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.

3. What is the mean of the sampling distribution of the sample means UX?

4. What is the standard deviation of 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:

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