Answer to Question #339730 in Statistics and Probability for Lleesh

Question

Consider the population of the values (1,2,3,4).

A. List all the possible samples of size 2 with replacement

B. Compute the mean of each sample.

C. Identify the probability of each sample.

D. Compute the mean of the sampling distribution of the means.

Answer 1

A.We have population values 1,2,3,4, population size N=5 and sample size n=2.

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

$N^n=4^2=16.$

$$S=\{(1,1), (1,2), (1,3), (1,4),$$

$$(2,1), (2,2), (2,3), (2,4),$$

$$(3,1), (3,2), (3,3), (3,4),$$

$$(4,1), (4,2), (4,3), (4,4)\}$$

B.

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

C.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 2/2 & 1/16 & 2/32 & 4/64 \\ \hdashline 3/2 & 2/16 & 6/32 & 18/64 \\ \hdashline 4/2 & 3/16 & 12/32 & 48/64 \\ \hdashline 5/2 & 4/16 & 20/32 & 100/64 \\ \hdashline 6/2 & 3/16 & 18/32 & 108/64 \\ \hdashline 7/2 & 2/16 & 14/32 & 98/64 \\ \hdashline 8/2 & 1/16 & 8/32 & 64/64 \\ \hdashline \end{array}$$

D.

Mean of population 

$$\mu=\dfrac{1+2+3+4}{4}=2.5$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=2.5=\mu$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{4}(2.25+0.25+0.25$$

$$+2.25)=1.25$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{440}{64}-(2.5)^2=0.625= \dfrac{\sigma^2}{n}$$