Question

A population consists of the five measurements 2, 6, 8, 0, and 1 with a sample size 4. Find the variance of the sample means

Answer 1

Mean

$$\mu=\dfrac{2+6+8+0+1}{5}=3.4$$

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

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

 We have population values $2,6,8,0,1$ population size $N=5$ and sample size $n=4.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{4}=5$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,6, 8, 0 & 4 \\ \hdashline 2 & 2,6,8,1 & 4.25 \\ \hdashline 3 & 2,6,0,1 & 2.25 \\ \hdashline 4 & 2,8,0,1 & 2.75 \\ \hdashline 5 & 6,8,0,1 & 3.75 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 9/4 & 1& 1/5 & 9/20 & 81/80 \\ \hdashline 11/4 & 1 & 1/5 & 11/20 & 121/80 \\ \hdashline 15/4 & 1 & 1/5 & 15/20 & 225/80 \\ \hdashline 4 & 1 & 1/5 & 16/20 & 256/80 \\ \hdashline 17/4 & 1& 1/5 & 17/20 & 289/80 \\ \hdashline Total & 5 & 1 & 68/20 & 972/80 \\ \hline \end{array}$$

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

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

the mean of the population.

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

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

$$=\dfrac{972}{80}-(\dfrac{68}{20})^2=\dfrac{472}{800}=\dfrac{59}{100}=0.59$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{59}{100}}\approx0.7681$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{4}(\dfrac{5-4}{5-1})$$ $$=0.59, True$$