Answer to Question #305599 in Statistics and Probability for Bentong

Question

A population consists of the four numbers 1,2,4 and 5 .list all the possible samples of size n=3 which can be drawn with repalcement from the population.find the following:

A .population mean b. Population variance c.population standard deviation d.mean of the sampling distribution of sample means e. Variance of the sampling distribution of sample mean f. Standard deviation of the sampling distribution of sample means

Answer 1

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

Thus, the number of possible samples which can be drawn with replacement is $4^3=64.$

a.

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

b.

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

$$=2.5$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{2.5}\approx1.58114$$

d.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 1,1,1 & 1\\ \hdashline 1,1,2 & 4/3\\ \hdashline 1,1,4 & 2\\ \hdashline 1,1,5 & 7/3\\ \hdashline 1,2,1 & 4/3\\ \hdashline 1,2,2 & 5/3\\ \hdashline 1,2,4 & 7/3\\ \hdashline 1,2,5 & 8/3\\ \hdashline 1,4,1 & 2\\ \hdashline 1,4,2 & 7/3\\ \hdashline 1,4,4 & 3\\ \hdashline 1,4,5 & 10/3\\ \hdashline 1,5,1 & 7/3\\ \hdashline 1,5,2 & 8/3\\ \hdashline 1,5,4 & 10/3\\ \hdashline 1,5,5 & 11/3\\ \hdashline 2,1,1 & 4/3\\ \hdashline 2,1,2 & 5/3\\ \hdashline 2,1,4 & 7/3\\ \hdashline 2,1,5 & 8/3\\ \hdashline 2,2,1 & 5/3\\ \hdashline 2,2,2 & 2\\ \hdashline 2,2,4& 8/3\\ \hdashline 2,2,5 & 3\\ \hdashline 2,4,1 & 7/3\\ \hdashline 2,4,2 & 8/3\\ \hdashline 2,4,4 & 10/3\\ \hdashline 2,4,5 & 11/3\\ \hdashline 2,5,1 & 8/3\\ \hdashline 2,5,2 & 3\\ \hdashline 2,5,4 & 11/3\\ \hdashline 2,5,5 & 4\\ \hdashline 4,1,1 & 2\\ \hdashline 4,1,2 & 7/3\\ \hdashline 4,1,4 & 3\\ \hdashline 4,1,5 & 10/3\\ \hdashline 4,2,1 & 7/3\\ \hdashline 4,2,2 & 8/3\\ \hdashline 4,2,4 & 10/3\\ \hdashline 4,2,5 & 11/3\\ \hdashline 4,4,1 & 3\\ \hdashline 4,4,2 & 10/3\\ \hdashline 4,4,4 & 4\\ \hdashline 4,4,5 & 13/3\\ \hdashline 4,5,1 & 10/3\\ \hdashline 4,5,2 & 11/3\\ \hdashline 4,5,4 & 13/3\\ \hdashline 4,5,5 & 14/3\\ \hdashline 5,1,1 & 7/3\\ \hdashline 5,1,2 & 8/3\\ \hdashline 5,1,4 & 10/3\\ \hdashline 5,1,5 & 11/3\\ \hdashline 5,2,1 & 8/3\\ \hdashline 5,2,2 & 3\\ \hdashline 5,2,4 & 11/3\\ \hdashline 5,2,5 &4\\ \hdashline 5,4,1 & 10/3\\ \hdashline 5,4,2 & 11/3\\ \hdashline 5,4,4 & 13/3\\ \hdashline 5,4,5 & 14/3\\ \hdashline 5,5,1 &11/3\\ \hdashline 5,5,2 & 4\\ \hdashline 5,5,4 & 14/3\\ \hdashline 5,5,5 & 5\\ \hdashline \end{array}$$

The sampling distribution of the sample mean $\bar{X}$ is

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 1 & 1 & 1/64 & 1/64 & 3/192\\ \hdashline & 4/3 & 3 & 3/64 & 4/64 & 16/192 \\ \hdashline & 5/3 & 3 & 3/64 & 5/64 & 25/192\\ \hdashline & 2 & 4 & 4/64 & 8/64 & 48/192 \\ \hdashline & 7/3 & 9 & 9/64 & 21/64 & 147/192 \\ \hdashline & 8/3 & 9 & 9/64 & 24/64 & 192/192 \\ \hdashline & 3 & 6 & 6/64 & 18/64 & 162/192 \\ \hdashline & 10/3 & 9 & 9/64 & 30/64 & 300/192 \\ \hdashline & 11/3 & 9 & 9/64 & 33/64 & 363/192 \\ \hdashline & 4 & 4 & 4/64 & 16/64 & 192/192 \\ \hdashline & 13/3 & 3 & 3/64 & 13/64 & 169/192 \\ \hdashline & 14/3 & 3 & 3/64 & 14/64 & 196/192 \\ \hdashline & 5 & 1 & 1/64 & 5/64 & 75/192 \\ \hdashline Sum= & & 64 & 1 & 3 & 59/6 \\ \hdashline \end{array}$$

e. The mean of the sample means is

$$\mu_{ \bar{X}}=E(\bar{X})=3$$

$$Var(\bar{X})=\sigma^2_{\bar{X}}=E(\bar{X}^2)-(E(\bar{X}))^2$$

$$=\dfrac{59}{6}-(3)^2=\dfrac{5}{6}$$

$$\mu_{\bar{X}}=\mu$$

$$\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}$$

f.

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{5}{6}}\approx0.91287$$