Question

A random sample is size n = 2 are drawn from a finite population consisting of the numbers 3,4,5,6 and 7

Answer 1

1. Mean

$$\mu=\dfrac{3+4+5+6+7}{5}=5$$

Variance

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

$$(6-5)^2+(7-5)^2\big)=2$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.4142$$

2. We have population values $3,4,5,6,7,$ population size $N=5$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{2}=10$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 3,4 & 3.5 \\ \hdashline 2 & 3,5 & 4 \\ \hdashline 3 & 3,6 & 4.5 \\ \hdashline 4 & 3,7 & 5 \\ \hdashline 5 & 4,5 & 4.5 \\ \hdashline 6 & 4,6 & 5 \\ \hdashline 7 & 4,7 & 5.5 \\ \hdashline 8 & 5,6 & 5.5 \\ \hdashline 9 & 5,7 & 6 \\ \hdashline 10 & 6,7 & 6.5 \\ \hline \end{array}$$

3. The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 7/2 & 1& 1/10 & 7/20 & 49/40 \\ \hdashline 4 & 1 & 1/10 & 8/20 & 64/40 \\ \hdashline 9/2 & 2 & 2/10 & 18/20 & 162/40 \\ \hdashline 5 & 2 & 2/10 & 20/20 & 200/40 \\ \hdashline 11/2 & 2 & 2/10 & 22/20 & 242/40 \\ \hdashline 6 & 1 & 1/10 & 12/20 & 144/40 \\ \hdashline 13/2 & 1& 1/10 & 13/20 & 169/40 \\ \hdashline Total & 10 & 1 & 100/20 & 1030/40 \\ \hline \end{array}$$

4.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{100}{20}=5$$

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

the mean of the population.

$$E(\bar{X})=5=\mu$$

5.

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

$$=\dfrac{1030}{40}-(\dfrac{100}{20})^2=\dfrac{3}{4}=0.75$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}\approx0.8660$$

Verification:

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

$$=\dfrac{3}{4}=0.75, True$$