Solution
= Solution
For every particular alternative mean $\mu_1>\mu_0$, the <likelihood ratio> is increasing in the same statistic $\overline x$. Its size-$\alpha$ threshold is fixed solely by the null distribution, so it is always the same $c_\alpha$ found in part (c), <independent> of $\mu_1$.
The single <rejection region> is therefore most powerful against every member of the composite alternative. Hence \b[the test is uniformly most powerful for $H_1:\mu>\mu_0$].