Answer to Question #340289 in Statistics and Probability for Imyn

Question

Consider the population consisting of the values 1, 3 and 4. List all the possible samples of size 2 that can be drawn from the population with replacement. Then, compute the mean x— for each sample. Lastly, find the mean of the sampling distribution of means and the mean of the population.

Answer 1

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

The number of possible samples which can be drawn with replacement is

$N^n=3^2=9.$

$$S=\{(1,1), (1,3), (1,4),(3,1), (3,3), (3,4),$$

$$(4,1), (4,3), (4,4)\}$$

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

C.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 2/2 & 1/9 & 2/18 & 4/36 \\ \hdashline 4/2 & 2/9 & 8/18 & 16/36 \\ \hdashline 5/2 & 2/9 & 10/18 & 25/36\\ \hdashline 6/2 & 1/9 & 6/18 & 36/36\\ \hdashline 7/2 & 2/9 & 14/18 & 49/36 \\ \hdashline 8/2 & 1/9 & 8/18 & 648/36 \\ \hdashline \end{array}$$

Mean of population 

$$\mu=\dfrac{1+3+4}{3}=\dfrac{8}{3}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{8}{3}=\mu$$