Mathematical Statistics and Data Analysis - Solutions

(a)

Estimate of the population mean, ${\overline{X}}_s=\frac{1}{N}\sum _{l=1}^L{N}_l{\overline{X}}_l$, which as proved in the book is an unbiased estimate of population mean.

Thus $\mu =\mathrm{E}({\overline{X}}_s)=\frac{1}{100,000+500,000}(100,000\mathrm{E}({\overline{X}}_H)+500,000\mathrm{E}({\overline{X}}_L))=\frac{1}{6}\mathrm{E}({\overline{X}}_H)+\frac{5}{6}\mathrm{E}({\overline{X}}_L)$.

(b)

Ignoring the finite population correction, We have $\mathrm{Var}({\overline{X}}_s)=\sum _{l=1}^L{W}_l^2\frac{1}{n_l}{\sigma }_l^2={\left(\frac{1}{6}\right)}_{}^2\frac{1}{100}{20}^2+{\left(\frac{5}{6}\right)}_{}^2\frac{1}{200}{10}^2=\frac{1}{9}+\frac{25}{72}=0.4583$. 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 $\mathrm{Var}({\overline{X}}_s)=\sum _{l=1}^L{W}_l^2\frac{1}{n_l}{\sigma }_l^2={\left(\frac{1}{6}\right)}_{}^2\frac{1}{200}{20}^2+{\left(\frac{5}{6}\right)}_{}^2\frac{1}{100}{10}^2=\frac{1}{18}+\frac{25}{36}=0.75$. 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:

$\mathrm{Var}({\overline{X}}_{SP})=\frac{1}{n}\sum _{l=1}^L{W}_l{\sigma }_l^2=\frac{1}{300}{(\frac{1}{6}{20}^2+\frac{5}{6}{10}^2)}_{}=\frac{900}{1800}=0.5$. 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.