### 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.

$$\tag*{$\blacksquare$} $$