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Fisher’s F-test

where we have introduced the variable

Ns 2 ∑ (x i − ¯x) 2
u = = i . (3.13)
σ 2 σ 2
0 0
We see that the rejection region λ < λ crit corresponds to a two-tailed rejection region on u
given by 0 < u < a and b < u < ∞, where a and b are such that λ crit (a) = λ crit (b). In practice,
however, it is difficult to determine a and b for a given significance level α, so a slightly different
rejection region, which we now describe, is usually adopted. The sampling distribution P(u|H 0 )
2
maybefoundstraightforwardlyfromthesamplingdistributionofs.LetusfirstdetermineP(s |H 0 )
by demanding that
2
2
P(s|H 0 )ds = P(s |H 0 )d(s ),
from which we find
2 (N−1)/2
P(s|H 0 ) (N/2σ ) ( Ns 2 )
2
2 (N−2)/2
P(s |H 0 ) = = 0 (s ) exp − . (3.14)
1
2s Γ( (N − 1)) 2σ 2
2 0
2
2
Thus, the sampling distribution of u = Ns /σ is given by
0
( )
1 1
P(u|H 0 ) = u (N−3)/2 exp − u .
1
2 (N−1)/2 Γ( (N − 1)) 2
2
We note, in passing, that the distribution of u is precisely that of an (N − 1)-th order chi-squared
variable, i.e. u ∼ χ 2 N−1 . Although it does not give quite the best test, one then takes the rejection
region to be 0 < u < a and b < u < ∞, with a and b chosen such that the two tails have equal
areas; the advantage of this choice is that tabulations of the chi-squared distribution make the size
of this region relatively easy to estimate. Thus, for a given significance level α, we have
∫ a α ∫ ∞ α
P(u|H 0 )du = and P(u|H 0 )du = .
0 2 b 2

Example 3.4. Ten independent sample values x i , i = 1, 2, . . . , 10, are drawn at
random from a Gaussian distribution with unknown mean µ and standard deviation σ.
The sample values are as follows:

2.22 2.56 1.07 0.24 0.18 0.95 0.73 − 0.79 2.09 1.81

Test the null hypothesis H 0 : σ = 2 at the 10% significance level. ,

2
Solution. For our null hypothesis σ = 2. Since for this sample s = 1.01 and N = 10, from
0
(3.13) we have u = 5.10. For α = 0.1 we find, either numerically or using tables, that a = 3.30
and b = 16.92. Thus, our rejection region is 0 < u < 3.33 and 16.92 < u < ∞. The value
u = 5.10 from our sample does not lie in the rejection region, and so we cannot reject the null
2
hypothesis H 0 : σ = 2.

We now turn to Fisher’s F-test. Let us suppose that two independent samples of sizes N 1 and N 2
2
2
are drawn from Gaussian distributions with means and variances µ 1 , σ and µ 2 , σ respectively,
1
2
2
2
2
and we wish to distinguish between the two hypotheses H 0 : σ = σ and H 1 : σ = σ . In this
2
1 2 1 2
case, the generalised likelihood ratio is found to be
(N 1 + N 2 ) (N 1 +N 2 )/2 [F(N 1 − 1)/(N 2 − 1)] N 1 /2
λ = · ,
N N 1 /2 N N 2 /2 [1 + F(N 1 − 1)/(N 2 − 1)] (N 1 +N 2 )/2
1 2
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