1. **Stating the problem:** We have a hypothesis test for a Normal distribution with unknown mean $M$ and known standard deviation $\sigma=1$. The null hypothesis is $H_0: M=0$ and the alternative is $H_1: M \neq 0$. The sample size is $N=100$ and the significance level is $\alpha=0.05$. 2. **Type I error probability:** The Type I error is the probability of rejecting $H_0$ when $H_0$ is true (i.e., $M=0$). By definition, this is the significance level $\alpha=0.05$. 3. **Test statistic:** The sample mean $\bar{X}$ is normally distributed with mean $M$ and standard deviation $\sigma/\sqrt{N} = 1/\sqrt{100} = 0.1$. 4. **Rejection region:** For a two-sided test at level $0.05$, the critical values for the standardized test statistic $Z = \frac{\bar{X} - 0}{0.1}$ are $\pm z_{\alpha/2} = \pm 1.96$. 5. **Calculating power at $M=1$:** The power is $P(\text{reject } H_0 | M=1) = P(|Z| > 1.96 | M=1)$. Since $\bar{X} \sim N(1, 0.1^2)$, the standardized variable under $M=1$ is $$Z = \frac{\bar{X} - 0}{0.1} = \frac{\bar{X} - 1 + 1}{0.1} = \frac{\bar{X} - 1}{0.1} + 10$$ Let $Z' = \frac{\bar{X} - 1}{0.1} \sim N(0,1)$, so $$Z = Z' + 10$$ The rejection region is $|Z| > 1.96$, so $$P(|Z| > 1.96) = P(Z < -1.96) + P(Z > 1.96)$$ Since $Z = Z' + 10$, $$P(Z < -1.96) = P(Z' < -1.96 - 10) = P(Z' < -11.96) \approx 0$$ $$P(Z > 1.96) = P(Z' > 1.96 - 10) = P(Z' > -8.04) \approx 1$$ 6. **Final answers:** - Probability of Type I error = $0.05$ - Power at $M=1$ = $\approx 1$ This means the test almost always rejects $H_0$ when $M=1$.