Question

Let \( X_1, X_2, X_3, X_4 \) be a random sample from \( N(\theta_1, \sigma^2) \) distribution and \( Y_1, Y_2, Y_3, Y_4 \) be a random sample from \( N(\theta_2, \sigma^2) \) distribution, where \( \theta_1, \theta_2 \in (-\infty, \infty) \) and \( \sigma>0 \). Further suppose that the two random samples are independent. For testing the null hypothesis \( H_0 : \theta_1 = \theta_2 \) against the alternative hypothesis \( H_1 : \theta_1 \neq \theta_2 \), suppose that a test \( \psi \) rejects \( H_0 \) if and only if \( \sum_{i=1}^{4} X_i>\sum_{i=1}^{4} Y_i \). The power of the test \( \psi \) at \( \theta_1 = 1 + \sqrt{2}, \theta_2 = 1 \) and \( \sigma^2 = 4 \) is

• A. 0.5987
• B. 0.7341
• C. 0.7612
• D. 0.8413

Hint:

To calculate the power of a test, find the probability of rejecting the null hypothesis under the alternative hypothesis.

Answer 1

The Correct Option is D

Step 1: Understanding the hypothesis test.
We are testing the hypothesis \( H_0 : \theta_1 = \theta_2 \) against the alternative hypothesis \( H_1 : \theta_1 \neq \theta_2 \) using the statistic \( \sum_{i=1}^{4} X_i - \sum_{i=1}^{4} Y_i \). Under the null hypothesis, \( \theta_1 = \theta_2 \), so the test statistic follows a normal distribution with mean 0.
Step 2: Finding the power of the test.
The power of the test is the probability of rejecting \( H_0 \) when the true value of the parameters is \( \theta_1 = 1 + \sqrt{2} \) and \( \theta_2 = 1 \). This involves calculating the cumulative distribution function (CDF) of the normal distribution for the given parameters.
Step 3: Final probability.
Using the power formula for hypothesis tests, we find that the power of the test at \( \theta_1 = 1 + \sqrt{2}, \theta_2 = 1 \) and \( \sigma^2 = 4 \) is approximately 0.8413.