Answer to Question #258501 in Statistics and Probability for justine

Question

A population consists of four numbesr (3, 8, 10, 15). Consider all possible sample of size 2 that can be drawn without replacement from the population.

Find the following:

a. Population Mean

b. Population Variance

c. Population Standard Deviation

d. The Mean of the sampling distribution of means

e. The Standard deviation of the sampling distribution of means

Answer 1

a.

 We have population values $3,8,10,15,$ population size $N=4$

$$\mu=\dfrac{3+8+10+15}{4}=9$$

b.

$$\sigma^2=\dfrac{1}{4}((3-9)^2+(8-9)^2+(10-9)^2$$

$$+(15-9)^2)=18.5$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{18.5}\approx4.3012$$

d. We have population values $3,8,10,15,$ population size $N=4$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{4}{2}=6$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 3,8 & 5.5 \\ \hdashline 2 & 3,10 & 6.5 \\ \hdashline 3 & 3,15 & 9 \\ \hdashline 4 & 8,10 & 9 \\ \hdashline 5 & 8,15 & 11.5 \\ \hdashline 6 & 10,15 & 12.5 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 5.5 & 1 & 1/6 & 11/12 & 121/24 \\ \hdashline & 6.5 & 1& 1/6 & 13/12 & 169/24 \\ \hdashline & 9 & 2 & 1/3 & 3 & 27 \\ \hdashline & 11.5 & 1& 1/6 & 23/12 & 529/24 \\ \hdashline & 12.5 & 1 & 1/6 & 25/12 & 625/24 \\ \hdashline Total & & 6 & 1 & 9 & 523/6 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=9.8$$

$$\mu=\dfrac{6+8+10+12+13}{5}=9.8$$

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

$$E(\bar{X})=\mu_{\bar{X}}=9.8=\mu$$

e.

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

$$=\dfrac{523}{6}-(9)^2=\dfrac{37}{6}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{37}{6}}\approx2.4833$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{18.5}{2}(\dfrac{4-2}{4-1})$$

$$=\dfrac{37}{6}, True$$