Suppose you have a sample from a normal population with mean mu, and known variance $\sigma^2$. What is the power function for $H_0: \mu = 0$ versus $H_a: \mu \ne 0$ at $\alpha = 0.05$?
Attempt: If we standardize, we get $Z= \bar{x} - \mu/(\sigma/\sqrt{n})$. So $\Pr(Z> c+ (\theta_0-\theta)/(\sigma/\sqrt{n}))$ or $\Pr(Z < -c- (\theta_0-\theta)/(\sigma/\sqrt{n}))$ is the power function.