Let $X_1,\dots,X_n$ be an iid sample from an $N(\theta,1)$ distribution. We want to test $H_0:\:\theta=0$ against the alternative $H_1\:\theta \neq 0$ using the test statistic $$T_n(X_1,\dots,X_n) = \bar{X}_n$$ and the corresponding test function $$\delta(X_1,\dots,X_n) = \begin{cases} 1, & \text{if } |T_n(X_1,\dots,X_n| \geq Q, \\ 0, & \mbox{otherwise}\end{cases}$$ where $Q\geq0$.

The goal is now to find the smallest value of $Q$ for which the significance level is $\alpha = 0.05$ and then to find the supremum over the paramater space of the probability of type II error for that value of $Q$.

For the Type I error, I got $$\mathbb{P}_\theta(\delta = 1) = 2\cdot\Phi(-\sqrt{n}\cdot Q), \quad \theta \in \Theta_0$$ In order to fix this error at level $\alpha = 0.05$, we need $2\cdot\Phi(-\sqrt{n}\cdot Q) = 0.05$, i.e. $Q=1.96/\sqrt{n}$, since $2\cdot\Phi(1.96)=2\cdot 0.025 = 0.05$

For the Type II error, I got $$\mathbb{P}_\theta(\delta = 0) = \Phi(\sqrt{n}\cdot (Q-\theta))-\Phi(-\sqrt{n} \cdot (Q+\theta)), \quad \theta \in \Theta_1$$

I now need to find the supremum of this error over the parameter space. Intuitively, the Type II error should increase the closer $\theta$ gets to 0, since that is the $\theta$-value of $H_0$. This would mean that $$\text{sup}_\theta \mathbb{P}_\theta(\delta = 0) = \Phi(\sqrt{n}\cdot Q)-\Phi(-\sqrt{n} \cdot Q), \quad \theta \in \Theta_1$$ Is this correct?


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