STATISTICS
whereαis the required significance level of the test. In our case we setα=0.05, and from
table 31.3 withn= 16 we find thattcrit=2.12. The rejection region is therefore
t<− 2. 12 and t> 2. 12.Sincet=− 2 .17 for our samples, we can reject the null hypothesisH 0 :μ 1 =μ 2 , although
only by a small margin. (Indeed, it is easily shown that one cannot rejectH 0 at the 2%
significance level). The 95% central confidence interval onω=μ 1 −μ 2 is given by
w ̄−σtˆcrit(
N 1 +N 2
N 1 N 2
) 1 / 2
<ω<w ̄+σtˆcrit(
N 1 +N 2
N 1 N 2
) 1 / 2
,
wheretcritis given by (31.120). Thus, we find
− 26. 1 <ω<− 0. 28 ,which, as expected, does not (quite) containω=0.
In order to apply Student’st-test in the above example, we had to make theassumption that the samples were drawn from Gaussian distributions possessing a
common variance, which is clearly unjustifieda priori.Wecan,however,perform
another test on the data to investigate whether the additional hypothesisσ 12 =σ 22
is reasonable; this test is discussed in the next subsection. If this additional test
shows that the hypothesisσ 12 =σ 22 may be accepted (at some suitable significance
level), then we may indeed use the analysis in the above example to infer that
the null hypothesisH 0 :μ 1 =μ 2 may be rejected at the 5% significance level.
If, however, we find that the additional hypothesisσ^21 =σ^22 must be rejected,
then we can only infer from the above example that the hypothesis that the two
samples were drawn from the same Gaussian distribution may be rejected at the
5% significance level.
Throughout the above discussion, we have assumed that samples are drawnfrom a Gaussian distribution. Although this is true for many random variables,
in practice it is usually impossible to knowa prioriwhether this is case. It can
be shown, however, that Student’st-test remains reasonably accurate even if the
sampled distribution(s) differ considerably from a Gaussian. Indeed, for sampled
distributions that differ only slightly from a Gaussian form, the accuracy of
the test is remarkably good. Nevertheless, when applying thet-test, it is always
important to remember that the assumption of a Gaussian parent population is
central to the method.
31.7.6 Fisher’sF-testHaving concentrated on tests for the meanμof a Gaussian distribution, we
now consider tests for its standard deviationσ. Before discussing Fisher’sF-test
for comparing the standard deviations of two samples, we begin by considering
the case when an independent samplex 1 ,x 2 ,...,xNis drawn from a Gaussian