Question

Let \( X_1, X_2, X_3 \) be a random sample from a Poisson distribution with mean \( \lambda \), \( \lambda > 0 \). For testing \( H_0: \lambda = \frac{1}{8} \) against \( H_1: \lambda = 1 \), a test rejects \( H_0 \) if and only if \( X_1 + X_2 + X_3 > 1 \). The power of this test is equal to __________ (round off to 2 decimal places).

Answer 1

Correct Answer: 0.79 - 0.81

1. Distribution Under \( H_1 \): - If \( X_1, X_2, X_3 \sim \text{Poisson}(\lambda) \), then \( T = X_1 + X_2 + X_3 \sim \text{Poisson}(3\lambda) \). - Under \( H_1: \lambda = 1 \), \( T \sim \text{Poisson}(3) \). 2. Power of the Test: - Power is the probability of rejecting \( H_0 \) under \( H_1 \): \[ \text{Power} = P(T > 1 \mid \lambda = 1) = 1 - P(T \leq 1 \mid \lambda = 1). \] 3. Compute \( P(T \leq 1 \mid \lambda = 1) \): - From the Poisson distribution: \[ P(T = 0) = \frac{e^{-3}3^0}{0!} = e^{-3}, \quad P(T = 1) = \frac{e^{-3}3^1}{1!} = 3e^{-3}. \] - Therefore: \[ P(T \leq 1) = e^{-3} + 3e^{-3} = 4e^{-3}. \] 
4. Compute the Power: Using \( e^{-3} \approx 0.0498 \): \[ P(T \leq 1) = 4 \cdot 0.0498 \approx 0.1992. \] - Power: \[ \text{Power} = 1 - P(T \leq 1) = 1 - 0.1992 = 0.8008. \]