Answer to Question #348297 in Statistics and Probability for Nalyn

Question

List all the possible samples and the corresponding mean

Answer 1

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

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

Variance of population 

$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{4+1+0+1+4}{5}=2$

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 $^{N}C_n=^{5}C_2=10.$

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

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 3/2 & 1/10 & 3/20 & 9/40 \\ \hdashline 4/2 & 1/10 & 4/20 & 16/40 \\ \hdashline 5/2 & 2/10 & 10/20 & 50/40 \\ \hdashline 6/2 & 2/10 & 12/20 & 72/40 \\ \hdashline 7/2 & 2/10 & 14/20 & 98/40 \\ \hdashline 8/2 & 1/10 & 8/20 & 64/40 \\ \hdashline 9/2 & 1/10 & 9/20 & 81/40 \\ \hdashline \end{array}$

Mean of sampling distribution 

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{60}{20}=3=\mu$

The variance of sampling distribution 

$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$ $=\dfrac{390}{40}-(3)^2=\dfrac{3}{4}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$