Answer to Question #266847 in Statistics and Probability for Krislyn

Question

A population consists of the numbers 2, 4, 9, 10 and 15. Let us list all possible sample size of 3 from this population and compute the mean and the variance of the sampling distribution of the sample means. Do the following tasks. 1. Compute the population Mean. 2. Compute the population Variance. 3. Determine the number of possible samples. 4. List all possible samples and their corresponding mean. 5. Construct the sampling distribution of the sample means. 6. Compute the mean of the sampling distribution of the sample means. 7. Compute the variance of the sampling distribution of the sample means. 8. Construct the Histogram.

Answer 1

1.

 We have population values $2,4,9,10,15$ population size $N=5$

$$\mu=\dfrac{2+4+9+10+15}{5}=8$$

2.

$$\sigma^2=\dfrac{1}{5}((2-8)^2+(4-8)^2+(9-8)^2$$

$$+(10-8)^2+(15-8)^2)=21.2$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{21.2}$$

3. We have population values $2,4,9,10,15,$ population size $N=5$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{3}=10$$

4.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,4,9 & 5 \\ \hdashline 2 & 2,4,10 & 16/3 \\ \hdashline 3 & 2,4,15 & 7 \\ \hdashline 4 & 2,9,10 & 7 \\ \hdashline 5 & 2,9,15 & 26/3 \\ \hdashline 6 & 2,10,15 & 9 \\ \hdashline 7 & 4,9,10 & 23/3 \\ \hdashline 8 & 4,9,15 & 28/3 \\ \hdashline 9 & 4,10,15 & 29/3 \\ \hdashline 10 & 9,10,15 &34/3 \\ \hline \end{array}$$

5. The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 5 & 1 & 1/10 & 1/2 & 5/2 \\ \hdashline & 16/3 & 1 & 1/10 & 8/15 & 128/45 \\ \hdashline & 7 & 2 & 1/5 & 7/5 & 49/5 \\ \hdashline & 23/3 & 1 & 1/10 & 23/30 & 529/90 \\ \hdashline & 26/3 & 1 & 1/10 & 13/15 & 338/45 \\ \hdashline & 9 & 1 & 1/10 & 9/10 & 81/10 \\ \hdashline & 28/3 & 1 & 1/10 & 14/15 & 392/45 \\ \hdashline & 29/3 & 1 & 1/10 & 29/30 & 841/90 \\ \hdashline & 34/3 & 1 & 1/10 & 17/15 & 578/45 \\ \hdashline Total & & 10 & 1 & 8 & 1013/15 \\ \hline \end{array}$$

6.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=8$$

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

$$E(\bar{X})=\mu_{\bar{X}}=8=\mu$$

7.

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

$$=\dfrac{1013}{15}-(8)^2=\dfrac{53}{15}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{53}{15}}$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{21.2}{3}(\dfrac{5-3}{5-1})$$

$$=\dfrac{53}{15}, True$$ $$\bar{X}\sim N(8, \dfrac{53}{15})$$

8.