Please solve

2. Let {X1, X2, ..., Any be the midterm scores from Section 1 and {Y1, 12, ...; I'm} be those from Section 2. We assume that the scores in Section 1 are Normal(#1, o') r.v.s and those of Section 2 are Normal(#2, o') r.v.s, where #1, #2, and the common variance o' are unknown parameters of interest. In addition, scores are independent r.v.s within and between sections. Our interest is to see whether there is a meaningful difference between the (population) averages, i.e., #1 - #2. (a) (3 pts) The most intuitive estimator for / - #2 is X -Y. Find the pivotal quantity of X - Y and its distribution. (b) (3 pts) The unknown variance o' is unknown, so we will replace it in the pivotal quantity in (a) with its good estimator (based on the principle of statistics). Since the unknown variance is the same for both sections, it makes sense to estimate of by pooling the two sample variances as follows: (n - 1)S* + (m - 1)S; n + m - 2 where SY = EL(X - X)?/(n - 1) and $, = EZ, (Y -Y)?/(m - 1). This pooled sample variance is our estimator of o'. Find the distribution of the pivotal quantity, (n + m -2)5,/o2. (c) (3 pts) Considering the distributions of the pivotal quantities in (a) and (b), find the distri- bution of (X - Y) - (ux - HY) SpViti which can be obtained by replacing the unknown common variance o in the pivotal quantity in (a) with the pooled variance defined in (b). (d) (3 pts) Derive the 95% confidence interval for #1 - /2, using the exact method. Let us assume that the notation 90.025 Or 90.975 indicate quantiles of the distribution in (c) that make the left and right tail areas equal to 0.025, respectively. Also note that 90.025 = -90.975 if the distribution in (c) is symmetric about 0. So (0 + 90.025 , + 90.975 ) is equivalent to the most common format for an interval estimator (0 - 90.975, 0 + 90.975)