Question

This item is 10 points. A population consists of the five measurements 1, 5, 3, 8, and 10 . Suppose samples of size 2 are drawn from this population. 

Compute and find the following:

1) Find the variance of the population.

2) Find the variance of the sampling distribution. 

Answer 1

1) We have population values 1,3,5,8,10, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{1+3+5+8+10}{5}=5.4$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(19.36+5.76$$ $$+0.16+6.76+21.16)=10.64$$

2) 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 $^{N}C_n=^{5}C_2=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,3 & 4/2 \\ \hdashline 2 & 1,5 & 6/2 \\ \hdashline 3 & 1,8 & 9/2 \\ \hdashline 4 & 1,10 & 11/2 \\ \hdashline 5 & 3,5 & 8/2 \\ \hdashline 6 & 3,8 & 11/2 \\ \hdashline 7 & 3,10 & 13/2 \\ \hdashline 8 & 5,8 & 13/2 \\ \hdashline 9 & 5,10 & 15/2 \\ \hdashline 10 & 8,10 & 18/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 4/2 & 1/10 & 4/20 & 16/40 \\ \hdashline 6/2 & 1/10 & 6/20 & 36/40 \\ \hdashline 8/2 & 1/10 & 8/20 & 64/40 \\ \hdashline 9/2 & 1/10 & 9/20 & 81/40 \\ \hdashline 11/2 & 2/10 & 22/20 & 242/40 \\ \hdashline 13/2 & 2/10 & 26/20 & 338/40 \\ \hdashline 15/2 & 1/10 & 15/20 & 225/40 \\ \hdashline 18/2 & 1/10 & 18/20 & 324/40 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{108}{20}=5.4=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1316}{40}-(5.4)^2=3.99= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$