Question

our hypothetical population contains the scores 5, 7, 9, and 11. determine the mean and variance of the sampling distribution of the sample mean., given that samples contain two scores drawn from the population with replacement.

Answer 1

We have population values 5,7,9,11, population size N=4 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{5+7+9+11}{4}=8$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{9+1+1+9}{4}=5$$

Select a random sample of size 2 with replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn with replacement is $N^n=4^2=16.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 5,5 & 5 \\ \hdashline 2 & 5,7 & 6 \\ \hdashline 3 & 5,9 & 7 \\ \hdashline 4 & 5,11 & 8 \\ \hdashline 5 & 7,5 & 6 \\ \hdashline 6 & 7,7 & 7 \\ \hdashline 7 & 7,9 & 8 \\ \hdashline 8 & 7,11 & 9 \\ \hdashline 9 & 9,5 & 7 \\ \hdashline 10 & 9,7 & 8 \\ \hdashline 11 & 9,9 & 9 \\ \hdashline 12 & 9,11 & 10 \\ \hdashline 13 & 11,5 & 8 \\ \hdashline 14 & 11,7 & 9 \\ \hdashline 15 & 11,9 & 10 \\ \hdashline 16 & 11,11 & 11 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 5 & 1/16 & 5/16 & 25/16 \\ \hdashline 6 & 2/16 & 12/16 & 72/16 \\ \hdashline 7 & 3/16 & 21/16 & 147/16 \\ \hdashline 8 & 4/16 & 32/16 & 256/16 \\ \hdashline 9 & 3/16 & 27/16 & 243/16 \\ \hdashline 10 & 2/16 & 20/16 & 200/16 \\ \hdashline 11 & 1/16 & 11/16 & 121/16 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{128}{16}=8=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1064}{16}-(8)^2=2.5= \dfrac{\sigma^2}{n}$$