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402 8. Tests of Hypotheses
is choosen small. In practice, one often chooses α = .10, .05 or .01 unless
otherwise stated. But the experimenter is free to choose any appropriate a
value.
Note that L(x; θ ), i = 0, 1, are two completely specified likelihood func-
i
tions. Intuitively speaking, a test for H versus H comes down to the com-
0
1
parison of L(x; θ ) with L(x; θ ) and figure out which one is significantly
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larger. We favor the hypothesis associated with the significantly larger likeli-
hood as the more plausible one. The following result gives a precise state-
ment.
Theorem 8.3.1 (Neyman-Pearson Lemma) Consider a test of H versus
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H stated in (8.3.1) with its rejection and acceptance regions for the null
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hypothesis H defined as follows:
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or equivalently, suppose that the test function has the form
where the constant k(≥ 0) is so determined that
Any test satisfying (8.3.2)-(8.3.3) is a MP level α test.
Proof We give a proof assuming that the Xs are continuous random vari-
ables. The discrete case can be disposed off by replacing the integrals with
the corresponding sums. First note that any test which satisfies (8.3.3) has
size α and hence it is level a too.
We already have a level a test function ψ(x) defined by (8.3.2)-(8.3.3). Let
ψ*(x) be the test function of any other level α test. Suppose that Q(θ), Q*(θ)
are respectively the power functions associated with the test functions ψ, ψ*.
Now, let us first verify that
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Suppose that x ∈ χ is such that ψ(x) = 1 which implies L(x; θ ) - kL(x;
1
θ ) > 0, by the definition of ψ in (8.3.2). Also for such x, one obviously
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has ψ(x) - ψ*(x) ≥ 0 since ψ*(x) ∈ (0, 1). That is, if x ∈ χ is such that
χ (x) = 1, we have verified (8.3.4). Next, suppose that x ∈ θ is such
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that χ (x) = 0 which implies L(x; θ ) - kL(x; θ ) < 0, by the definition
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