Answer to Question #303181 in Statistics and Probability for Kimberly

Question

A population consist of the values 2,4,6 and 8

a) List all the possible sample size 2 when drawing with replacement and compute the mean of each sample.

b)Construct the sampling distribution of the mean.

c) Compute the mean, variance and standard deviation of the sample mean then compare these with the mean, Variance and standard deviation of the population.

Answer 1

We have population values $2,4,6,8$ population size $N=4$ and sample size $n=2.$

Thus, the number of possible samples which can be drawn with replacement is $4^2=16.$

a.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 2,2 & 2\\ \hdashline 2,4 & 3\\ \hdashline 2,6 & 4\\ \hdashline 2,8 & 5\\ \hdashline 4,2 & 3\\ \hdashline 4,4 & 4\\ \hdashline 4,6 & 5\\ \hdashline 4,8 & 6\\ \hdashline 6,2 & 4\\ \hdashline 6,4 & 5\\ \hdashline 6,6 & 6\\ \hdashline 6,8 & 7\\ \hdashline 8,2 & 5\\ \hdashline 8,4 & 6\\ \hdashline 8,6 & 7\\ \hdashline 8,8 & 8\\ \hdashline \end{array}$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 2 & 1 & 1/16 & 2/16 & 4/16\\ \hdashline & 3 & 2 & 2/16 & 6/16 & 18/16 \\ \hdashline & 4 & 3 & 3/16 & 12/16 & 48/16\\ \hdashline & 5 & 4 & 4/16 & 20/16 & 100/16 \\ \hdashline & 6 & 3 & 3/16 & 18/16 & 108/16 \\ \hdashline & 7 & 2 & 2/16 & 14/16 & 98/16 \\ \hdashline & 8 & 1 & 1/16 & 8/16 & 64/16\\ \hdashline Sum= & & 16 & 1 & 5 & 55/2\\ \hdashline \end{array}$$

c.

$$\mu=\dfrac{2+4+6+8}{4}=5$$

$$\sigma^2=\dfrac{1}{4}((2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2)$$

$$=5$$

Check

The mean of the sample means is

$$\mu_{\bar{X}}=E(\bar{X})=5=\mu$$

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

$$=\dfrac{55}{2}-(5)^2=\dfrac{5}{2}$$

The standard deviation of the sample means is

$$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{5/2}$$

$$\dfrac{\sigma^2}{n}=\dfrac{5}{2}=\sigma^2_{\bar{X}}$$

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

$$\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}$$