ANOVA 13 (Comparison of Three or more means)
Previously we looked at 2-sample Z and 2-sample T tests, which can be used to test if two samples
came from populations with the same mean. In this chapter, we will learn a test to check if three or
more samples are from populations with the same mean. As an example we will use the following
data. Assume the treatments are three antiseptics and the observed values are bacterial counts.
Treatment
1
2
3
20
24
35
observed values
32
27
25
30
20
22
22
28
27
mean
26
24
28
sx
4.9666
4.3205
5.3541
=
SS
74
56
86
( x x)
2
The ANOVA (ANalysis Of VAriance) test may seem a little surprising at first since the statistic
does not explicitly contain any means, but rather it uses a comparison of two estimates of variation and
thus has an F distributiom. This is the same distribution we used when testing if two samples came
from populations with the same variance (the 2sample F test). The F statistic will be equivalent to the
following:
F
sample variance based on variation among the sample means
=
sample variance based on pooling the sample variances
=
ns
s
2
x
2
p
The following is a brief description of the reasoning used in forming this ratio
______________________________
________________________________________
Assuming all three samples come from a single population we have two ways to estimate this 2.
1) Average of the three sample variances.
s
2
p
2
S12 + S 2 + S 32
3
=
s
2
p
is called the within sample or pooled estimate of 2.
2) Use the sample variance of the three sample means. We will assume x is distributed as a
2
x = 2/n .
N(, / n ) , so the variance of x is given by:
n x = 2 . The population variance of x
can be approximated the by sample variance of the x s, which is:
2
Multiplying both sides of this equation by n yields:
2
2
2
2
S x = (x 1 - x ) + (x 2 - x ) + (x 3 - x )
3 -1
with
x
=
x 1 + x 2 + x3
3
2
Thus n S x 2.
This second estimate is called the between sample estimate of 2
When we take the samples from the same population both the within and between sample estimates
are unbiased estimates of 2 and should be approximately equal to each other; thus their ratio,
ns
s
2
2
x
,
p
Should be close to one. .
For the above data x = 26
Thus the ratio F =
ns
s
2
s
s
x
2
p
= [ (24-26)2 + (26-26)2 + (28-26)2 ]/2 = 4
=
(74 + 56 + 86) / 3
3
=
216
9
= 24
2
x
2
= (4*4)/24 = 2/3. This F statistic is not statistically significant at either the
p
1% or 5% levels.
Pvalue = Fcdf(2/3,100,2,9) = .5370
These calculations can be put into a table as follows
Source of
variation
Degrees
of Freedom
Sum of
Squares
Treatments
2
32
16
Error
9
216
24
Total
11
248
Requirements of the ANOVA test
1) samples are random and independent
2) sampled populations are normally distributed
3) population variances are equal
Mean
Square
F
2/3
Pval
.5370 = p(F2/3) with 2/9 df
Practice ANOVA Question
A department store manager wants to compare the sales of three clerks so she finds the number of
sales that each clerk has on four consecutive days
Clerk
A
B
C
Number of Sales
44
50
53
55
53
54
48
54
57
45
55
56
Perform an Analysis of Variance on this data to test:
Ho: A = B = C
Ha: the means are not all equal
Use = .05
State if you accept or reject Ho
Source of
variation
Degrees
of Freedom
Sum of
Squares
Mean
Square
Treatments
2
104
52
Error
9
98
Total
11
202
10.89
F
Pval
4.776 .0386