Question

Given $\Phi(1.645)=0.95$ and $\Phi(2.33)=0.99$. For a random sample $X_1,\ldots,X_n\sim N(\mu,2^2)$ with $\mu$ unknown, test $H_0:\mu=10$ vs $H_1:\mu=12$ using a rule that rejects $H_0$ when $\overline X$ is large. Find the smallest $n$ (in integer) so that Type I error is $0.05$ and Type II error is at most $0.01$.

Hint:

For known variance $\,\sigma^2\,$ in one–sided tests, solve the size constraint for the critical value, then impose the desired power under the alternative using the $z$–transform.

Answer 1

Correct Answer: 16

Under $H_0$, $\overline X\sim N(10,4/n)$. The level-$0.05$ critical value $c$ satisfies \[ P_{H_0}(\overline X>c)=0.05 \;\;\Longleftrightarrow\;\; \frac{\overline X-10}{2/\sqrt n}>1.645 \;\;\Longrightarrow\;\; c=10+\frac{2\cdot1.645}{\sqrt n}. \] Power at $\mu=12$ is \[ P_{12}(\overline X>c) = P\!\left(\frac{\overline X-12}{2/\sqrt n}>\frac{c-12}{2/\sqrt n}\right) = P\!\left(Z> -\sqrt n+1.645\right). \] Type II error $\le 0.01$ means $P_{12}(\overline X>c)\ge 0.99$, i.e. \[ \Phi(\sqrt n-1.645)\ge 0.99 \;\Longrightarrow\; \sqrt n-1.645\ge 2.33 \;\Longrightarrow\; \sqrt n\ge 3.975. \] Hence $n\ge 15.80$, so the smallest integer is $\boxed{16}$.