Answer to Question #241176 in Statistics and Probability for Ali

Question

Draw all possible samples of size 2 from samples  10   20  30   40   50 with replacement .

Prove that mean of samples mean is same as the population mean and form empirical relation between samples variance and population variance.

Answer 1

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} Sample & Mean & & Sample & Mean & & \\ \hline (10,10) & 10 & & (10,20) & 15 \\ \hdashline (10,30) & 20 & & (10,40) & 25 \\ \hdashline (10,50) & 30 & & (20,10) & 15 \\ \hdashline (20,20) & 20 & & (20,30) & 25 \\ \hdashline (20,40) & 30 & & (20,50) & 35 \\ \hdashline (30,10) & 20 & & (30,20) & 25 \\ \hdashline (30,30) & 30 & & (30,40) & 35 \\ \hdashline (30,50) & 40 & & (40,10) & 25 \\ \hdashline (40,20) & 30 & & (40,30) & 35 \\ \hdashline (40,40) & 40 & & (40,50) & 45 \\ \hdashline (50,10) & 30 & & (50,20) & 35 \\ \hdashline (50,30) & 40 & & (50,40) & 45 \\ \hdashline (50,50) & 50 & & \\ \end{array}$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c} \bar{x} & p(\bar{x}) \\ \hline 10 & 1/25 \\ 15 & 2/25 \\ 20 & 3/25 \\ 25 & 4/25 \\ 30 & 5/25 \\ 35 & 4/25 \\ 40 & 3/25 \\ 45 & 2/25 \\ 50 & 1/25 \\ \end{array}$$

$$\mu_{\bar{X}}=\sum_i\bar{x}_ip(\bar{x}_i)=10(\dfrac{1}{25})+15(\dfrac{2}{25})+20(\dfrac{3}{25})$$

$$+25(\dfrac{4}{25})+30(\dfrac{5}{25})+35(\dfrac{4}{25})+40(\dfrac{3}{25})+45(\dfrac{2}{25})$$

$$+50(\dfrac{1}{25})=30$$

$$\sum_i\bar{x}_i^2p(\bar{x}_i)=(10)^2(\dfrac{1}{25})+(15)^2(\dfrac{2}{25})+(20)^2(\dfrac{3}{25})$$

$$+(25)^2(\dfrac{4}{25})+(30)^2(\dfrac{5}{25})+(35)^2(\dfrac{4}{25})+(40)^2(\dfrac{3}{25})$$

$$+(45)^2(\dfrac{2}{25})+(50)^2(\dfrac{1}{25})=1000$$

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

$$=1000-(30)^2=100$$

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

$$\mu=\dfrac{10+20+30+40+50}{5}=30$$

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

$$\mu_{\bar{X}}=30=\mu$$

$$\sigma^2=\dfrac{1}{5}((10-30)^2+(20-30)^2+(30-30)^2$$

$$+(40-30)^2+(50-30)^2)=200$$

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

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

$$100=\dfrac{200}{2}$$