Answer to Question #174109 in Statistics and Probability for Denisse Bisuña

Question

Given the population: 1, 3, 4, 6, and 8; and suppose samples of size 3 are drawn from this population:

1. What is the mean and standard deviation of the population?

2. How many different samples of size n=3 can be drawn from the population? List them with their corresponding means.

3. Construct the sampling distribution of the sample means.

4. What is the mean of the sampling distribution of the sample means? Compare this to the mean of the population.

5. What is the standard deviation of the sampling distribution of the sample means? Compare this to the standard deviation of the population.

Answer 1

1. Mean

$\mu=\dfrac{1+3+4+6+8}{5}=4.4$

Variance

$\sigma^2=\dfrac{1}{5}\big((1-4.4)^2+(3-4.4)^2+(4-4.4)^2$

$(6-4.4)^2+(8-4.4)^2\big)=5.84$

Standard deviation

$\sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166$

2. We have population values $1,3,4,6,8,$ population size $N=5$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$\dbinom{N}{n}=\dbinom{5}{2}=10$ $\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,3,4 & 8/3 \\ \hdashline 2 & 1,3,6 & 10/3 \\ \hdashline 3 & 1,3,8 & 4 \\ \hdashline 4 & 1,4,6 & 11/3 \\ \hdashline 5 & 1,4,8 & 13/3 \\ \hdashline 6 & 1,6,8 & 5 \\ \hdashline 7 & 3,4,6 & 13/3 \\ \hdashline 8 & 3,4,8 & 5 \\ \hdashline 9 & 3,6,8 & 17/3 \\ \hdashline 10 & 4,6,8 & 6 \\ \hline \end{array}$

3. The sampling distribution of the sample means.

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 8/3 & 1& 1/10 & 8/30 & 64/90 \\ \hdashline 10/3 & 1& 1/10 & 10/30 & 100/90 \\ \hdashline 11/3 & 1& 1/10 & 11/30 & 121/90 \\ \hdashline 4 & 1& 1/10 & 12/30 & 144/90 \\ \hdashline 13/3 & 2& 2/10 & 26/30 & 338/90 \\ \hdashline 5 & 2& 2/10 & 30/30 & 450/90 \\ \hdashline 17/3 & 1& 1/10 & 17/30 & 289/90 \\ \hdashline 6 & 1& 1/10 & 18/30 & 324/90 \\ \hdashline Total & 10 & 1 & 132/30 & 1830/90 \\ \hline \end{array}$

4.

$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{132}{30}=4.4$

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

the mean of the population.

$E(\bar{X})=4.4=\mu$

5.

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

$=\dfrac{1830}{90}-(\dfrac{132}{30})^2=\dfrac{876}{900}=\dfrac{73}{75}$

$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{73}{75}}\approx0.9866$

Verification:

$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.84}{3}(\dfrac{5-3}{5-1})$

$=\dfrac{5.84}{6}=\dfrac{73}{75}, True$