Question

A population consists of the four numbers 1,2,8 and 9. List all the possible sample of size n=2 which can be drawn without replacement from the population find the following:

a. Population mean

b. Population variance

c. Population standard deviation

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means

Answer 1

We have population values $1,2,8,9$ population size $N=4,$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{4}{2}=6$$

a.

$$mean=\mu=\dfrac{1+2+8+9}{4}=5$$

b.

$$Variance=\sigma^2$$

$$=\dfrac{1}{4}((1-5)^2+(2-5)^2+(8-5)^2$$

$$+(9-5)^2)=12.5$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{12.5}=2.5\sqrt{2}\approx3.535534$$

d.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} No & Sample & Mean \\ \hline 1 & (1,2) & 1.5 \\ \hdashline 2 & (1,8) & 4.5 \\ \hdashline 3 & (1,9) & 5 \\ \hdashline 4 & (2,8) & 5 \\ \hdashline 5 & (2,9) & 5.5 \\ \hdashline 6 & (8,9) & 8.5 \\ \hdashline \end{array}$$

The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 1.5 & 1 & 1/6 & 3/12 & 9/24 \\ \hdashline 4.5 & 1 & 1/6 & 9/12 & 81/24 \\ \hdashline 5 & 2 & 2/6 & 20/12 & 200/24 \\ \hdashline 5.5 & 1 & 1/6 & 11/12 & 121/24 \\ \hdashline 8.5 & 1 & 1/6 & 17/12 & 289/24 \\ \hdashline Total & 6 & 1 & 5 &175 /6 \\ \hdashline \end{array}$$

$$E(\bar{X})=\sum\bar{x}f(\bar{x})=5$$

e.

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

$$=\dfrac{175}{6}-5^2=\dfrac{25}{6}$$

f.

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{25}{6}}=\dfrac{5\sqrt{6}}{6}\approx2.041241$$

Check

$$E(\bar{X})=5=\mu$$

$$Var(\bar{X})=\dfrac{25}{6}=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{12.5}{2}(\dfrac{4-2}{4-1})$$