The question:
Let $X_1,...,X_n$ be a random sample from the $N(\mu, 1)$ distribution. Consider testing $H_0: \mu = 0$ vs $H_1: \mu = 1$. Let the rejection region be $$R = \{(X_1, ..,X_n): \frac{1}{n} \sum_i X_i > c\}. $$ Find $c$ so that the test has size $\alpha$. Then find the power of the test under $H_1$, i.e. power(1).
My attempt:
For the first part, I formed the equation: $\alpha = P(Z > \frac{c - 0}{1/\sqrt{n}})$ to get $c = \frac{z_{1-\alpha}}{\sqrt{n}}$. But I'm unsure how to find the power(1) of the test. Thanks for any help.
The power of a test is defined by $$1 - \beta = \Pr[\text{reject } H_0 \mid H_1],$$ that is to say, it is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. In your case, your test statistic is $\bar X = \frac{1}{n}\sum_{i=1}^n X_i$, the sample mean of observations, and if it exceeds some criterion $c$, you reject the null $H_0$. From your earlier calculation, you determined that $c = z_{1-\alpha}/\sqrt{n}$, where $z_{1-\alpha}$ is the $1-\alpha$ quantile of the standard normal distribution; i.e., it is a number such that $\Pr[Z \le z_{1-\alpha}] = 1-\alpha$.
So, for this choice of $c$, and now assuming that observations are in fact drawn from a normal distribution with mean $1$ and variance $1$ (rather than drawn from a standard normal), what is the probability that the sample mean $\bar X$ exceeds $c$? Here, the quantity $$\left. \frac{\bar X}{\sqrt{n}} \,\right|\, H_1 \sim \operatorname{Normal}(1,1).$$ What is $$\Pr\left[ \left.\frac{\bar X}{\sqrt{n}} > \frac{z_{1-\alpha}}{\sqrt{n}} \,\right|\, H_1\right]?$$ Hint: the event in this probability is equivalent to $$\frac{\bar X - 1}{\sqrt{n}} > \frac{z_{1-\alpha} - 1}{\sqrt{n}},$$ and the random variable on the left-hand side of this inequality is standard normal under $H_1$.