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:

population mean

population variance and standard deviation

mean of each sample and the mean of the sampling distributions of means

variance of the sampling distributions of means

standard deviation of the sampling distribution of the means.

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.

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