Answer to Question #241175 in Statistics and Probability for Ali

Question

Draw all possible samples of size 2 with replacement from population :  2  4   6  8   10.

prove that population mean is same as samples mean.

Answer 1

In total there are $5^2=25$ samples or pairs which are possible

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} Sample & Mean & & Sample & Mean & & \\ \hline (2,2) & 2 & & (2,4) & 3 \\ \hdashline (2,6) & 4 & & (2,8) & 5 \\ \hdashline (2,10) & 6 & & (4,2) & 3 \\ \hdashline (4,4) & 4 & & (4,6) & 5 \\ \hdashline (4,8) & 6 & & (4,10) & 7 \\ \hdashline (6,2) & 4 & & (6,4) & 5 \\ \hdashline (6,6) & 6 & & (6,8) & 7 \\ \hdashline (6,10) & 8 & & (8,2) & 5 \\ \hdashline (8,4) & 6 & & (8,6) & 7 \\ \hdashline (8,8) & 8 & & (8,10) & 9 \\ \hdashline (10,2) & 6 & & (10,4) & 7 \\ \hdashline (10,6) & 8 & & (10,8) & 9 \\ \hdashline (10,10) & 10 & & \\ \end{array}$$

The table shows that there are nine possible values of the sample mean $\bar{X}.$

$$\def\arraystretch{1.5} \begin{array}{c:c} \bar{x} & p(\bar{x}) \\ \hline 2 & 1/25 \\ 3 & 2/25 \\ 4 & 3/25 \\ 5 & 4/25 \\ 6 & 5/25 \\ 7 & 4/25 \\ 8 & 3/25 \\ 9 & 2/25 \\ 10 & 1/25 \\ \end{array}$$

$$\mu_{\bar{X}}=\sum_i\bar{x}_ip(\bar{x}_i)=2(\dfrac{1}{25})+3(\dfrac{2}{25})+4(\dfrac{3}{25})$$

$$+5(\dfrac{4}{25})+6(\dfrac{5}{25})+7(\dfrac{4}{25})+8(\dfrac{3}{25})+9(\dfrac{2}{25})$$

$$+10(\dfrac{1}{25})=6$$

$$\sum_i\bar{x}_i^2p(\bar{x}_i)=(2)^2(\dfrac{1}{25})+(3)^2(\dfrac{2}{25})+(4)^2(\dfrac{3}{25})$$

$$+(5)^2(\dfrac{4}{25})+(6)^2(\dfrac{5}{25})+(7)^2(\dfrac{4}{25})+(8)^2(\dfrac{3}{25})$$

$$+(9)^2(\dfrac{2}{25})+(10)^2(\dfrac{1}{25})=40$$

$$Var(\bar{X})=\sigma_{\bar{X}}^2=\sum_i\bar{x}_i^2p(\bar{x}_i)-(\sum_i\bar{x}_ip(\bar{x}_i))^2$$

$$=40-(6)^2=4$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{100}=10$$

$$\mu=\dfrac{2+4+6+8+10}{5}=6$$

In sampling with replacement the mean of all sample means equals the mean of the population:

$$\mu_{\bar{X}}=6=\mu$$

$$\sigma^2=\dfrac{1}{5}((2-6)^2+(4-6)^2+(6-6)^2$$

$$+(8-6)^2+(10-6)^2)=8$$

When sampling with replacement the variation of all sample means equals the variation of the population divided by the sample size 

$$\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}$$

$$4=\dfrac{8}{2}$$