Question

Let \( X_1, X_2, X_3 \) be a random sample from \( N(\mu_1, \sigma_1^2) \) distribution and \( Y_1, Y_2, Y_3 \) be a random sample from \( N(\mu_2, \sigma_2^2) \) distribution. Also, assume that \( (X_1, X_2, X_3) \) and \( (Y_1, Y_2, Y_3) \) are independent. Let the observed values of \( \sum_{i=1}^{3} \left[ X_i - \frac{1}{3} (X_1 + X_2 + X_3) \right]^2 \) and \( \sum_{i=1}^{3} \left[ Y_i - \frac{1}{3} (Y_1 + Y_2 + Y_3) \right]^2 \) be 1 and 5, respectively. Then the solution for the 90% confidence interval for \( \mu_1 - \mu_2 \) equals ................

Hint:

For confidence intervals involving differences of means, use the formula involving the sample standard deviations and appropriate z or t values, depending on the sample size and distribution.

Answer 1

Correct Answer: 4.2

Step 1: Apply the properties of the sample variance.
Using the sample variances of the two random samples, we apply the formula for the confidence interval for the difference between two means: \[ CI = \left( \hat{\mu_1} - \hat{\mu_2} \right) \pm z_{\alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}. \] Step 2: Conclusion.
Using the given observed values and degrees of freedom, we find the appropriate z-value and compute the confidence interval for \( \mu_1 - \mu_2 \). The final value is: \[ \boxed{0.15}. \]