Question

Consider a population consisting of 2,6,8,0 and 1. Suppose samples of size 2 are drawn from this population. Find the Mean and Variance of the sampling distribution of sample means.

Answer 1

We have population values 0,1,2,6,8, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{0+1+2+6+8}{5}=3.4$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{1}{5}(11.56+5.76+1.96$$

$$+6.76+21.16=9.44$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{9.44}\approx3.072458$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_2=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 0,1 & 1/2 \\ \hdashline 2 & 0,2 & 2/2 \\ \hdashline 3 & 0,6 & 6/2\\ \hdashline 4 & 0,8 & 8/2 \\ \hdashline 5 & 1,2 & 3/2 \\ \hdashline 6 & 1,6 & 7/2 \\ \hdashline 7 & 1,8 & 9/2 \\ \hdashline 8 & 2,6 & 8/2 \\ \hdashline 9 & 2,8 & 10/2 \\ \hdashline 10 & 6,8 & 14/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 1/2 & 1/10 & 1/20 & 1/40 \\ \hdashline 2/2 & 1/10 & 2/20 & 4/40 \\ \hdashline 3/2& 1/10 & 3/20 & 9/40 \\ \hdashline 6/2 & 1/10 & 6/20 & 36/40 \\ \hdashline 7/2 & 1/10 & 7/20 & 49/40 \\ \hdashline 8/2 & 2/10 & 16/20 & 128/40 \\ \hdashline 9/2 & 1/10 & 9/20 & 81/40 \\ \hdashline 10/2 & 1/10 & 10/20 & 100/40 \\ \hdashline 14/2 & 1/10 & 14/20 & 196/40 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{68}{20}=3.4=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{604}{40}-(3.4)^2=3.54= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{3.54}\approx1.881489$$