TESTING THE EQUALITY OF TWO VARIANCES (THE F TEST)

advertisement
TESTING THE EQUALITY OF
TWO VARIANCES: THE F TEST
Application
test assumption of equal variances that
was made in using the t-test
 interest in actually comparing the variance
of two populations

fdist
1
The F-Distribution
Assume we repeatedly select a random
sample of size n from two normal
populations.
 Consider the distribution of the ratio of
two variances:
F = s12/s12.
 The distribution formed in this manner
approximates an F distribution with the
following degrees of freedom:
v1 = n1 - 1 and
v1 = n1 - 1

fdist
2
Assumptions
Random, independent samples from 2
normal populations
 Variability

fdist
3
F-Table
The F table can be found on the appendix
of our text. It gives the critical values of the
F-distribution which depend upon the
degrees of freedom.
fdist
4
Example 1

Assume that we have two samples with:
n2 = 7
df = 7-1= 6

and
and
n1 =10
df = 10-1= 9
Let v = F(6,9)
where 6 is the df from the numerator and 9 is
the df of the denominator.

Using the table with the appropriate df,
we find :
P(v < 3.37) = 0.95.
fdist
5
Example 2: Hypothesis Test
to Compare Two Variances
1. Formulate the null and alternate hypotheses.
H 0:
s12= s12
s12> s22
H a:
[Note that we might also use s12 < s22 or s12 =/ s22]
2. Calculate the F ratio.
F = s12/s12
[where s1 is the largest or the two variances]
3. Reject the null hypothesis of equal population variances if
F(v1-1, v2-1) > Fa
[or Fa/2 in the case of a two tailed test]
fdist
6
Example 2
The variability in the amount of impurities present in
a batch of chemicals used for a particular process
depends on the length of time that the process is in
operation.
Suppose a sample of size 25 is drawn from the
normal process which is to be compared to a sample
of a new process that has been developed to reduce
the variability of impurities.
n
s2
Sample 1
25
1.04
Sample 2
25
0.51
fdist
7
Example 2 continued
H0:
Ha:
s1 2 = s2 2
s12 > s22
F(24,24) = s12/s22 = 1.04/.51 = 2.04
Assuming a = 0.05
cv = 1.98 < 2.04
Thus, reject H0 and conclude that the variability in the new
process (Sample 2) is less than the variability in the
original process.
fdist
8
Try This
A manufacturer wishes to determine
whether there is less variability in the silver
plating done by Company 1 than that done
by Company 2. Independent random
samples yield the following results. Do the
populations have different variances?
[solution: reject H0 since 3.14 > 2.82]
fdist
9
Download
Study collections