Question

given: random sample of size n= 2 are drawn from a finite population consisting of the numbers 5,6,7,8, and 9

1. How many possible samples are there
2. List all the possible samples and the corresponding mean for each sample.
3. Construct the sampling distribution of the sample means.
4. Construct the histogram for the sampling distribution of the sample mean.

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Answer 1

1.

We have population values $5,6,7,8,9$ population size $N=5$ and sample size $n=2.$Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{2}=10$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 5,6 & 5.5 \\ \hdashline 2 & 5,7 & 6.0 \\ \hdashline 3 & 5,8 & 6.5 \\ \hdashline 4 & 5,9 & 7.0\\ \hdashline 5 & 6,7 & 6.5 \\ \hdashline 6 & 6,8 & 7.0 \\ \hline 7 & 6,9 & 7.5 \\ \hline 8 & 7,8 & 7.5 \\ \hline 9 & 7,9 & 8.0 \\ \hline 10 & 8,9 & 8.5 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 5.5 & 1 & 0.1 & 0.55 & 3.025 \\ \hdashline 6.0 & 1 & 0.1 & 0.60 & 3.600 \\ \hdashline 6.5 & 2 & 0.2 & 1.30 & 8.450 \\ \hdashline 7.0 & 2 & 0.2 & 1.40 & 9.800 \\ \hdashline 7.5 & 2 & 0.2 & 1.50 & 11.250 \\ \hdashline 8.0 & 1& 0.1 & 0.80 & 6.400\\ \hdashline 8.5 & 1& 0.1 & 0.85 & 7.225 \\ \hdashline Total & 10 & 1 & 7 & 49.75 \\ \hline \end{array}$$

3.

Mean

$$\mu=\dfrac{5+6+7+8+9}{5}=7$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=7$$

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

$$\sigma^2=\dfrac{1}{5}\big((5-7)^2+(6-7)^2+(7-7)^2$$ $$(8-7)^2+(9-7)^2\big)=2$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.41421356$$

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

$$=49.75-(7)^2=0.75$$

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

Verification:

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

$$=0.75, True$$