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The Neyman-Pearson test

We consider first the choice of rejection region. Even in the general case, in which the test
statistic t is a multi-dimensional (vector) quantity, the Neyman-Pearson lemma states that, for a
given significance level α, the rejection region for H 0 giving the highest power for the test is the
region of t-space for which
P(t|H 0 )
> c, (3.2)
P(t|H 1 )

where c is some constant determined by the required significance level.
In the case where the test statistic t is a simple scalar quantity, the Neyman-Pearson lemma
is also useful in deciding which such statistic is the ’best’ in the sense of having the maximum
power for a given significance level α. From (3.2), we can see that the best statistic is given by the
likelihood ratio
P(x|H 0 )
t(x) = (3.3)
P(x|H 1 )
and that the corresponding rejection region for H 0 is given by t < t c rit. In fact, it is clear that any
statistic u = f(t) will be equally good, provided that f(t) is a monotonically increasing function
of t. The rejection region is then u < f(t crit ). Alternatively, one may use any test statistic v = g(t)
where g(t) is a monotonically decreasing function of t; in this case the rejection region becomes
v > g(t crit ). To construct such statistics, however, one must know P(x|H 0 ) and P(x|H 1 ) explicitly,
and such cases are rare.

Example 3.1. Ten independent sample values x i , i = 1, 2, . . . , 10, are drawn at
random from a Gaussian distribution with standard deviation σ = 1. The mean µ of
the distribution is known to equal either zero or unity. 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 : µ = 0 at the 10% significance level. ,

Solution. The restricted nature of the hypothesis space means that our null and alternative
hypotheses are H 0 : µ = 0 and H 1 : µ = 1 respectively. Since H 0 and H 1 are both simple
hypotheses, the best test statistic is given by the likelihood ratio (3.3). Thus, denoting the
means by µ 0 and µ 1 , we have
1 ∑ 2 1 ∑ 2 2
exp[− (x i − µ 0 ) ] exp[− (x − 2µ 0 x i + µ )]
i
0
t(x) = 2 1 ∑ i = 2 ∑ i =
1
2
2
2
exp[− (x i − µ 1 ) ] exp[− (x − 2µ 1 x i + µ )]
2 i 2 i i 1
∑ 1
2
2
= exp[(µ 0 − µ 1 ) x i − N(µ − µ )].
2 0 1
i
1
Inserting the values, µ 0 = 0, and µ 1 = 1, yields t = exp(−N ¯x + N), where ¯x is the sample
2
mean. Since − ln t is a monotonically decreasing function of t, however, we may equivalently
use as our test statistic
1 1
ν = − ln t + = ¯x,
N 2
where we have divided by the sample size N and added 1 for convenience. Thus we may take
2
the sample mean as our test statistic. We know that the sampling distribution of the sample
2
mean under our null hypothesis H 0 is the Gaussian distribution N(µ 0 , σ /N), where µ 0 = 0,
2
σ = 1 and N = 10. Thus ¯x ∼ N(0, 0.1).
Since ¯x is a monotonically decreasing function of t, our best rejection region for a given
significance α is ¯x > ¯x crit , where ¯x crit depends on α. Thus, in our case, ¯x crit is given by
( )
¯ x crit − µ 0
α = 1 − Φ = 1 − Φ(10¯x crit ),
σ
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