Answer to Question #260529 in Statistics and Probability for Ella

Question

A population consists of the five numbers: 1, 3, 5, 7, and 9.

Consider samples of size 3 that can be drawn from this population

2 points for every Sample and 1 point for the corresponding Mean of that Sample

Answer 1

1. Mean

$\mu=\dfrac{1+3+5+7+9}{5}=5$

Variance

$\sigma^2=\dfrac{1}{5}\big((2-3.4)^2+(6-2.4)^2+(8-3.4)^2$ $(0-3.4)^2+(1-3.4)^2\big)=9.44$

Standard deviation

$\sigma=\sqrt{\sigma^2}=\sqrt{9.44}\approx3.0725$

2. We have population values $2,6,8,0,1$ population size $N=5$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$\dbinom{N}{n}=\dbinom{5}{2}=10$ $\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,6 & 4 \\ \hdashline 2 & 2,8 & 5 \\ \hdashline 3 & 2,0 & 1 \\ \hdashline 4 & 2,1 & 1.5 \\ \hdashline 5 & 6,8 & 7 \\ \hdashline 6 & 6,0 & 3 \\ \hdashline 7 & 6,1 & 3.5 \\ \hdashline 8 & 8,0 & 4 \\ \hdashline 9 & 8,1 & 4.5 \\ \hdashline 10 & 0,1 & 0.5 \\ \hline \end{array}$

3. The sampling distribution of the sample means.

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 1/2 & 1& 1/10 & 1/20 & 1/40 \\ \hdashline 1 & 1 & 1/10 & 2/20 & 4/40 \\ \hdashline 3/2 & 1 & 1/10 & 3/20 & 9/40 \\ \hdashline 3 & 1 & 1/10 & 6/20 & 36/40 \\ \hdashline 7/2 & 1 & 1/10 & 7/20 & 49/40 \\ \hdashline 4 & 2 & 2/10 & 16/20 & 128/40 \\ \hdashline 9/2 & 1& 1/10 & 9/20 & 81/40 \\ \hdashline 5 & 1& 1/10 & 10/20 & 100/40 \\ \hdashline 7 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline Total & 10 & 1 & 68/20 & 604/40 \\ \hline \end{array}$

4.

$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{68}{20}=3.4$

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

the mean of the population.

$E(\bar{X})=3.4=\mu$

5.

$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$ $=\dfrac{604}{40}-(\dfrac{68}{20})^2=\dfrac{1416}{400}=\dfrac{354}{100}=3.54$ $\sqrt{Var(\bar{X})}=\sqrt{\dfrac{354}{100}}\approx1.8815$

Verification:

$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{2}(\dfrac{5-2}{5-1})$ $=3.54, True$