Answer to Question #345111 in Statistics and Probability for jn cepeda

Question

Consider a population consist of 9, 5, 6, 12, and 15. Suppose that sample size of 2were drawn from this population (without replacement), describe the sampling distribution of the sample means

Answer 1

We have population values 5,6,9,12,15, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{5+6+9+12+15}{5}=9.4$

Variance of population 

$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(19.36+11.56$

$+0.16+6.76+31.36)=13.84$

$\sigma=\sqrt{13.84}\approx3.720215$

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 $^{N}C_n=^{5}C_2=10.$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 5,6 & 11/2 \\ \hdashline 2 & 5,9 & 14/2 \\ \hdashline 3 & 5,12 & 17/2 \\ \hdashline 4 & 5,15 & 20/2 \\ \hdashline 5 & 6,9 & 15/2 \\ \hdashline 6 & 6,12 & 18/2 \\ \hdashline 7 & 6,15 & 21/2 \\ \hdashline 8 & 9,12 & 21/2 \\ \hdashline 9 & 9,15 & 24/2 \\ \hdashline 10 & 12,15 & 27/2 \\ \hdashline \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 11/2 & 1/10 & 11/20 & 121/40 \\ \hdashline 14/2 & 1/10 & 14/20 & 196/40 \\ \hdashline 15/2 & 1/10 & 15/20 & 225/40 \\ \hdashline 17/2 & 1/10 & 17/20 & 289/40 \\ \hdashline 18/2 & 1/10 & 18/20 & 324/40 \\ \hdashline 20/2 & 1/10 & 20/20 & 400/40 \\ \hdashline 21/2 & 2/10 & 42/20 & 882/40 \\ \hdashline 24/2 & 1/10 & 24/20 & 576/40 \\ \hdashline 27/2 & 1/10 & 27/20 & 729/40 \\ \hdashline \end{array}$

Mean of sampling distribution 

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{188}{20}=9.4=\mu$

The variance of sampling distribution 

$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$ $=\dfrac{3742}{40}-(5.4)^2=5.19= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$