Question

consider a papulation consisting 2,5,8,11,13 and 25. Suppose sample of size 5 are drawn from this population.

Answer 1

We have population values $2,5,8,11,13,25$ population size $N=6$ and sample size $n=5.$

Thus, the number of possible samples which can be drawn without replacements is $\dbinom{6}{5}=6.$

1.

$$mean=\mu=\dfrac{2+5+8+11+13+25}{6}=\dfrac{32}{3}$$

2

$$Variance=\sigma^2=\dfrac{1}{6}((2-\dfrac{32}{3})^2+(5-\dfrac{32}{3})^2$$

$$+(8-\dfrac{32}{3})^2+(11-\dfrac{32}{3})^2+(13-\dfrac{32}{3})^2$$

$$+(25-\dfrac{32}{3})^2)=\dfrac{488}{9}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{488}{9}}=\dfrac{2\sqrt{122}}{3}\approx7.363574$$

3.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean\ (\bar{x}) \\ \hline 2,5,8,11,13 & 7.8 \\ \hdashline 2,5,8,11,25 & 10.2 \\ \hdashline 2,5,8,13,25 & 10.6 \\ \hdashline 2,5,11,13,25 & 11.2\\ \hdashline 2,8,11,13,25 & 11.8 \\ \hdashline 5,8,11,13,25 & 12.4 \\ \hdashline \end{array}$$

5. The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{x} & f & f(\bar{x}) & \bar{x}f(\bar{x}) & \bar{x}^2f(\bar{x})\\ \hline & 7.8 & 1 & 1/6 & 7.8/6 & 60.84/6 \\ \hdashline & 10.2 & 1 & 1/6 & 10.2/6 & 104.04/6 \\ \hdashline & 10.6 & 1 & 1/6 & 10.6/6 & 112.36/6 \\ \hdashline & 11.2 & 1 & 1/6 & 11.2/6 & 125.44/6 \\ \hdashline & 11.8 & 1 & 1/6 & 11.8/6 & 139.24/6 \\ \hdashline & 12.4 & 1 & 1/6 & 12.4/6 & 153.76/6 \\ \hdashline Sum= & & 6 & 1 & 32/3 & 695.68/6 \\ \hdashline \end{array}$$

6.

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{x}f(\bar{x})=\dfrac{32}{3}$$

7.

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

$$=\dfrac{347.84}{3}-(\dfrac{32}{3})^2=\dfrac{19.52}{9}$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{19.52}{9}}\approx1.472715$$

Check

$$\mu_{\bar{X}}=E(\bar{X})=\dfrac{32}{3}=\mu$$

$$Var(\bar{X})=\dfrac{19.52}{9}=\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}=\dfrac{488}{9(5)}\cdot\dfrac{6-5}{6-1}$$