Let us take a sample from population X from a normal distribution with unknown mean an, and unknown variance oi, and population Y from normal distribution with unknown mean ny and unknown variance 03,. Suppose that we want to nd the interval estimation of differences between two population means a: ,rry. Let 171,172,--- ,1:,,,, denote the random sample of size nz form the population X and let y1, y2, - - - , yny denote the random sample of size 1a,, form the population Y. Assume that the populations X and Y or: = or: = 02. We have used in our class the usual estimator of the common variance :32 pooling the sample data to obtain the pooled variance estimator si: lid-5'1; if]? + 2:1(14'2' E32 a3 + as, 2 is obtained by a = Let s: and s: be the sample variances of sample X and sample Y respectively. a) Show that sf: is an unbiased estimator for the common variance o2. 13] Show that if um = y, sEF is simply the average of s: and 3:. C} Show that if my, 72 my and \"a: 2::- ny, 312:. is the weighted average of a: and 5:, with larger weight given to sample X. d] A study has been made to compare the nicotine contents of two brands of cigarettes. Ten cigarettes of Brand A has an average nicotine content of 3.1 milligrams with standard deviation of 0.5 milligrams, while eight cigarettes of Brand B had an average nicotine content to 2.? milligrams with a standard deviation [1.7 milligrams. Assuming that the two sets of data are independent random samples from normal populations with equal variances, construct a a 95% condence interval for leg #3, the true difference between the mean nicotine contents of the two brands of cigarettes. You mayr use ZTestU or t.test() to verify your answer, but replicate the values mathematically yourself and show your work as well. 9} RBpeat part d under the assumption that the variances of each brand are not equal. This time, you may use ZTestO or t.testl[) on its own. Explain how this problem is different from part d, but no need to mathematically verify the values