Discussion
If, for a given \(\alpha\), \(0 \lt \alpha \lt 1\), non-randomized Neyman-Pearson and likelihood ratio tests of a simple hypothesis against a simple alternative exists, then :
Answer: A
The tests are equivalent.
For a simple null against a simple alternative, the Neyman-Pearson most powerful test rejects when the likelihood ratio \(f_{1}(x)/f_{0}(x)\) exceeds a constant \(k\) chosen to give size \(\alpha\). The likelihood ratio test rejects when \(\lambda(x) = f_{0}(x)/\max(f_{0}(x), f_{1}(x))\) is small, which for \(k \gt 1\) is the same event. When a non-randomised test of exact size \(\alpha\) exists, both reduce to the same critical region.
They are built differently (one from an optimality theorem, the other from a general principle), which is why the key prefers 'equivalent' to 'one and the same'. They are certainly not opposite.
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