Question

Direction: Solve the problem below.

1. A population consists of three numbers (3, 6, 9). Consider all possible samples of size

2 which can be drawn without replacement from the population. Find the following:

a. Population mean

b. Population variance

c. Populations standard deviation

d. Mean of the sampling distribution of the means

e. Variance of the sampling distribution of the means

f. Standard deviation of the sampling distributions of the means.

Answer 1

a)

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

b)

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

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

c)

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

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

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

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

e)

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

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

f)

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

Check

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

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

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

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