1

Let $X_1, \dots, X_n$ be iid $\mathcal N(\theta, 1)$ and consider $H_0: \theta=0, H_1: \theta=1$. We are now asked to suppose that the rejection region of a test $\psi$ has the form $R=\{\bar X_n : \bar X_n >c\}$. We further suppose that the test $\psi$ has level $\alpha=0.05$. What is the power of $\psi$ ? What is the power as $n\rightarrow \infty$?

Now it seems to me that I will have to work with the finite sample distribution of $\bar X_n$ to really solve $\beta(\theta)=P[\bar X_n > c]$. Is that right? How should I go about doing this?

manav
  • 253

1 Answers1

1

In fact, the power of the test is the conditional probability of rejecting the null hypothesis, given that the alternative is true; i.e., $$1 - \beta = \Pr[\text{reject } H_0 \mid H_1] = \Pr[\bar X_n > c \mid \theta = 1]. \tag{1}$$ Consequently, the solution involves two steps: first, find the critical value $c$ for which the test has level $\alpha$; then second, compute the conditional probability in Equation $(1)$ above.

To understand how to compute $c$, we note that the Type I error is the conditional probability of rejecting the null hypothesis given that the null is true; i.e. $$\alpha = \Pr[\text{reject } H_0 \mid H_0] = \Pr[\bar X_n > c \mid \theta = 0]. \tag{2}$$ Since under the assumption $\theta = 0$ each observation $X_i$ is standard normal, what is the distribution of $\bar X_n$? What is the value of $c$ that satisfies Equation $(2)$, as a function of the sample size $n$?

heropup
  • 135,869