Question

A population consists of the four members 6,9,15,18. Consider all possible samples of

size two which can be drawn with replacement from the population. Find population

mean, the standard deviation and the mean of the sampling distribution of means and

standard deviation of sampling distribution of means.

Answer 1

We have population values $6,9,15,18,$ population size $N=4$

$$\mu=\dfrac{6+9+15+18}{4}=12$$ $$\sigma^2=\dfrac{1}{4}((6-12)^2+(9-12)^2+(15-12)^2$$ $$+(18-12)^2)=\dfrac{45}{2}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{45}{2}}=3\sqrt{2.5}\approx4.7434$$

Sample size is $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{4}{2}=6$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 6,9 & 7.5 \\ \hdashline 2 & 6,15 & 10.5 \\ \hdashline 3 & 6,18 & 12 \\ \hdashline 4 & 9,15 & 12 \\ \hdashline 5 & 9,18 & 13.5 \\ \hdashline 6 & 15,18 & 16.5 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 7.5 & 1 & 1/6 & 15/12 & 75/8\\ \hdashline & 10.5 & 1& 1/6 & 21/12 & 147/8 \\ \hdashline & 12 & 2 & 1/3 & 4 & 48 \\ \hdashline & 13.5 & 1& 1/6 & 27/12 & 243/8 \\ \hdashline & 16.5 & 1 & 1/6 & 33/12 & 363/8\\ \hdashline Total & & 6 & 1 & 12 & 303/2 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=12$$

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

$$E(\bar{X})=\mu_{\bar{X}}=12=\mu$$

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

$$=\dfrac{303}{2}-(12)^2=\dfrac{15}{2}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{15}{2}}\approx2.7386$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{45}{2(2)}(\dfrac{4-2}{4-1})$$ $$=\dfrac{15}{2}, True$$