.238.
Machine 1
Y1
Machine 2
(Y1 - Y1 )2
Y2
Machine 3
(Y2 - Y2 )2
Y3
10
6
11
6
7
8
8
9
13
12
4
10
6
6
10
10
12
5
6
8
6
42
Y1 
42
 8 .4
5
67
Y2 
67
 6.7
10
64
Y3 
64
 10.67
6
(Y3 - Y3 )2
We next look at the variation in output by computing sums of squared deviations.
Machine 1
Machine 3
Y1
(Y1 - Y1 )2
Y2
(Y2 - Y2 )2
Y3
(Y3 - Y3 )2
10
2.56
6
0.49
11
0.1111
6
5.76
7
0.09
8
7.1111
8
0.16
9
5.29
13
5.4444
12
12.96
4
7.29
10
0.4444
6
5.76
6
0.49
10
0.4444
10
10.89
12
1.7778
5
2.89
6
0.49
8
1.69
6
0.49
67
30.10
64
15.33
42
Y1 
Machine 2
42
 8 .4
5
27.2
Y2 
67
 6.7
10
Y3 
64
 10.67
6
The sum of squared variation within groups or SS within
is the sum of the sums of squared deviations for the 3 groups
= 27.2 + 30.10 + 15.33 = 72.63 .
Machine 1
Machine 3
Y1
(Y1 - Y1 )2
Y2
(Y2 - Y2 )2
Y3
(Y3 - Y3 )2
10
2.56
6
0.49
11
0.1111
6
5.76
7
0.09
8
7.1111
8
0.16
9
5.29
13
5.4444
12
12.96
4
7.29
10
0.4444
6
5.76
6
0.49
10
0.4444
10
10.89
12
1.7778
5
2.89
6
0.49
8
1.69
6
0.49
67
30.10
64
15.33
42
Y1 
Machine 2
42
 8 .4
5
27.2
Y2 
67
 6.7
10
Y3 
64
 10.67
6
The mean squared variation within groups
MS within = (SS Within) / (n – c).
= 72.63/(21 – 3) = 72.63/18 = 4.035
The sum of squared variation between groups or SS Between is
SS Between  n1( Y1  Y )2  n 2 ( Y2  Y )2  n 3 ( Y3  Y )2
The mean square variation between groups or MS Between is
MS Between 
In our example,
n1=5
Y1  8.4
SS Between
c 1
n2=10
Y2  6.7
n3=6
Y3  10.67
c=3
Y  8.238
So, SS Between  5(8.4  8.238)2 10(6.7  8.238)2  6(10.67  8.238)2
 5(0.026)  10(2.365)  6(5.898)
MS Between 
SS Between 59.18

 29.59
c 1
2
 59.18
The sum of squares total or SS Total
= SS Between + SS Within
The mean squared total or MS Total
= SS Total /( n – 1)
We compile our information in a table called an ANOVA table.
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
square
Between
Groups
c–1
3–1=2
59.18
29.59
Within
Groups
n–c
21 – 3 = 18
72.63
4.035
Total
n–1
21 – 1 = 20
131.81
6.591
Notice that the degrees of freedom column adds up to the
total at the bottom, and the sum of squares column adds
up to the total at the bottom, but the mean square column
does NOT add up to the total at the bottom.
Remember that the purpose of all the work we have been doing
is to test whether there is a “treatment effect,” that is a
difference in the means of the various groups.
The hypotheses are
H0: there is no difference in the means, and
H1: there is a difference in the means.
In our current example, we want to know if there is a
difference in the mean output level of the various machines.
The statistic we use here is the F-statistic:
Fc 1, n  c
MS Between

.
MS Within
The subscripts are the degrees of freedom of the F-statistic
and they are the degrees of freedom that are associated
with the numerator and the denominator of the statistic.
Like the 2, the F distribution is skewed to the right,
and the critical region for the test is the right tail.
f(Fc-1, n-c)
acceptance
region
crit. reg.
Fc-1, n-c
For our current example, we have:
Fc 1, n  c  F31, 21 3
MS Between 29.59
 F2, 18 
 7.33

MS Within
4.035
The F table shows that for
2 and 18 degrees of freedom,
the 5% critical value is 3.55.
f(F2, 18)
Since our F has a value of 7.33,
we reject the null hypothesis and
conclude that there is a
difference in the mean output
levels of the 3 machines.
acceptance
region
0.05
crit. reg.
3.55
7.33
F2, 18
Example 2: In order to compare the average tread life of
3 brands of tires, random samples of 6 tires of each brand
are tested on a machine which simulates road conditions.
The tread life for each tire is measured.
Complete the analysis of variance table and test at the
1% level whether there is a difference in the average
tread life of the 3 tire brands.
Source of
variation
Sum of
squares
Between
224.78
Within
118.83
Total
Degrees of
freedom
Mean
square
First we can calculate SS Total by adding SS Between
and SS Within.
Source of
variation
Sum of
squares
Between
224.78
Within
118.83
Total
343.61
Degrees of
freedom
Mean
square
The degrees of freedom for the SS Between is
c – 1 = 3 – 1 = 2,
since there are three groups (3 tire brands).
Source of
variation
Sum of
squares
Degrees of
freedom
Between
224.78
2
Within
118.83
Total
343.61
Mean
square
The degrees of freedom for the SS Within is
n – c = 18 – 3 = 15,
since there are 18 observations (6 of each of the 3 brands).
Source of
variation
Sum of
squares
Degrees of
freedom
Between
224.78
2
Within
118.83
15
Total
343.61
Mean
square
The degrees of freedom for the SS total is
n – 1 = 18 – 1 = 17.
Source of
variation
Sum of
squares
Degrees of
freedom
Between
224.78
2
Within
118.83
15
Total
343.61
17
Mean
square
The MS Between is
(SS Between) / (dof Between)
= 224.78 / 2 = 112.39
Source of
variation
Sum of
squares
Degrees of
freedom
Mean
square
Between
224.78
2
112.39
Within
118.83
15
Total
343.61
17
The MS Within is
(SS Within) / (dof Within)
= 118.83 / 15 = 7.922
Source of
variation
Sum of
squares
Degrees of
freedom
Mean
square
Between
224.78
2
112.39
Within
118.83
15
7.922
Total
343.61
17
The MS Total is
(SS Total ) / (dof Total )
= 343.61 / 17 = 20.212
Source of
variation
Sum of
squares
Degrees of
freedom
Mean
square
Between
224.78
2
112.39
Within
118.83
15
7.922
Total
343.61
17
20.212
So,
Fc 1, n  c
MS Between 112.39
 F2, 15 
 14.19

MS Within
7.922
The F table shows that for
2 and 15 degrees of freedom,
the 1% critical value is 6.36.
f(F2, 15)
Since our F has a value of 14.19,
we reject the null hypothesis and
conclude that there is a
difference in the mean tread life
of the 3 tire brands.
acceptance
region
0.01
crit. reg.
6.36
14.19
F2, 15
Two-Factor ANOVA with Interaction Effects
Now, we have two factors of interest, A and B, and
the factors may interact to influence a particular
variable Y with which we are concerned.
For example, we may want to explore the effects of
different teachers and different teaching methods on
student performance. So,
student performanc e
 (effect of having a particular teacher)
 (effect of a particular teaching method)
 (effect of the interactio n of the teacher and the method)
 (random error associated with the particular student)
We have 3 sets of hypotheses.
H0: factor A has no effect on Y.
H1: factor A has an effect on Y.
H0: factor B has no effect on Y.
H1: factor B has an effect on Y.
H0: the interaction of factors A and B has no
effect on Y.
H1: the interaction of factors A and B has an
effect on Y.
For each A and B possibility, we have r
replications (or observations).
For example, suppose we have a = 4 different
teachers, b = 3 different methods, and r = 3
replications.
There are 12 different teaching possibilities: You
could have teacher A, B, C, or D, and that
instructor could be using method 1, 2, or 3.
For each one of these 12 situations, we have 3
replications, or 3 observations, or exam scores of
3 students.
Our ANOVA table now looks like this:
Source of
Variation
Sum of Squares
Degrees of
Freedom
Mean Square
Factor A
SSA
a–1
MSA
Factor B
SSB
b–1
MSB
Interaction
between A & B
SSAB
(a – 1)(b – 1)
MSAB
Random Error
SSE
ab(r – 1 )
MSE
Total
SST
n–1
(or abr – 1)
MST
Test Statistics for Two-Way ANOVA
Testing for the effect of factor A:
Fa 1,
ab ( r 1)
MS Factor A

MS Error
Testing for the effect of factor B:
Fb 1,
ab ( r 1)
Notice that in all three
cases, the denominator
is the same; it’s the
Mean Squared Error.
MS Factor B

MS Error
Testing for the effect of the interaction of A and B:
F( a 1)( b 1),
ab ( r 1)
MS Interactio n

MS Error
Example 3: Consider the following ANOVA table
relating student performance to 4 teachers, 3 teaching
methods, and the interaction of those 2 factors, using 3
replications.
Source of
Variation
Sum of Squares
Teacher
300
Method
400
Interaction
900
Error
1200
Total
Degrees of
Freedom
Mean Square
Complete the table and then test at the 5% level whether
student performance depends on (1) the teacher, (2) the
teaching method, and (3) the interaction of the teacher
and the method.
Source of
Variation
Sum of Squares
Teacher
300
Method
400
Interaction
900
Error
1200
Total
Degrees of
Freedom
Mean Square
First, we complete the table.
Source of
Variation
Sum of Squares
Degrees of
Freedom
Mean Square
Teacher
300
4–1=3
100
Method
400
3–1=2
200
Interaction
900
(4 – 1)(3 – 1) = 6
150
Error
1200
ab(r – 1)
= (4)(3)(3 – 1)
= 24
50
Total
2800
n – 1 = abr – 1
= 36 – 1 = 35
80
Now we test the hypotheses. We’ll begin with the teachers.
F3, 24
MS Teachers 100


2
MS Error
50
From the F table, we find that the 5%
critical value for 3 and 24 degrees of
freedom is 3.01.
f(F3, 24)
acceptance
region
0.05
crit. reg.
2.00
3.01
Source of
Variation
SS
dof
MS
Teacher
300
3
100
Method
400
2
200
Interaction
900
6
150
Error
1200
24
50
Total
2800
35
80
Since our statistic, 2.00, is in
the acceptance region, we
accept H0 that the teacher has
no effect on student
performance.
F3, 24
Next we look at the teaching methods.
F2, 24
MS Method 200


4
MS Error
50
Source of
Variation
SS
dof
MS
Teacher
300
3
100
Method
400
2
200
Interaction
900
6
150
Error
1200
24
50
Total
2800
35
80
From the F table, we find that the 5%
critical value for 2 and 24 degrees of
freedom is 3.40.
f(F2, 24)
Since our statistic, 4.00, is in
the critical region, we reject
H0 and accept H1 that the
method does affect student
performance.
acceptance
region
0.05
crit. reg.
3.40
4.00
F2, 24
Last we look at the interaction of teachers and methods.
F6, 24
MS Interactio n 150


3
MS Error
50
From the F table, we find that the 5%
critical value for 6 and 24 degrees of
freedom is 2.51.
f(F6, 24)
acceptance
region
0.05
crit. reg.
2.51
3.00
Source of
Variation
SS
dof
MS
Teacher
300
3
100
Method
400
2
200
Interaction
900
6
150
Error
1200
24
50
Total
2800
35
80
Since 3 is in the critical region,
we reject H0 & accept H1: there
is an interaction effect of
teacher & method on student
performance.
F6, 24