Answer to Question #302606 in Statistics and Probability for Ralph

Question

Random samples with size 4 are drawn from the population containing the values 14, 19, 26, 31, 48, and 53

a. Construct a sampling distribution of the sample means.

b. Find the mean of the sample means.

c. Compute the standard error of the sample means.

Answer 1

We have population values $14, 19, 26, 31, 48, 53$ population size $N=6$ and sample size $n=4.$

Thus, the number of possible samples which can be drawn without replacement is $\dbinom{6}{4}=15.$

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 14, 19, 26, 31 & 22.5\\ \hdashline 14, 19, 26, 48 & 26.75\\ \hdashline 14, 19, 26, 53 & 28\\ \hdashline 14, 19, 31, 48 & 28\\ \hdashline 14, 19, 31, 53 & 29.25\\ \hdashline 14, 19, 48, 53 & 33.5\\ \hdashline 14, 26, 31, 48 & 29.75\\ \hdashline 14, 26, 31, 53 & 31\\ \hdashline 14, 26, 48, 53 & 35.25\\ \hdashline 14, 31, 48, 53 & 36.5\\ \hdashline 19, 26, 31, 48 & 31\\ \hdashline 19, 26, 31, 53 & 32.25\\ \hdashline 19, 26, 48, 53 & 36.5\\ \hdashline 19, 31, 48, 53 & 37.75\\ \hdashline 26, 31, 48, 53 & 39.5\\ \hdashline \end{array}$$

a.

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 22.5 & 1 & 1/15 & 90/60 & 8100/240\\ \hdashline & 26.75 & 1 & 1/15 & 107/60 & 11449/240 \\ \hdashline & 28 & 2 & 2/15 & 224/60 & 25088/240\\ \hdashline & 29.25 & 1 & 1/15 & 117/60 & 13689/240 \\ \hdashline & 29.75 & 1 & 1/15 & 119/60& 14161/240 \\ \hdashline & 31 & 2 & 2/15 & 248/60 & 30752/240 \\ \hdashline & 32.25 & 1 & 1/15 & 129/60 & 16641/240\\ \hdashline & 33.5 & 1 & 1/15 & 134/60 & 17956/240 \\ \hdashline & 35.25 & 1 & 1/15 & 141/60 & 19881/240 \\ \hdashline & 36.5 & 2 & 2/15 & 292/60 & 42632/240 \\ \hdashline & 37.75 & 1 & 1/15 & 151/60 & 22801/240 \\ \hdashline & 39.5 & 1 & 1/15 & 158/60 & 24964/240 \\ \hdashline Sum= & & 16 & 1 & 191/6 & 124057/120\\ \hdashline \end{array}$$

b.The mean of the sample means is

$$\mu_{\bar{X}}=E(\bar{X})=191/6=\mu$$

c.

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

$$=\dfrac{124057}{120}-(\dfrac{191}{6})^2=\dfrac{7361}{360}$$

The standard error of the mean

$$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{\dfrac{7361}{360}}\approx4.52186$$