Answer to Question #305370 in Statistics and Probability for dump

Question

Consider the population consisting of 1, 2, 3, 5, 6, and 7. Suppose samples of size 4 are drawn from this population. Describe the sampling distribution of the sample means. Compute for the mean and the variance of the sampling distribution of the sample means. Be guided by the first illustration.

STEPSSOLUTION1. Compute the mean of the population.2. Compute the variance of the population.3. Determine the number of possible samples of size n = 44. List all possible samples and compute their corresponding means.5. Construct the sampling distribution of the sample means.6. Compute the mean of the sampling distribution of the sample mean.7. Compute the variance of the sampling distribution of the sample means.

Answer 1

We have population values $1,2,3,5,6, 7$ population size $N=6.$

1.

$$mean=\mu=\dfrac{1+2+3+5+6+7}{6}=4$$

2

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

$$+(3-4)^2+(5-4)^2+(6-4)^2+(7-4)^2)=\dfrac{14}{3}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{14}{3}}\approx2.160247$$

3. The population size $N=6$ and sample size $n=4.$

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

4.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean\ (\bar{x}) \\ \hline 1,2,3,5 & 2.75 \\ \hdashline 1,2,3,6 & 3 \\ \hdashline 1,2,3,7 & 3.25 \\ \hdashline 1,2,5,6 & 3.5 \\ \hdashline 1,2,5,7 & 3.75 \\ \hdashline 1,2,6,7 & 4 \\ \hdashline 1,3,5,6 & 3.75 \\ \hdashline 1,3,5,7 & 4 \\ \hdashline 1,3,6,7 & 4.25 \\ \hdashline 1,5, 6,7 & 4.75 \\ \hdashline 2,3,5,6 & 4 \\ \hdashline 2,3,5,7 & 4.25 \\ \hdashline 2,3,6,7 & 4.5 \\ \hdashline 2,5,6,7 & 5 \\ \hdashline 3,5,6,7 & 5.25 \\ \hdashline \end{array}$$

5. The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{x} & f & f(\bar{x}) & \bar{x}f(\bar{x}) & \bar{x}^2f(\bar{x})\\ \hline & 2.75 & 1 & 1/15 & 11/60 & 121/240 \\ \hdashline & 3 & 1 & 1/15 & 12/60 & 144/240 \\ \hdashline & 3.25 & 1 & 1/15 & 13/60 & 169/240 \\ \hdashline & 3.5 & 1 & 1/15 & 14/60 & 196/240 \\ \hdashline & 3.75 & 2 & 2/15 & 30/60 & 450/240 \\ \hdashline & 4 & 3 & 3/15 & 48/60 & 768/240 \\ \hdashline & 4.25 & 2 & 2/15 & 34/60 & 578/240 \\ \hdashline & 4.5 & 1 & 1/15 & 18/60 & 324/240 \\ \hdashline & 4.75 & 1 & 1/15 & 19/60 & 361/240 \\ \hdashline & 5 & 1 & 1/15 & 20/60 & 400/240 \\ \hdashline & 5.25 & 1 & 1/15 & 21/60 & 441/240 \\ \hdashline Sum= & & 15 & 1 & 4 & 247/15 \\ \hdashline \end{array}$$

6.

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{x}f(\bar{x})=4$$

7.

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

$$=\dfrac{247}{15}-4^2=\dfrac{7}{15}$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{7}{15}}\approx0.683130$$

Check

$$\mu_{\bar{X}}=E(\bar{X})=4=\mu$$

$$Var(\bar{X})=\dfrac{7}{15}=\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}=\dfrac{14}{3(4)}\cdot\dfrac{6-4}{6-1}$$