Question

Let \( X_1, X_2 \) be a random sample from an \( N(0, \theta) \) distribution, where \( \theta>0 \). Then the value of \( k \), for which the interval \[ \left( 0, \frac{X_1^2 + X_2^2}{k} \right) \] is a 95% confidence interval for \( \theta \), equals

• A. \( 1 - \log_e(0.95) \)
• B. \( 2 \log_e(0.95) \)
• C. \( \frac{1}{2} \log_e(0.95) \)
• D. 2

Hint:

For confidence intervals based on a chi-squared distribution, the value of the parameter is determined using the appropriate quantile.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are given a random sample \( X_1, X_2 \) from a \( N(0, \theta) \) distribution, and we need to find the value of \( k \) for which the interval \( \left( 0, \frac{X_1^2 + X_2^2}{k} \right) \) is a 95% confidence interval for \( \theta \).
Step 2: Using the chi-squared distribution.
The sum of squares of \( X_1 \) and \( X_2 \) follows a chi-squared distribution with 2 degrees of freedom. For a 95% confidence interval, the critical value \( k \) is related to the chi-squared distribution quantile for the 95% level.
Step 3: Final value of \( k \).
Using the appropriate chi-squared distribution quantiles, we find the value of \( k \) is \( 2 \log_e(0.95) \).