Suppose that $X_1,\ldots,X_n\sim N(\theta, \sigma^2)$ and assume that $\sigma^2$ is known. We consider testing $H_0:\theta \leq \theta_0$ versus $H_1:\theta > \theta_0$.
Exercise: Derive an expression for the $p$-value using a frequentist test with test statistic $T =\sqrt{n}(\bar{X}_n - \theta_0)/\sigma$.
What I've tried: We're using a frequentist test $(T,C)$ where $T$ is a test statistic and $C$ a critical region. The $p$-value is the smallest value for which $T$ lies in $C$, or equivalently, the smallest value for which we reject the null hypothesis when observing the realisation $T$. This means that $p = \operatorname{Pr}(X\leq T) = \operatorname{Pr}(X\leq\sqrt{n}(\bar{X}_n - \theta_0)/\sigma)$.
Question: How can I further specify an expression for the $p$-value? Is my answer correct?
Thanks in advance!
You have one sided hypothesis and monotone likelihood ratio w.r.t to the minimial sufficient statistic $\bar{X}_n$, then you can construct the most powerful test for a given test size $\alpha$ using the likelihood ratio (Neyman-Pearson test). Then, using Karlin-Rubin theorem to deduce that this test is also the UMP test for every pair of $(\theta_1, \theta_0)$ for a given test size $\alpha$. So, let us build a MP test for $H_0: \theta = \theta_0$, $H_0: \theta = \theta_1$ , $\theta_1 > \theta_0$, $$ LR = \frac{\mathcal{L}(\theta_1)}{\mathcal{L}(\theta_0)}\propto\exp{\left\{ \frac{\bar{X}_n(\theta_1 - \theta_0)}{\sigma^2} \right\} } \ge k, $$ hence the UMPT is $$ \psi(X) = I\{\bar{X}_n \ge k^*\}. $$ Thus, as under $H_0$, $\bar{X}_n \sim \mathcal{N}(\theta_0, \sigma^2/n)$ the p.value is $$ p.value = \mathbb{P}_{\theta_0}(\bar{X}_n \ge \bar{x}_n) = 1-\Phi\left(\frac{\sqrt{n}(\bar{x}_n - \theta_0)}{\sigma} \right). $$