Question

Solve. Given a population of the scores of BSHS honor students in Mathematics as 1, 3, 4, 6, and 8. Suppose a sample size of 3 have to be drawn from it for the first, second and third rank;

Answer 1

a.

 We have population values $1,3,4,6,8,$ population size $N=5$

$$\mu=\dfrac{1+3+4+6+8}{5}=4.4$$

b.

$$\sigma^2=\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2$$

$$+(6-4.4)^2)+(8-4.4)^2)=5.84$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166$$

d. We have population values $1,3,4,6,8$ population size $N=5$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{3}=10$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,3,4 & 8/3 \\ \hdashline 2 & 1,3,6 & 10/3 \\ \hdashline 3 & 1,3,8 & 4 \\ \hdashline 4 & 1,4,6 & 11/3 \\ \hdashline 5 & 1,4,8 & 13/3\\ \hdashline 6 & 1,6,8 & 5 \\ \hdashline 7 & 3,4,6 & 13/3 \\ \hdashline 8 & 3,4,8 & 5 \\ \hdashline 9 & 3,6,8 & 17/3 \\ \hdashline 10 & 4,6,8 & 6 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 8/3 & 1 & 1/10 & 8/30 & 64/90 \\ \hdashline & 10/3 & 1 & 1/10 & 10/30 & 100/90 \\ \hdashline & 11/3 & 1 & 1/10 & 11/10 & 121/90 \\ \hdashline & 4 & 1& 1/10 & 12/30 & 144/90 \\ \hdashline & 13/3 & 2 & 2/10 & 26/30 & 338/90 \\ \hdashline & 5 & 2 & 2/10 & 30/30 & 450/90 \\ \hdashline & 6 & 1 & 1/10 & 18/30 & 324/90 \\ \hdashline & 17/3 & 1 & 1/10 & 17/30 & 289/90 \\ \hdashline Total & & 10 & 1 & 4.4 & 61/3 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=4.4$$

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

$$E(\bar{X})=\mu_{\bar{X}}=4.4=\mu$$

e.

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

$$=\dfrac{61}{3}-(4.4)^2=\dfrac{2.92}{3}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{2.92}{3}}\approx0.9866$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.84}{3}(\dfrac{5-3}{5-1})$$

$$=\dfrac{2.92}{3}, True$$