Answer to Question #341210 in Statistics and Probability for Melquian

Question

Consider a population consisting of 2, 4, 6, 8 and 10. Suppose samples of size 3 are drawn from this population.

a. Describe the sampling distribution of the sample means

b. What are the mean and variance of the sampling distribution of the sample means?

c. Construct a histogram for the sampling distribution.

Answer 1

We have population values 2,4,6,8,10, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{2+4+6+8+10}{5}=6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{16+4+0+4+16}{5}=8$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2}\approx2.8284$$

a. Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,4,6 & 12/3 \\ \hdashline 2 & 2,4,8 & 14/3 \\ \hdashline 3 & 2,4,10 & 16/3 \\ \hdashline 4 & 2,6,8 & 16/3 \\ \hdashline 5 & 2,6,10 & 18/3 \\ \hdashline 6 & 2,8,10 & 20/3 \\ \hdashline 7 & 4,6,8 & 18/3 \\ \hdashline 8 & 4,6,10 & 20/3 \\ \hdashline 9 & 4,8,10 & 22/3 \\ \hdashline 10 & 6,8,10 & 24/3 \\ \hdashline \end{array}$$

b.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 12/3 & 1/10 & 12/30 & 144/90 \\ \hdashline 14/3 & 1/10 & 14/30 & 196/90 \\ \hdashline 16/3 & 2/10 & 32/30 & 512/90 \\ \hdashline 18/3 & 2/10 & 36/30 & 648/90 \\ \hdashline 20/3 & 2/10 & 40/30 & 800/90 \\ \hdashline 22/3 &1/10 & 22/30 & 484/90 \\ \hdashline 24/3 & 1/10 & 24/30 & 576/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{180}{30}=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{3360}{90}-(6)^2=\dfrac{4}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{4}{3}}\approx1.1547$$

c.