Question

We are looking to calculate the power of a one-sided test from n independent observations Xi from a N(μ, σ2) distribution with a null hypothesis of Ho : μ-μ0 and an alternative H, : μ 〉 μ0. Supposing that we know σ2, we can form a test statistic T= and reject the null hypothesis when T 〉 1.645. This test has level α 0.05. We want a formula for the power of this test against the alternative that μ-74-This power formula can be written as equal to PtZ> a] where Z is a standard normal. You need to find the a that is a function of n, σ2, 140, and μ1-

Answer 1

$Power~at~\mu=\mu_1=P\left[T>1.645|\mu=\mu_1 \right ]=P\left[\frac{\sqrt{n}(\bar{X}-\mu_0)}{\sigma}>1.645|\mu=\mu_1 \right ]\\ =P\left[ \frac{\sqrt{n}(\bar{X}-\mu_1+\mu_1-\mu_0)}{\sigma}>1.645|\mu=\mu_1\right ]=1-\Phi\left(1.645-\frac{\sqrt{n}(\mu_1-\mu_0)}{\sigma} \right )\\ where~\Phi(.)~is~cdf~of~standard~normal~distribution.\\ Hence~a=1.645-\frac{\sqrt{n}(\mu_1-\mu_0)}{\sigma}$