Question

A population consists of the five numbers 2,5,6, 8,and 11. Consider samples of size 2 that can be drawn from the population.

A. List the possible samples and the corresponding mean.

Answer 1

We have population values 2,5,6,8,11, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{2+5+6+8+11}{5}=6.4$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(19.36+1.96+0.16$$

$$+2.56+21.16)=9.04$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{9.04}\approx3.00666$$

A. 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 $^{N}C_n=^{5}C_2=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,5 & 7/2 \\ \hdashline 2 & 2,6 & 8/2 \\ \hdashline 3 & 2,8 & 10/2 \\ \hdashline 4 & 2,11 & 13/2 \\ \hdashline 5 & 5,6 & 11/2 \\ \hdashline 6 & 5,8 & 13/2 \\ \hdashline 7 & 5,11 & 16/2 \\ \hdashline 8 & 6,8 & 14/2 \\ \hdashline 9 & 6,11 & 17/2 \\ \hdashline 10 & 8,11 & 19/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 7/2 & 1/10 & 7/20 & 49/40 \\ \hdashline 8/2 & 1/10 & 8/20 & 64/40 \\ \hdashline 10/2 & 1/10 & 10/20 & 100/40 \\ \hdashline 11/2 & 1/10 & 11/20 & 121/40 \\ \hdashline 13/2 & 2/10 & 26/20 & 338/40 \\ \hdashline 14/2 & 1/10 & 14/20 & 196/40 \\ \hdashline 16/2 & 1/10 & 16/20 & 256/40 \\ \hdashline 17/2 & 1/10 & 17/20 & 289/40 \\ \hdashline 19/2 & 1/10 & 19/20 & 361/40 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{128}{20}=6.4=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1774}{40}-(6.4)^2=3.39= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{3.39}\approx1.8412$$