Answer to Question #147362 in Statistics and Probability for Rehan Tahir

Question

A population consists of 10, 10, 14, 14, 16, 18 and 20.

i. Calculate the sample means for all possible random samples of size 2, that can be drawn from this population, without replacement.
ii. Verify that
a. = µ
b. = σ2 / n (N-n/N-1)
iii. Also draw all possible samples of size 2 with replacement and verify the following properties:-
a. = µ
b. = σ2 / n

Answer 1

$$\mu=\dfrac{10+10+14+14+16+18+20}{7}=\dfrac{102}{7}$$

$$\sigma^2=\dfrac{1}{7}\bigg((10-\dfrac{102}{7})^2+(10-\dfrac{102}{7})^2$$

$$+(14-\dfrac{102}{7})^2+(14-\dfrac{102}{7})^2+(16-\dfrac{102}{7})^2$$

$$+(18-\dfrac{102}{7})^2+(20-\dfrac{102}{7})^2\bigg)$$

$$=\dfrac{600}{49}$$

i. The number of possible samples which can be drawn without replacement is

$$\dbinom{7}{2}=21$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample\ values & Sample\ mean (\bar{X}) & f \\ \hline 10,10 & 10 & 1 \\ 10,14 & 12 & 4 \\ 10,16 & 13 & 2 \\ 10,18 & 14 & 2 \\ 10,20 & 15 & 2 \\ 14,14 & 14 & 1 \\ 14,16 & 15 & 2 \\ 14,18 & 16 & 2 \\ 14,20 & 17 & 2 \\ 16,18 & 17 & 1 \\ 16,20 & 18 & 1 \\ 18,20 & 19 & 1 \end{array}$$

a.

$$E(\bar{X})=\dfrac{1}{21}\bigg(10(1)+12(4)+13(2)+14(3)$$

$$+15(4)+16(2)+17(3)+18(1)+19(1)\bigg)$$

$$=\dfrac{102}{7}$$

b.

$$E(\bar{X}^2)=\dfrac{1}{21}\bigg(10^2(1)+12^2(4)+13^2(2)+14^2(3)$$

$$+15^2(4)+16^2(2)+17^2(3)+18^2(1)+19^2(1)\bigg)$$

$$=\dfrac{1522}{7}$$

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

$$=\dfrac{1522}{7}-(\dfrac{102}{7})^2=\dfrac{250}{49}$$

$$\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{\dfrac{600}{49}}{2}(\dfrac{7-2}{7-1})$$

$$=\dfrac{250}{49}=Var(\bar{X})$$

$E(\bar{X})=\mu=\dfrac{102}{7}$

$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{250}{49}$

ii.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample\ values & Sample\ mean (\bar{X}) & Probability \\ \hline 10,10 & 10 & \dfrac{2}{7}(\dfrac{2}{7})\\ \\ 10,14 & 12 & \dfrac{2}{7}(\dfrac{2}{7}) \\ \\ 10,16 & 13 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 10,18 & 14 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 10,20 & 15 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 14,10 & 12 & \dfrac{2}{7}(\dfrac{2}{7}) \\ \\ 14,14 & 14 & \dfrac{2}{7}(\dfrac{2}{7}) \\ \\ 14,16 & 15 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 14,18 & 16 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 14,20 & 17 & \dfrac{2}{7}(\dfrac{1}{7}) \\ \\ 16,10 & 13 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 16,14 & 15 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 16,16 & 16 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 16,18 & 17 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 16,20& 18 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 18,10 & 14 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 18,14 & 16 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 18,16 & 17 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 18,18 & 18 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 18,20 & 19 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 20,10 & 15 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 20,14 & 17 & \dfrac{1}{7}(\dfrac{2}{7}) \\ \\ 20,16 & 18 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 20,18 & 19 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ 20,20 & 20 & \dfrac{1}{7}(\dfrac{1}{7}) \\ \\ \end{array}$$ $$E(\bar{X})=\dfrac{1}{49}\bigg(10(4)+12(8)+13(4)+14(8)$$

$$+15(8)+16(5)+17(6)+18(3)+19(2)+20(1)\bigg)$$

$$=\dfrac{102}{7}$$

b.

$$E(\bar{X}^2)=\dfrac{1}{49}\bigg(10^2(4)+12^2(8)+13^2(4)+14^2(8)$$

$$+15^2(8)+16^2(5)+17^2(6)+18^2(3)+19^2(2)+(20^2(1)\bigg)$$ $$=\dfrac{10704}{49}$$

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

$$=\dfrac{10704}{49}-(\dfrac{102}{7})^2=\dfrac{300}{49}$$

$$\dfrac{\sigma^2}{n}=\dfrac{\dfrac{600}{49}}{2}=\dfrac{300}{49}=Var(\bar{X})$$

$E(\bar{X})=\mu=\dfrac{102}{7}$

$Var(\bar{X})=\dfrac{\sigma^2}{n}=\dfrac{300}{49}$