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Student’s t-test

2
2
Since t is a monotonically decreasing function of λ, the corresponding rejection region is t > c,
where c is a positive constant depending on the required significance level α. It is conventional,
however, to use t itself as our test statistic, in which case our rejection region becomes two-tailed
and is given by
t < −t crit and t > t crit , (3.6)
where t crit is the positive square root of the constant c.
The definition (3.5) and the rejection region (3.6) form the basis of Student’s t-test. It only
remains to determine the sampling distribution P(t|H 0 ). At the outset, it is worth noting that if
√
2
we write the expression (3.5) for t in terms of the standard estimator ˆσ = Ns /(N − 1) of the
standard deviation then we obtain
¯ x − µ 0
t = √ . (3.7)
ˆ σ/ N
If, in fact, we knew the true value of σ and used it in this expression for t then it is clear that t would
follow a Gaussian distribution with mean 0 and variance 1, i.e. t ∼ N(0, 1). When σ is not known,
however, we have to use our estimate ˆσ in (3.7), with the result that t is no longer distributed as the
standard Gaussian. As one might expect from the central limit theorem, however, the distribution
of t does tend towards the standard Gaussian for large values of N.
As noted earlier, the exact distribution of t, valid for any value of N, was first discovered by
William Gosset. If the hypothesis H 0 is true then the joint sampling distribution of ¯x and s is given
by
( 2 ) [ 2 ]
Ns N(¯x − µ)
P(¯x, s|H 0 ) = Cs N−2 exp − exp − , (3.8)
2σ 2 2σ 2
where C is a normalisation constant. We can use this result to obtain the joint sampling
distribution of s and t by demanding that

P(¯x, s|H 0 )d¯xds = P(t, s|H 0 )dtds.
√
Using (3.5) to substitute for ¯x − µ in (3.8), and noting that d¯x = (s/ N − 1)dt, we find

[ 2 ( 2 )]
Ns t
P(¯x, s|H 0 )d¯xds = As N−1 exp − 1 + dtds.
2σ 2 N − 1
where A is another normalisation constant. In order to obtain the sampling distribution of t
alone, we must integrate P(t, s|H 0 ) with respect to s over its allowed range, from 0 to ∞. Thus,
the required distribution of t alone is given by

∫ ∫ [ ( )]
2
∞ ∞ Ns | t 2
P(t|H 0 ) = P(t, s|H 0 )ds = A s N−1 exp − 2 1 + ds. (3.9)
0 0 2σ N − 1
2
To perform this integration, we make the change of variable y = s1 + [t /(N − 1)] 1/2 , which on
substitution into (3.9) yields

( 2 ) −N/2 ∫ ( 2 )
t ∞ Ny
P(t|H 0 ) = A 1 + y N−1 exp − dy.
N − 1 0 2σ 2
Since the integral over y does not depend on t, it is simply a constant. We thus find that that the
sampling distribution of the variable t is

1
1 Γ( N) ( t 2 ) −N/2
2 1 + , (3.10)
P(t|H 0 ) = √ 1
(N − 1)π Γ( (N − 1)) N − 1
2
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