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Hypothesis testing

The rejection region for H 0 is given by (3.6), where t crit satisfies

C N 1 +N 2 −2 (t crit ) = 1 − α/2, (3.12)

where α is the required significance level of the test. In our case we set α = 0.05, and with n
= 16 we find that t crit = 2.12. The rejection region is therefore t < −2.12 and t > 2.12. Since
t = −2.17 for our samples, we can reject the null hypothesis H 0 : µ 1 = µ 2 , although only by
a small margin. (Indeed, it is easily shown that one cannot reject H 0 at the 2% significance
level). The 95% central confidence interval on ω = µ 1 − µ 2 is given by

( ) 1/2 ( ) 1/2
N 1 + N 2 N 1 + N 2
¯ w − ˆσt crit < ω < ¯w + ˆσt crit ,
N 1 N 2 N 1 N 2
where t crit is given by (3.12). Thus, we find −26.1 < ω < −0.28, which, as expected, does not
(quite) contain ω = 0.

InordertoapplyStudent’st-testintheaboveexample,wehadtomaketheassumptionthatthe
samples were drawn from Gaussian distributions possessing a common variance, which is clearly
unjustified a priori. We can, however, perform another test on the data to investigate whether the
2
2
additional hypothesis σ = σ is reasonable; this test is discussed in the next subsection. If this
1 2
2
2
additional test shows that the hypothesis σ = σ may be accepted (at some suitable significance
2
1
level), then we may indeed use the analysis in the above example to infer that the null hypothesis
H 0 : µ 1 = µ 2 may be rejected at the 5% significance level. If, however, we find that the additional
2
2
hypothesis σ = σ must be rejected, then we can only infer from the above example that the
2
1
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 drawn from a Gaussian
distribution. Although this is true for many random variables, in practice it is usually impossible
to know a priori whether this is case. It can be shown, however, that Student’s t-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 the t-test, it is always important to
remember that the assumption of a Gaussian parent population is central to the method.

Fisher’s F-test

Having concentrated on tests for the mean µ of a Gaussian distribution, we now consider tests for
its standard deviation σ. Before discussing Fisher’s F-test for comparing the standard deviations
of two samples, we begin by considering the case when an independent sample x 1 , x 2 , . . . , x N is
drawn from a Gaussian distribution with unknown µ and σ, and we wish to distinguish between
2
2
2
2
2
the two hypotheses H 0 : σ = σ , −∞ < µ < ∞ and H 1 : σ = σ , −∞ < µ < ∞, where σ is a
0
0
0
2
given number. Here, the parameter space A is the half-plane −∞ < µ < ∞, 0 < σ < ∞, whereas
2
2
the subspace S characterized by the null hypothesis H 0 is the line σ = σ , −∞ < µ < ∞.
0
The likelihood function for this situation is given by
[ ∑ ]
1 (x i − µ) 2
2
L(x; µ, σ ) = exp − i .
2 N/2
(2πσ ) 2σ 2
2
2
The maximum of L in A occurs at µ = ¯x and σ = s , whereas the maximum of L in S is at µ = ¯x
2
2
and σ = σ . Thus, the generalised likelihood ratio is given by
0
2
L(x; ¯x, σ ) ( u ) N/2 [ 1 ]
λ(x) = 0 = exp − (u − N) ,
2
L(x; ¯x, s ) N 2
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