Question

You are given a population of 3 elements , which are 3,4,5. Suppose all possible samples of size 2 are drawn from the population with replacement, compute for the following;

Mean of the sample means;

Variance of the sample means; and standard deviation of the sample means

Answer 1

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

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

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1+0+1}{3}=\dfrac{2}{3}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{2}{3}}\approx0.8165$$

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

The number of possible samples which can be drawn with replacement is $N^n=3^2=9$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 6/2 & 1/9 & 6/18 & 36/36 \\ \hdashline 7/2 & 2/9 & 14/18 & 98/36\\ \hdashline 8/2 & 3/9 & 24/18 & 192/36 \\ \hdashline 9/2 & 2/9 & 18/18 & 162/36 \\ \hdashline 10/2 & 1/9 & 10/18 & 100/36 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{72}{18}=4=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{588}{36}-(4)^2= \dfrac{1}{3}= \dfrac{\sigma^2}{n}$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{1}{3}}\approx0.57735$$