Answer to Question #283669 in Statistics and Probability for Lehri

Question

For the population of 2,5,4,3,2

Construct the sampling distribution of the mean x of size 3 without replacement

Answer 1

Mean

$\mu=\dfrac{2+5+3+4+2}{5}=3.2$

Variance

$\sigma^2=\dfrac{1}{5}\big((2-3.2)^2+(5-3.2)^2+(3-3.2)^2$ $+(4-3.2)^2+(2-3.2)^2\big)=1.36$

Standard deviation

$\sigma=\sqrt{\sigma^2}=\sqrt{1.36}\approx1.1662$

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

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

The sampling distribution of the sample means.

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 7/3 & 1& 1/10 & 7/30 & 49/90 \\ \hdashline 8/3 & 1& 1/10 & 8/30 & 64/90 \\ \hdashline 3 & 3 & 3/10 & 27/30 & 243/90 \\ \hdashline 10/3 & 2 & 2/10 & 20/30 & 200/90\\ \hdashline 11/3 & 2 & 2/10 & 22/30 & 242/90 \\ \hdashline 4 & 1 & 1/10 & 12/30 & 144/90 \\ \hdashline Total & 10 & 1 & 96/30 & 942/90 \\ \hline \end{array}$

$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{96}{30}=3.2$

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

the mean of the population.

$E(\bar{X})=3.2=\mu$

$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$ $=\dfrac{942}{90}-(\dfrac{96}{30})^2=\dfrac{68}{300}=\dfrac{17}{75}$ $\sqrt{Var(\bar{X})}=\sqrt{\dfrac{17}{75}}\approx0.4761$

Verification:

$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{1.36}{3}(\dfrac{5-3}{5-1})$ $=\dfrac{0.68}{3}=\dfrac{17}{75}, True$