STAT 410 96-2 Assignment 6 Solutions

5.1: For this population: Ybar1 = 1, Ybar2 = 7, S_1^2=1 and S_2^2 =13. You also have N1=N2=3 so W1=W2=1/2. Neyman allocation gives n1=4 x 3 x sqrt(1) / (3 x sqrt(1) + 3 x sqrt(13)) and n2=4-n1 which round off to 1 and 3 respectively.

For the Neyman allocation there are only 3 possible samples: you always take all 3 from stratum 2 and there are 3 possibilities from stratum 1. Here is a table of sample and corresponding statistic values. Note that all three samples have probability 1/3 of being drawn.

Sample	ybar 1	ybar 2	ybar_st	ybar_st - Ybar
0; 4,6,11	0	7	3.5	-0.5
1; 4,6,11	1	7	4	0
2; 4,6,11	2	7	4.5	0.5

Averaging the values of ybar_st gives E(ybar_st) = (3.5+4+4.5)/3 = 4 = Ybar. Then square and average the last column to get var(ybar_st) = 1/6.

For proportional allocation n1=n2=2 and there are 3C2 x 3C2 = 9 possible samples all having probability 1/9. The corresponding table is:

Sample	ybar 1	ybar 2	ybar_st	ybar_st - Ybar
0,1; 4,6	0.5	5	2.75	-1.25
0,1; 4,11	0.5	7.5	4	0
0,1; 6,11	0.5	8.5	4.5	0.5
0,2; 4,6	1	5	3	-1
0,2; 4,11	1	7.5	4.25	0.25
0,2; 6,11	1	8.5	4.75	0.75
1,2; 4,6	1.5	5	3.25	-0.75
1,2; 4,11	1.5	7.5	4.5	0.5
1,2; 6,11	1.5	8.5	5	1

Averaging the values of ybar_st gives E(ybar_st) = 4 = Ybar. Then square and average the last column to get var(ybar_st) = 85/36.

The rest of this problem is just plugging into formulas.

5.2: Part a asks for the Neyman allocation. You know that S_1 = c sqrt(Ybar1) and S_2 = c sqrt(Ybar2). You also are given Ybar1 = 9 Ybar2. Now plug these in to 5.26 and you will see that c and the Ybars all cancel out. You get n1=n x N1 x c x sqrt(Ybar 1) / ( N1 x c x sqrt(Ybar 1) + N2 x c x sqrt(Ybar 2)) = n x N1 x 3 x sqrt(Ybar 2) / ( N1 x 3 sqrt(Ybar 2) + N2 x sqrt (Ybar 2) ) = 1000 x 4000 x 3 / ( 4000 x 3 + 20000)=3000/8=375. Then n2=625.

For b you must minimize var(ybar1 - ybar2 ) = S_1^2/n1+S_1^2/n2 subject to n1+n2=1000. Use the equations in a above to see that S1 = 3 x S2 and see that you must minimize S2^2 x (9/n1 + 1/(1000-n1)). Differentiate with respect to n1 and set the result equal to 0. You will see the answer does not depend on the actual value of S2. Alternatively see that this is just the same as minimizing var(ybarst) = W_1^2 S_1^2/n1 + W_2^2 S_1^2/n2 with the special choice W_1 = W_2 =1 . So you can use the Neyman allocation formulas with W_1=W_2=1.

5.3 Note: my copy of the text has a typo: "compute the same sizes" should be "compute the sample sizes". This question just calls for you to plug into 5.26 and work out the corresponding standard error of ybar st as well as doing the same for proportional allocation. To compare with the precision of SRS you must find S^2 for the whole population: see 5.32.

5.5: For a given cost C and allocations w1 and w2 the total sample size will be n=C/(c1 w1 + c2 w2). For proportional allocation the w's are the same as the W's and the variance of ybar_st is W1^2 S^2/(nW1) + W2^2 S^2/(nW2) which simplifies to S^2(W1+W2)/n = S^2 / n(using S to denote the common population SD in the two strata). Putting in the value of n gives a variance for proportional allocation of S(c1 W1 + c2 W2)/C.
For optimal allocation wi is given by [Wi S / sqrt(ci)] / [ W1 S / sqrt(c1) + W2 S / sqrt(c2) ] which simplifies to [Wi / sqrt(ci)] / [ W1 / sqrt(c1) + W2 / sqrt(c2) ]. The variance of the stratified estimate is then S^2 [W1 sqrt(c1) +W2 sqrt(c2) ] [ W1 / sqrt(c1) + W2 / sqrt(c2) ] / n. Now n=C/(c1 w1 + c2 w2) simplifies to n = C [ W1 / sqrt(c1) + W2 / sqrt(c2) ] / [W1 sqrt(c1) + W2 sqrt(c2) ] and the variance becomes S^2 [W1 sqrt(c1) +W2 sqrt(c2) ]^2 / C Taking the ratio of these two variance formulae the C and S terms cancel and we are left with [c1 W1 + c2 W2]/ [W1 sqrt(c1) +W2 sqrt(c2) ]^2. If W1=W2 then both are 1/2 and the formula is 2(c1+c2)/(sqrt(c1)+sqrt(c2))^2 so just plug in c2 = 2 c1 and c2 = 4 c1.

5.8: For part a you must minimize n1+n2 subject to (0.8)^2 2^2 / n1 + (0.2)^2 4^2 / n2 = (0.1)^2. This is Neyman allocation so the solution has n1/n = 0.8 x 2 / ( 0.8 x 2 + 0.2 x 4)=2/3. Plug n1=2n/3 and n2=n/3 into the variance formula and solve for n.
For part b you must minimize n=n1+n2 subject to S1^2 / n1 + S2^2/n2 = (0.1)^2. The solution is found from the Neyman allocation formula with W1=W2=1 (because the constraint is a special case of the allocation constraint formula with w1=W2=1) and so n1/n = S1/(S1+S2) = 1/3. Put n1=n/3 and n2=2n/3 in the variance formula and solve for n.

5.10: This is a Lagrange multiplier problem: minimize sum (W_h^2S_h^2 (1-f_h)/n_h) subject to sum t_h sqrt(n_h) = C-C0. The derivative of the Lagrangian f-lambda g wrt n_h is -W_h^2S_h^2/n_h^2 - lambda t_h / (2 sqrt(n_h)) which gives n_h = (-W_h^2 S_h^2/(lambda t_h))^(2/3). The question does not ask you to find lambda, but merely to note that the answer is proportional to (W_h^2 S_h^2 / t_h)^(2/3).

5.14: For part a use Neyman allocation with the Pi's in 5.60 taken as the midpoints of the ranges given. For part b plug in the P_h values given to formula 5.57. For part c the answer is sqrt(PQ/400) where P = (.62)(.48)+(.31)(.21) + (.07)(.04) and Q=1-P.

The questions.