Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample Poisson distribution with mean \(\theta\). Then the uniformly most powerful test for testing \(H_{0} : \theta = 1\) against \(H_{1} : \theta \gt 1\) :
Answer: C
The UMP test rejects \(H_{0}\) if \(\bar{X} \gt c\).
The Poisson family has a monotone likelihood ratio in \(\sum X_{i}\) (equivalently \(\bar{X}\)): for \(\theta_{1} \gt \theta_{0}\),
\[ \begin{aligned} &\frac{L(\theta_{1})}{L(\theta_{0})} \\ &= e^{-n(\theta_{1} - \theta_{0})}\left(\frac{\theta_{1}}{\theta_{0}}\right)^{\sum x_{i}} \end{aligned} \]
increases with \(\sum x_{i}\). By the Neyman-Pearson lemma and the Karlin-Rubin theorem, the UMP test of \(\theta = 1\) against \(\theta \gt 1\) rejects for large \(\bar{X}\) (with randomisation if needed, since the distribution is discrete).
Rejecting for small \(\bar{X}\) suits \(\theta \lt 1\), and two-sided or interval regions are not UMP for a one-sided alternative.
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