Chapter 7, Survey Sampling
Solution 55
(a)
Estimate of the population mean, $\, \bar X_s = \frac 1 N \sum_{l=1}^{L} N_l \bar X_l \,$, which as proved in the book is an unbiased estimate of population mean.
Thus $\, \mu = \Exp(\bar X_s) = \frac 1 {100,000 + 500,000} (100,000 \Exp(\bar X_H) + 500,000 \Exp(\bar X_L)) = \frac 1 6 \Exp(\bar X_H) + \frac 5 6 \Exp(\bar X_L) \,$.
(b)
Ignoring the finite population correction, We have $\, \Var(\bar X_s) = \sum^{L}_{l=1} W_l^2 \frac 1 {n_l} \sigma_l^2 = \Prn{\frac 1 6}^2 \frac 1 {100} {20}^2 + \Prn{\frac 5 6}^2 \frac 1 {200} {10}^2 = \frac {1} {9} + \frac {25} {72} = 0.4583 \,$. Thus standard error = $\, \sqrt { 0.4583 } = 0.6770 \,$.
(c)
If the standard error is less than the previous alllocation than the new allocation will be better. Thus, we first find standard error:
Ignoring the finite population correction, We have $\, \Var(\bar X_s) = \sum^{L}_{l=1} W_l^2 \frac 1 {n_l} \sigma_l^2 = \Prn{\frac 1 6}^2 \frac 1 {200} {20}^2 + \Prn{\frac 5 6}^2 \frac 1 {100} {10}^2 = \frac {1} {18} + \frac {25} {36} = 0.75 \,$. Thus standard error = $\, \sqrt { 0.75 } = 0.866 \,$.
The standard error for the present allocation is higher than that of the previous allocation, thus this allocation is not better than the previous allocation.
(d)
We first find the standard error for the proportional allocation:
$\, \Var(\bar X_{SP}) = \frac 1 n \sum_{l=1}^{L} W_l \sigma_l^2 = \frac 1 {300} \Prn{\frac 1 6 {20}^2 + \frac 5 6 {10}^2 } = \frac {900} {1800} = 0.5 \,$. Thus standard error is $\, \sqrt{0.5} = 0.707 \,$.
Here also standard error is more than the original allocation. Thus this allocation is also not better compared to the original allocation.
$$\tag*{$\blacksquare$} $$