Let $X_1,...,X_n$ be i.i.d from a normal distribution with expectation $\theta$ and variance 1. The test for testing $H_0:\theta=\theta_0 $ v.s. $H_1:\theta=\theta_1$ where $\theta_0<\theta_1$ has rejection region $CR=$ {$\mathbf{x}:x_{(n)}>a$}, where $x_{(n)}=max(x_1,...,x_n)$. Also, a is such that $$P(X_{(n)}>a|\theta_0)=\alpha,0<\alpha<1$$.
Find the value of $a$ in terms of $\theta$ and a quantile of a standard normal distribution. Then compute $P(\mathbf{X}\in CR|\theta_1).$
The distribution of $X_{(n)}$ is $F(x)=[\Phi(x-\theta)]^n$. If $q_{\alpha}$ is the $\alpha$-quantile of that distribution under $H_0$, then $$ \mathsf{P}_{\theta_0}\left(X_{(n)}>q_{1-\alpha}\right)=\alpha. $$ In order to find $q_{1-\alpha}$ in terms the quantiles of the standard normal distibution, let $q_{\alpha}:=\Phi^{-1}(\alpha^{1/n})+\theta_0$. Then $$ F(q_{\alpha})=[\Phi(q_{\alpha}-\theta_0)]^n=(\alpha^{1/n})^n=\alpha. $$