Question

A population consists of the five measurements 2, 6, 8, 0, and 1.

1. How many different samples of size 3 can be drawn from the population (no replacement)?

2. Construct the sampling distribution of the sample means.

Answer 1

 1.

We have population values $2,6,8,0,1$ population size $N=5$ and sample size $n=3.$Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{3}=10$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,6, 8 & 16/3 \\ \hdashline 2 & 2,6,0 & 8/3 \\ \hdashline 3 & 2,6,1 & 9/3 \\ \hdashline 4 & 2,8,0 & 10/3\\ \hdashline 5 & 2,8,1 & 11/3 \\ \hdashline 6 & 2,1,0 & 3/3 \\ \hline 7 & 6,8,1 & 15/3 \\ \hline 8 & 6,8,0 & 14/3 \\ \hline 9 & 6,1,0 & 7/3 \\ \hline 10 & 8,1,0 & 9/3 \\ \hline \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 3/3 & 1& 1/10 & 3/30 & 9/90 \\ \hdashline 7/3 & 1 & 1/10 & 7/30 & 49/90 \\ \hdashline 8/3 & 1 & 1/10 & 8/30 & 64/90 \\ \hdashline 9/3 & 2 & 2/10 & 18/30 & 162/90 \\ \hdashline 10/3 & 1 & 1/10 & 10/30 & 100/90 \\ \hdashline 11/3 & 1& 1/10 & 11/30 & 121/90 \\ \hdashline 14/3 & 1& 1/10 & 14/30 & 196/90 \\ \hdashline 15/3 & 1& 1/10 & 15/30 & 225/90 \\ \hdashline 16/3 & 1& 1/10 & 16/30 & 256/90 \\ \hdashline Total & 10 & 1 & 102/30 & 1182/90 \\ \hline \end{array}$$

3.

Mean

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

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{102}{30}=3.4$$

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

the mean of the population.

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

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

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

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

$$=\dfrac{1182}{90}-(\dfrac{102}{30})^2=\dfrac{118}{75}\approx1.573333$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{118}{75}}\approx1.254326$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{3}(\dfrac{5-3}{5-1})$$ $$=\dfrac{9.44}{6}=\dfrac{118}{75}\approx1.573333, True$$