Chapter 5, Problem 87E

(a)

Section 1:

To determine

To find: The distribution of the proportion of working females.

Answer to Problem 87E

Solution: The proportion of females who were engaged in work follows normal distribution with mean ${\mu }_{{\widehat{p}}_F}=0.82$ and standard deviation ${\sigma }_{{\widehat{p}}_F}=0.0192$.

Explanation of Solution

Calculation: The sampling distribution of the sample proportion of success $\widehat{p}$ is distributed normally with mean ${\mu }_{\widehat{p}}=p$ and standard deviation ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$, where p is the success proportion in the population and n is the size of sample.

Therefore, for the given problem,

$\begin{array}{c}{\mu }_{{\widehat{p}}_F}=p_F\\ =0.82\end{array}$

And,

$\begin{array}{c}{\sigma }_{{\widehat{p}}_F}=\sqrt{\frac{p_F\left(1-p_F\right)}{n_F}}\\ =\sqrt{\frac{0.82\times \left(1-0.82\right)}{400}}\\ =0.0192\end{array}$

Section 2:

To determine

To find: The distribution of the proportion of working males.

Answer to Problem 87E

Solution: The proportion of males who were engaged in work follows normal distribution with average ${\mu }_{{\widehat{p}}_M}=0.88$ and standard deviation ${\sigma }_{{\widehat{p}}_M}=0.01625$.

Explanation of Solution

Calculation: The sampling distribution of the sample proportion of success $\widehat{p}$ is distributed normally with mean ${\mu }_{\widehat{p}}=p$ and standard deviation ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$. Therefore, for the given problem,

$\begin{array}{c}{\mu }_{{\widehat{p}}_M}=p_M\\ =0.88\end{array}$

And,

$\begin{array}{c}{\sigma }_{{\widehat{p}}_M}=\sqrt{\frac{p_M\left(1-p_M\right)}{n_M}}\\ =\sqrt{\frac{0.88\times \left(1-0.88\right)}{400}}\\ =0.01625\end{array}$

(b)

To determine

To find: The sampling distribution of the difference of sample proportions.

Answer to Problem 87E

Solution: The sampling distribution of $\left({\widehat{p}}_M-{\widehat{p}}_F\right)$ is distributed normally with mean ${\mu }_{{\widehat{p}}_M-{\widehat{p}}_F}=0.06$ and standard deviation ${\sigma }_{{\widehat{p}}_M-{\widehat{p}}_F}=0.02516$.

Explanation of Solution

Calculation: The proportion of females who were engaged in work follows normal distribution with mean ${\mu }_{{\widehat{p}}_F}=0.82$ and standard deviation ${\sigma }_{{\widehat{p}}_F}=0.01921$.

The proportion of males who were engaged in work follows normal distribution with mean ${\mu }_{{\widehat{p}}_M}=0.88$ and standard deviation ${\sigma }_{{\widehat{p}}_M}=0.01625$.

The sampling distribution of ${\widehat{p}}_M-{\widehat{p}}_F$ is distributed normally with mean ${\mu }_{{\widehat{p}}_M-{\widehat{p}}_F}={\widehat{p}}_M-{\widehat{p}}_F$ and standard deviation ${\sigma }_{{\widehat{p}}_M-{\widehat{p}}_F}=\sqrt{{\sigma }_{p_M}^2+{\sigma }_{p_F}^2}$. Therefore,

$\begin{array}{c}{\mu }_{{\widehat{p}}_M-{\widehat{p}}_F}={\widehat{p}}_M-{\widehat{p}}_F\\ =0.88-0.82\\ =0.06\end{array}$

And,

$\begin{array}{c}{\sigma }_{{\widehat{p}}_M-{\widehat{p}}_F}=\sqrt{{\sigma }_{p_M}^2+{\sigma }_{p_F}^2}\\ =\sqrt{{0.01921}^2+{0.01625}^2}\\ =0.02516\end{array}$

(c)

To determine

To find: The probability that a higher proportion of females worked than males.

Answer to Problem 87E

Solution: The required probability is $\underset{\_}{P\left({\widehat{p}}_F>{\widehat{p}}_M\right)=0.0087}$.

Explanation of Solution

Calculation: The following probability needs to be calculated for the given problem:

$P\left({\widehat{p}}_M<{\widehat{p}}_F\right)=P\left({\widehat{p}}_M-{\widehat{p}}_F<0\right)$

The sampling distribution of ${\widehat{p}}_M-{\widehat{p}}_F$ is distributed normally with mean ${\mu }_{{\widehat{p}}_M-{\widehat{p}}_F}=0.06$ and standard deviation ${\sigma }_{{\widehat{p}}_M-{\widehat{p}}_F}=0.02516$.

The probability that ${\widehat{p}}_M-{\widehat{p}}_F$ is smaller than 0, which is the area to the left of the point 0 under the normal curve. To find the area under the normal curve, first convert the value of sample mean ${\widehat{p}}_M-{\widehat{p}}_F=0$ to a Z-score. The area left to the particular Z-score can be obtained using Table A provided in the book.

The Z-score for the value ${\widehat{p}}_M-{\widehat{p}}_F=0$ is calculated as:

$\begin{array}{c}Z=\frac{\left({\widehat{p}}_M-{\widehat{p}}_F\right)-{\mu }_{{\widehat{p}}_M-{\widehat{p}}_F}}{{\sigma }_{{\widehat{p}}_M-{\widehat{p}}_F}}\\ =\frac{0-0.06}{0.02516}\\ =-2.38\end{array}$

The area below the standard normal curve to the left of $Z=-2.38$ is obtained as 0.0087 from the standard normal table.