Question

A population consists of five numbers 1, 2, 3, 4, 5 and 6. Suppose samples of size 2 are drawn from this

population.

(a) Find the mean and variance of the population

(b) Describe the sampling distribution of the sample means.

(c) Find the mean and variance of the sampling distribution of the sample means.

Answer 1

(a) We have population values 1,2,3,4,5,6, population size N=6 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{1+2+3+4+5+6}{6}=3.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{1}{6}(6.25+2.25+0.25$$

$$+0.25+2.25+6.25)=\dfrac{17.5}{6}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{17.5}{6}}\approx1.707825$$

(b) The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_2=15.$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 3/2 & 1/15 & 3/30 & 9/60 \\ \hdashline 4/2 & 1/15 & 4/30 & 16/60 \\ \hdashline 5/2 & 2/15 & 10/30 & 50/60 \\ \hdashline 6/2 & 2/15 & 12/30 & 72/60 \\ \hdashline 7/2 & 3/15 & 21/30 & 147/60 \\ \hdashline 8/2 & 2/15 & 16/30 & 128/60 \\ \hdashline 9/2 & 2/15 & 18/30 & 162/60 \\ \hdashline 10/2 & 1/15 & 10/30 & 100/60 \\ \hdashline 11/2 & 1/15 & 11/30 & 121/60 \\ \hdashline \end{array}$$

(c) Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{105}{30}=3.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{805}{60}-(3.5)^2=\dfrac{7}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{7}{6}}\approx1.080123$$