Answer to Question #345126 in Statistics and Probability for Gmtheone1

Question

2. Consider all samples of size 2 from this population:

2 5 6 8 10 12

a. Compute the mean () and variance () of the population.

b. List all samples of size 2 and compute the mean for each sample.

c. Construct the sampling distribution of the sample means.

d. Calculate the mean of the () of the sampling distribution of the sample means.

e. Calculate the variance of the () of the sampling distribution of the sample means.

Answer 1

1.

 We have population values $2,5,6,8, 10, 12$ population size $N=6$

$$\mu=\dfrac{2+5+6+8+10+12}{6}=7.17$$

$$\sigma^2=\dfrac{1}{6}((2-7.17)^2+(5-7.17)^2+(6-7.17)^2$$ $$+(8-7.17)^2+(10-7.17)^2+(12-7.17)^2)=10.8$$

2.

The number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{6}{2}=15$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,5 & 3.5 \\ \hdashline 2 & 2,6 & 4 \\ \hdashline 3 & 2,8 & 5\\ \hdashline 4 & 2,10 & 6 \\ \hdashline 5 & 2,12 & 7 \\ \hdashline 6 & 5,6 & 5.5 \\ \hdashline 7 & 5,8 & 6.5 \\ \hdashline 8 & 5,10 & 7.5 \\ \hdashline 9 & 5,12 & 8.5 \\ \hdashline 10 & 6,8 &7 \\ \hdashline 11 & 6,10 & 8 \\ \hdashline 12 & 6,12 & 9 \\ \hdashline 13&8,10 & 9 \\ \hdashline 14 & 8,12 & 10 \\ \hdashline 15 & 10,12 &11 \\ \hline \end{array}$$

3.

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 3.5 & 1 & 1/15 &0.23& 0.81 \\ \hdashline & 4 & 1 & 1/15 & 0.26 & 1.06 \\ \hdashline & 5 & 1 & 1/15 & 0.33& 1.66 \\ \hdashline & 5.5 & 1 & 1/15 & 0.36& 2.02 \\ \hdashline & 6 & 1 & 1/15 & 0.4 & 2.4 \\ \hdashline & 6.5 & 1 & 1/15& 0.43 & 2.81 \\ \hdashline & 7 & 2 & 2/15 & 0.93 & 1.86 \\ \hdashline & 7.5 & 1 & 1/15 & 0.5 & 3.75 \\ \hdashline & 8 & 1 & 1/15 & 0.53 & 4.26\\ \hdashline & 8.5 & 1 & 1/15& 0.56& 4.82 \\ \hdashline & 9& 2 & 2/15 & 1.2 & 10.8 \\ \hdashline & 10 & 1 & 1/15 & 0.66 & 6.66 \\ \hdashline & 11 & 1 & 1/15 & 0.73& 8.1 \\ \hdashline Total & & 15 & 1 & 7.17 & 55.68 \\ \hline \end{array}$$

4.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=7.17$$

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

$$E(\bar{X})=\mu_{\bar{X}}=7.17=\mu$$

5.

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

$$={55.68}-(7.17)^2=4.27$$