Answer to Question #342106 in Statistics and Probability for Joseph

Question

A group of the following students got the following score in a test:6,9,12,15,and18 compute the mean of the sample mean

Answer 1

We have population values 6,9,12,15,18, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+18}{5}=12$

Variance of population 

$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{36+9+0+9+36}{5}=18$

$\sigma=\sqrt{\sigma^2}=\sqrt{18}\approx4.24264$

Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,12 & 9 \\ \hdashline 2 & 6,9,15 & 10 \\ \hdashline 3 & 6,9,18 & 11 \\ \hdashline 4 & 6,12,15 & 11 \\ \hdashline 5 & 6,12,18 & 12 \\ \hdashline 6 & 6,15,18 & 13 \\ \hdashline 7 & 9,12,15 & 12 \\ \hdashline 8 & 9, 12,18 & 13 \\ \hdashline 9 & 9,15,18 & 14 \\ \hdashline 10 & 12, 15,18 & 15 \\ \hdashline \end{array}$

B.

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9 & 1/10 & 9/10 & 81/10 \\ \hdashline 10 & 1/10 & 10/10 & 100/10 \\ \hdashline 11 & 2/10 & 22/10 & 242/10 \\ \hdashline 12 & 2/10 & 24/10 & 288/10 \\ \hdashline 13 & 2/10 & 26/10 & 338/10 \\ \hdashline 14 & 1/10 & 14/10 & 196/10 \\ \hdashline 15 & 1/10 & 15/10 & 225/10 \\ \hdashline \end{array}$

Mean of sampling distribution 

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{120}{10}=12=\mu$

The variance of sampling distribution 

$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$ $=\dfrac{1470}{10}-(12)^2=3= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$

$\sigma_{\bar{X}}=\sqrt{3}\approx1.73205$