Question

A population consists of five members. The marital status of each member is given below, where M and S stand for married and single respectively.

 

Member

1

2

3

4

5

Marital Status

M

S

M

S

S

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Marital status} & M & S & M & S & S \\ \hdashline \end{array}$$

Find the proportion $P$ of married members in population.

$$P=\dfrac{2}{5}$$

Select all possible sample of 2 members from the population without replacement.

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 1 & 2 \\ \hline \text{Marital status} & M & S \\ \hdashline \end{array}, P_{12}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 1 & 3 \\ \hline \text{Marital status} & M & M \\ \hdashline \end{array}, P_{13}=1$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 1 & 4 \\ \hline \text{Marital status} & M & S \\ \hdashline \end{array}, P_{14}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 1 & 5 \\ \hline \text{Marital status} & M & S \\ \hdashline \end{array}, P_{15}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 2 & 3 \\ \hline \text{Marital status} & S & M \\ \hdashline \end{array}, P_{23}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 2 & 4 \\ \hline \text{Marital status} & S & S \\ \hdashline \end{array}, P_{24}=0$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 2 & 5 \\ \hline \text{Marital status} & S & S \\ \hdashline \end{array}, P_{25}=0$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 3 & 4 \\ \hline \text{Marital status} & M & S \\ \hdashline \end{array}, P_{34}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 3 & 5 \\ \hline \text{Marital status} & M & S \\ \hdashline \end{array}, P_{35}=\dfrac{1}{2}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} \text{Members} & 4 & 5 \\ \hline \text{Marital status} & S & S \\ \hdashline \end{array}, P_{45}=0$$

Compute the mean $\mu_P$ of a sample proportion

$$\mu_P=\dfrac{1}{10}(\dfrac{1}{2}+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+0+0$$

$$+\dfrac{1}{2}+\dfrac{1}{2}+0)=\dfrac{2}{5}$$

Check

$$\mu_P=\dfrac{2}{5}=P$$

Compute the standard deviation $\sigma_P$ of a sample proportion

$$\sigma^2_P=\dfrac{1}{10}((\dfrac{1}{2}-\dfrac{2}{5})^2+(1-\dfrac{2}{5})^2$$

$$+(\dfrac{1}{2}-\dfrac{2}{5})^2+(\dfrac{1}{2}-\dfrac{2}{5})^2+(\dfrac{1}{2}-\dfrac{2}{5})^2$$

$$+(0-\dfrac{2}{5})^2+(0-\dfrac{2}{5})^2+(\dfrac{1}{2}-\dfrac{2}{5})^2$$

$$+(\dfrac{1}{2}-\dfrac{2}{5})^2+(0-\dfrac{2}{5})^2)=\dfrac{9}{100}$$

$$\sigma_P=\sqrt{\sigma^2}=\sqrt{\dfrac{9}{100}}=\dfrac{3}{10}$$

Check

$$\sqrt{\dfrac{P(1-P)}{n}\cdot\dfrac{N-n}{N-1}}=\sqrt{\dfrac{\dfrac{2}{5}(1-\dfrac{2}{5})}{2}\cdot\dfrac{5-2}{5-1}}$$

$$=\dfrac{3}{10}=\sigma_P$$

$$\sigma_P=\dfrac{3}{10}=\sqrt{\dfrac{P(1-P)}{n}\cdot\dfrac{N-n}{N-1}}$$