Question

Given a population consists of three numbers (2,5,9). Consider all possible samples of size 2 which can be drawn from the population. Find the standard deviation of the sampling distribution of the sample means.

Answer 1

a.

 We have population values $2,5,9$ population size $N=3$

$$\mu=\dfrac{2+5+9}{3}=\dfrac{16}{3}$$ $$\sigma^2=\dfrac{1}{3}((2-16/3)^2+(5-16/3)^2+(9-16/3)^2$$

$$=\dfrac{74}{9}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{74}{9}}=\dfrac{\sqrt{74}}{3}$$

We have population values $2,5,9$ population size $N=3$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{3}{2}=3$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,5 & 7/2 \\ \hdashline 2 & 2,9 & 11/2 \\ \hdashline 3 & 5,9 & 7 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 7/2 & 1 & 1/3 & 7/6 & 49/12 \\ \hdashline & 11/2 & 1 & 1/3 & 11/6 & 121/12 \\ \hdashline & 7 & 1 & 1/3 & 7/3 & 49/3 \\ \hdashline Total & & 3 & 1 & 16/3 & 183/6 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{16}{3}$$

The mean of the sampling distribution of the sample means is equal to the the mean of the population.

$$E(\bar{X})=\mu_{\bar{X}}=\dfrac{16}{3}=\mu$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$

$$=\dfrac{183}{6}-(\dfrac{16}{3})^2=\dfrac{37}{18}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{37}{18}}=\dfrac{\sqrt{74}}{6}$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{\dfrac{74}{9}}{2}(\dfrac{3-2}{3-1})$$

$$=\dfrac{37}{18}, True$$ $$\bar{X}\sim N(\dfrac{16}{3}, \dfrac{37}{18})$$