Answer to Question #206729 in Statistics and Probability for Maria Leonora Racs

Question

Consider all the possible samples of size 2 that can be drawn with replacement from the population 1, 2, 3, 4, 5, and 6. Create a sampling distribution of the sample mean

Answer 1

We have population values $1,2,3,4,5,6$ population size $N=6$ and sample size $n=2.$Thus, the number of possible samples which can be drawn with replacement is

$$(N)^n=6^2=36$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,1 & 1.0 \\ \hdashline 2 & 1,2 & 1.5 \\ \hdashline 3 & 1,3 & 2.0 \\ \hdashline 4 & 1,4 & 2.5\\ \hdashline 5 & 1,5 & 3.0 \\ \hdashline 6 & 1,6 & 3.5 \\ \hline 7 & 2,1 & 1.5 \\ \hline 8 & 2,2 & 2.0 \\ \hline 9 & 2,3 & 2.5 \\ \hline 10 & 2,4 & 3.0 \\ \hline 11 & 2,5 & 3.5 \\ \hline 12 & 2,6 & 4.0 \\ \hline 13 & 3,1 & 2.0 \\ \hline 14 & 3,2 & 2.5 \\ \hline 15 & 3,3 & 3.0 \\ \hline 16 & 3,4 & 3.5 \\ \hline 17 & 3,5 & 4.0 \\ \hline 18 & 3,6 & 4.5 \\ \hline 19 & 4,1 & 2.5 \\ \hline 20 & 4,2 & 3.0 \\ \hline 21 & 4,3 & 3.5 \\ \hline 22 & 4,4 & 4.0 \\ \hline 23 & 4,5 & 4.5 \\ \hline 24 & 4,6 & 5.0 \\ \hline 25 & 5,1 & 3.0 \\ \hline 26 & 5,2 & 3.5 \\ \hline 27 & 5,3 & 4.0 \\ \hline 28 & 5,4 & 4.5 \\ \hline 29 & 5,5 & 5.0 \\ \hline 30 & 5,6 & 5.5 \\ \hline 31 & 6,1 & 3.5 \\ \hline 32 & 6,2 & 4.0 \\ \hline 33 & 6,3 & 4.5 \\ \hline 34 & 6,4 & 5.0 \\ \hline 35 & 6,5 & 5.5 \\ \hline 36 & 6,6 & 6.0 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 1.0& 1 & 1/36 & 2/72 & 4/144 \\ \hdashline 1.5 & 2 & 2/36 & 6/72 & 18/144 \\ \hdashline 2.0 & 3 & 3/36 & 12/72 & 48/144 \\ \hdashline 2.5 & 4 & 4/36 & 20/72 & 100/144 \\ \hdashline 3.0 & 5 & 5/36& 30/72 & 180/144 \\ \hdashline 3.5 & 6 & 6/36 & 42/72 & 294/144 \\ \hdashline 4.0 & 5 & 5/36 & 40/72 & 320/144\\ \hdashline 4.5 & 4 & 4/36 & 36/72 & 324/144 \\ \hdashline 5.0 & 3 & 3/36 & 30/72 & 300/144\\ \hdashline 5.5 & 2 & 2/36 & 22/72 & 242/144 \\ \hdashline 6.0 & 1& 1/36 & 12/72 & 144/144\\ \hdashline Total & 36 & 1 & 3.5 & 329/24 \\ \hline \end{array}$$

Mean

$$\mu=\dfrac{1+2+3+4+5+6}{6}=3.5$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=3.5$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=3.5=\mu$$

Variance

$$\sigma^2=\dfrac{1}{6}\big((1-3.5)^2+(2-3.5)^2+(3-3.5)^2$$ $$(4-3.5)^2+(5-3.5)^2+(6-3.5)^2\big)=\dfrac{8.75}{3}$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{8.75}{3}}\approx1.7078$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$

$$=\dfrac{329}{24}-(\dfrac{7}{2})^2=\dfrac{35}{24}$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{35}{24}}\approx1.2076$$

If a population is infinite and the sampling is random or if the population is finite and the sampling is with replacement, then the variance of the sampling distribution of means, $Var(\bar{X})$ is given by  

$$Var(\bar{X})=\dfrac{\sigma^2}{n}$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}=\dfrac{8.75}{3(2)}=\dfrac{8.75}{6}=\dfrac{35}{24}, True$$