120 VARIANCES: ESTIMATION AND TESTS
In entering the F tables, the d.f.’s for the numerator are 721 - 1 = 11 and the d.f.’s for
the denominator are n2 - 1 = 18. Since there is a different F distribution for every
pair of sample sizes, F tables become quite extensive. In Table AS, just four areas
under the curve have been given: ,025, .95, .975, and .99. Degrees of freedom for
the numerator are denoted in the table by v1 , for the denominator by v2. The column
headings of Table AS give v1, v2 is listed by rows.
For a = .05 and a two-sided test, we will use the area of .975 (or the 97.5 percentile)
of the F table. We do not have to check the lower percentile since we placed the larger
variance in the numerator. In Table AS, there are entries for v1 = 11, but there are
no entries for v2 = 18. When there are no entries for a particular d.f., a conservative
approach is to use the F value from Table AS for the closest lower d.f. For example,
for v2 = 18 the results tabled for v2 = 15 can be used. For v1 = 11 and vz = 15,
the 97.5 percentile is given as F = 3.01. The correct value for F must lie between
F = 3.01 for 15 d.f. and F = 2.72 for 20 d.f. Since our computed F = 1.96 is less
than the correct value of F, we will not be able to reject the null hypothesis of equal
variances.
If our computed F had been equal to 2.80, we might want to find the percentile
for v1 = 11 and v2 = 18, since by looking at Table AS it would not be clear whether
we could accept or reject the null hypothesis. In this case, we could either perform
the test using a statistical program that gave the actual P value or interpolate in Table
AS.
As a second example of the use of the F test suppose that with the data from
two samples in Table 7.1, we wish to prove the assertion that the observed gains in
weight under the standard diet are less variable than under the supplemented diet.
Now, we will make a one-sided test of the hypothesis of 5 02”. The variance for the
supplemented diet is s: = 20,392 and the variance for the standard diet is si = 7060.
We will reject the null hypothesis only if s: is very large compared to si. It clearly
is larger, but without performing the test it is not clear if it is enough larger.
Under the null hypothesis, the F statistic computed is
s: 20,392
si 7060F=-=- = 2.89
The two sample sizes are 721 = 16 and 722 = 9, so Table AS is entered with v1 = 15
d.f. for the numerator and v2 = 8 d.f. for the denominator. To test the null hypothesis
with a = .05, we will compare the computed F to the 95th percentile in the F table
since a one-sided test is being performed and the entire rejection region is in the
right tail. The 95th percentile from Table AS is 3.22, so we cannot reject the null
hypothesis; we cannot conclude that the population variance for the supplemented
diet is larger than that under the standard diet.
Note that in most cases a two-sided F test is performed since often the researcher
does not know ahead of time which population is likely to have the larger variance.
It is important to screen the data for outliers before testing that two variances are
equal using the F test. This test is not robust with respect to the assumption of
normality or outliers. Outliers have a greater effect on the variance than they do
on the mean since the difference from the mean for each observation is squared in