The designer of a sample survey stratifies a population into two strata, H and L
ID: 3233104 • Letter: T
Question
The designer of a sample survey stratifies a population into two strata, H and L. H contains 10,000 people, and L contains 20,000. He decides to allocate 50 samples to stratum H and 50 to stratum L, taking a simple random sample in each stratum. Suppose that the population standard deviation in stratum H is sigma_H = 10 and the standard deviation in stratum L is sigma_L = 5. (a) How should the designer estimate the population mean? The stratified estimate of mu is X_st^- = Sigma_t = 1^L W_l X_l^- = 1/3X_H^- + 2/3 X_L^- (b) What will be the standard error of his estimate? Var(X_st^-) = Sigma_l = 1^L (1 - n_l -1/N_l - 1) approximately 0.221 + 0.222 = 0.443 implies sigma_X_st approximately 0.665 (c) Would it be better to use proportional allocation? In proportional allocation: n_H = 5 times 100 approximately 33, n_L = 2/3 times 100 approximately 67: implies Var(X_sp^-) approximately 1/n Sigma W_l sigma_l^2 = 0.5 > Var (X_st^-) = 0.443 (d) Find the optimal allocation for estimating the population mean (ignore the finite population correction). n_H = 1/3 times 10/1/3 times 10 + 2/3 times 5 times 100 = 50, n_L = n - n_H = 50Explanation / Answer
1) This is the correct estimate of population mean. We take a proportional average.
2) THe correct formula is used since we are combining two populations.
3) We prefer a method with lower variance. Hence proportional allocation is used.
4) The optimal allocation is one in which both have the same standard deviation.
Hence, Standard deviation of H = 1/3*10
Total standard deviation = 1/3*10+2/3*5 = 20/3
hence, nL and nH can be calculated.