Question

A population consist of three numbers (2, 4, 6). Consider all possible samples of size 2 which can be drawn without

replacement from the population. Find the following:

Answer 1

1. We have population values 2,4,6, population size N=3 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{2+4+6}{3}=4$

2.Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{4+0+4}{3}=\dfrac{8}{3}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{8}{3}}\approx1.633$$

Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{3}C_2=3.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,4 & 3 \\ \hdashline 2 & 2,6 & 4 \\ \hdashline 3 & 4,6 & 5 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 3 & 1/3 & 3/3 & 9/3 \\ \hdashline 4 & 1/3 & 4/3 & 16/3 \\ \hdashline 5 & 1/3 & 5/3 & 25/3 \\ \end{array}$$

3. Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{12}{3}=4=\mu$$

4. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{50}{3}-(4)^2=\dfrac{2}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

5.

$$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{\dfrac{2}{3}}\approx0.8165$$