A general method for finding a confidence interval for the difference between two means of normal populations is given by expression (5.21) on page 365. A pooled method that can be used when the variances of the populations are known to be equal is given by expression (5.22) on page 367. This exercise is designed to compare the coverage probabilities of these methods under a variety of conditions. A fair amount of coding may be required, depending on the software used.

a. Let n_X = 10, n_Y = 10, σ_X = 1, and σ_Y = 1. Generate 10,000 pairs of samples: X^∗₁, . . . , X^∗_nX from a N(0, σ²_X ) distribution, and Y^∗₁ , . . . , Y^∗_nY from a N(0, σ²_Y ) distribution. For each pair of samples, compute a 95% confidence interval using the general method, and a 95% confidence interval using the pooled method. Note that each population has mean 0, so the true difference between the means is 0. Estimate the coverage probability for each method by computing the proportion of confidence intervals that cover the true value 0. 

b. Repeat part (a), using n_X = 10, n_Y = 10, σ_X = 1, and σ_Y = 5.

c. Repeat part (a), using n_X = 5, n_Y = 20, σ_X = 1, and σ_Y = 5.

d. Repeat part (a), using n_X = 20, n_Y = 5, σ_X = 1, and σ_Y = 5.

e. Does the coverage probability for the general method differ substantially from 95% under any of the conditions in parts (a) through (d)? (It shouldn’t.)

f. Based on the results in parts (a) through (d), under which conditions does the pooled method perform most poorly?

i. When the sample sizes are equal and the variances differ.

ii. When both the sample sizes and the variances differ, and the larger sample comes from the population with the larger variance.

iii. When both the sample sizes and the variances differ, and the smaller sample comes from the population with the larger variance.