Question

A population consists of three numbers (3, 6 ,9). Consider all possible samples of sizes 2 which can be drawn without replacement from the population. Find the following

Answer 1

I.

$$mean=\mu=\dfrac{3+6+9}{3}=6$$

$$variance=\sigma^2=\dfrac{1}{3}((3-6)^2+(6-6)^2$$

$$+(9-6)^2)=6$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{6}$$

II. There are $\dbinom{3}{2}=3$ samples of size two which can be drawn without replacement: 

$$\begin{matrix} Sample & Sample\ mean \\ (3,6) & 4.5 \\ (3,9) & 6 \\ (6,9) & 7.5 \\ \end{matrix}$$

III.

$$\begin{matrix} \bar{X} & P(\bar{X}) \\ 4.5 & 1/3 \\ 6 & 1/3 \\ 7.5 & 1/3 \\ \end{matrix}$$

IV.

$$\mu_{\bar{X}}=4.5(1/3)+6(1/3)+7.5(1/3)=6$$

$$\sum_i\bar{X}_i^2P(\bar{X_i})=4.5^2(1/3)+6^2(1/3)+7.5^2(1/3)$$ $$=37.5$$

$$\sigma_{\bar{X}}^2=\sum_i\bar{X}_i^2P(\bar{X_i})-\mu_{\bar{X}}^2=37.5-6^2=1.5$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{1.5}$$

$$\mu_{\bar{X}}=6, \sigma_{\bar{X}}=\sqrt{1.5}$$

V.

The mean $\mu_{\bar{X}}$ and standard deviation $\sigma_{\bar{X}}$ of the sample mean $\bar{X}$ satisfy

$$\mu_{\bar{X}}=6=\mu,$$

$$\sigma_{\bar{X}}=\sqrt{1.5}=\dfrac{\sqrt{6}}{\sqrt{2}}\sqrt{\dfrac{3-2}{3-1}}=\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}$$