Q1. Consider a hypothetical population of size 8 with an auxiliary variable x that has known values given by x1 = 3, x2 = 7, x3 =16, x4 = 20, x5 = 26, x6 = 33, x7 = 40, x8 = 55. To select a sample of size 3 from this population, a unit is first drawn with probability proportional to the value of x1, and then two more units are drawn from the remaining 7 units using simple random sampling.
(a) What is the probability of selecting the sample {2, 5, 7} under this sampling scheme? Hint: There are 3 possibilities, depending on which unit is drawn first. For subsequent draws, you can use the facts that the first and second order inclusion probabilities under simple random sampling are given by n/N and n(n -1)/N(N-1), where n is the sample size, and N the population size.
(b) What is the probability that unit 7 is included in the sample?
(c) What is the probability that units 5 and 7 are included in the sample?
Q2. A first phase simple random sample S of size n (with sample mean x- for variable x) is drawn from a finite population U of size N with population mean x-U and population variance S2x for variable x. At phase 2, a subsample S2 of size n2 < n is drawn from the phase 1 samples using S simple random sampling. It follows naturally that S2 is a simple random sample of size n2 from population U (no need to prove this).
(a) Let x-2 be the mean of x calculated from subsample S2, write down E(x-2) and var(x-2).
(b) Show that E((x-2 - x-)x-)= 0 . You may use the fact that E((x-2- x-)x-)= E{E2((x-2-x-)x-)} where E2 denote the conditional expectation under the second phase sampling given the phase 1 sample.
(c) Use (b) to show that cov(x-2 - x-,x-) = 0 and hence cov(x-2, x-) = var(x-) .
(d) Use (c) to show that var(x-2 - x-) = var(x-2) - var(x-) and simplify it to (1/n2 - 1/n)S2x.