Discussion
Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid random sample of size \(n\) from exponential distribution with mean \(\theta\). The MP test of size \(\alpha\) for testing \(H_{0} : \theta = \theta_{0}\) against \(H_{1} : \theta = \theta_{1} \lt \theta_{0}\), is :
where \(T = \sum_{1}^{n} X_{i}\)
Answer: B
The MP test rejects when \(T \lt \theta_{0}\chi^{2}_{2n, 1-\alpha}/2\).
For the exponential with mean \(\theta\), the likelihood ratio \(f_{\theta_{1}}/f_{\theta_{0}}\) is a decreasing function of \(T = \sum X_{i}\) when \(\theta_{1} \lt \theta_{0}\), so by Neyman-Pearson the MP test rejects for small T. Under \(H_{0}\), \(2T/\theta_{0}\) has a chi-square distribution with \(2n\) degrees of freedom, so
\[ \begin{aligned} P\left(T \lt \tfrac{\theta_{0}}{2}c\right) &= P(\chi^{2}_{2n} \lt c) \\ &= \alpha \end{aligned} \]
needs \(c\) to be the lower \(\alpha\) point, which in the upper-tail notation \(\chi^{2}_{\nu, p}\) (the point with area \(p\) to its right) is \(\chi^{2}_{2n, 1-\alpha}\). Options A and D put \(\theta_{0}\) in the denominator, and C uses the upper \(\alpha\) point, which would give size \(1 - \alpha\).
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