= Solution
{c}
Let $S=\sum_iX_i$. For the <exponential distribution> with <mean> $\theta$, the <likelihood function> is $L(\theta)=\theta^{-n}e^{-S/\theta}$ on positive observations. Fix any $0<\theta_1<\theta_0$. The alternative-to-null <likelihood ratio> is
$$
\frac{L(\theta_1)}{L(\theta_0)}
=\left(\frac{\theta_0}{\theta_1}\right)^n
\exp\!\left\{-\left(\frac1{\theta_1}-\frac1{\theta_0}\right)S\right\}.
$$
Since $1/\theta_1-1/\theta_0>0$, this is strictly decreasing in $S$. The <Neyman-Pearson lemma> says to reject for small $S$. Under the <null hypothesis>, the given sum rule yields $S\sim\operatorname{Gamma}(n,1/\theta_0)$, where the second argument is a rate, and hence
$$
\frac{2S}{\theta_0}\sim\chi^2_{2n}.
$$
Let $q_\alpha$ be the lower $\alpha$ <quantile> of this <chi-squared distribution>. The required <critical region> is
$$
\boxed{\frac{2\sum_iX_i}{\theta_0}<q_\alpha,\qquad
q_\alpha=F^{-1}_{\chi^2_{2n}}(\alpha).}
$$
Its <test size> is exactly $\alpha$ because the <null distribution> is continuous. Its cutoff is the same for every $\theta_1<\theta_0$, so it is simultaneously most powerful against every such value: it is a <uniformly most powerful test>. Equivalently, reject if $S$ is below the lower $\alpha$ <quantile> of the specified <gamma distribution>.
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