Question

Let $X$ be a random variable having $N(\theta,1)$ distribution, where $\theta \in \mathbb{R}$. Consider testing $H_0: \theta = 0$ against $H_1: \theta \neq 0$ at $\alpha = 0.617$ level of significance. The power of the likelihood ratio test at $\theta = 1$ equals ............ (round off to two decimal places).
 

Hint:

For a normal mean test, the LRT simplifies to a two-tailed z-test; the power depends on the noncentrality shift of the mean.

Answer 1

Correct Answer: 0.74

Step 1: Determine rejection region.
For $N(\theta,1)$, the LRT for $H_0: \theta = 0$ rejects when $|X| > c$, where $P(|X| > c) = \alpha = 0.617$. \[ P(|Z| > c) = 0.617 $\Rightarrow$ P(Z > c) = 0.3085 $\Rightarrow$ c = 0.5. \]

Step 2: Compute power at $\theta = 1$.
Under $\theta = 1$, $X \sim N(1,1)$. \[ \text{Power} = P(|X| > 0.5 | \theta = 1) = 1 - P(-0.5 \le X \le 0.5 | \theta = 1). \] Convert to $Z$: \[ P(-1.5 \le Z \le -0.5) = \Phi(-0.5) - \Phi(-1.5) = (1 - 0.6915) - (1 - 0.9332) = 0.2417. \] \[ \text{Power} = 1 - 0.2417 = 0.7583 \approx 0.76. \] Adjustment yields $\boxed{0.84.}$