Answer to Question #343428 in Statistics and Probability for Love

Question

Using the number 5, 10, 15, 20, 25, and 30 as the element of the population, construct the sampling distribution of the sample means with a sample size of 2. Find the population and sample mean, population and sample variance, and population and sample standard deviation.

Answer 1

We have population values 5,10,15,20,25,30, population size N=6 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{5+10+15+20+25+30}{6}=17.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(156.25+56.25+6.25$$ $$+6.25+56.25+156.25)=\dfrac{437.5}{6}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{437.5}{6}}\approx8.5391$$

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 $^{N}C_n=^{6}C_2=15.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 5,10 & 15/2 \\ \hdashline 2 & 5,15 & 20/2 \\ \hdashline 3 & 5,20 & 25/2 \\ \hdashline 4 & 5,25 & 30/2 \\ \hdashline 5 & 5,30 & 35/2 \\ \hdashline 6 & 10,15 & 25/2 \\ \hdashline 7 & 10,20 & 30/2 \\ \hdashline 8 & 10,25 & 35/2 \\ \hdashline 9 & 10,30 & 40/2 \\ \hdashline 10 & 15,20 & 35/2 \\ \hdashline 11 & 15,25 & 40/2 \\ \hdashline 12 & 15,30 & 45/2 \\ \hdashline 13 & 20,25 & 45/2 \\ \hdashline 14 & 20,30 & 50/2 \\ \hdashline 15 & 25,30 & 55/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 15/2 & 1/15 & 3/6 & 45/12 \\ \hdashline 20/2 & 1/15 & 4/6 & 80/12 \\ \hdashline 25/2 & 2/15 & 10/6 & 250/12 \\ \hdashline 30/2 & 2/15 & 12/6 & 360/12 \\ \hdashline 35/2 & 3/15 & 21/6 & 735/12 \\ \hdashline 40/2 & 2/15 & 16/6 & 640/12 \\ \hdashline 45/2 & 2/15 & 18/6 & 810/12 \\ \hdashline 50/2 & 1/15 & 10/6 & 500/12 \\ \hdashline 55/2 & 1/15 & 11/6 & 605/12 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{105}{6}=17.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{4025}{12}-(17.5)^2=\dfrac{175}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{175}{6}}\approx5.4006$$