Chapter 7.2, Problem 34E

(a)

To determine

To Explain: the mean of the sampling distribution of $\widehat{p}$ .

Answer to Problem 34E

0.59

Explanation of Solution

Given:

  $\begin{array}{c}p=\text{59\% =0}\text{.59}\\ n\text{=50}\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportions is same to the population proportion $p$ .

  ${\mu }_{\widehat{p}}=p=59\%=0.59$

(b)

To determine

To Calculate:

Answer to Problem 34E

0.0696

Explanation of Solution

Given:

  $\begin{array}{c}p=\text{59\% =0}\text{.59}\\ n\text{=50}\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The mean of the sampling distribution of the sample proportions is same to the population proportion $p$

  ${\mu }_{\widehat{p}}=p=59\%=0.59$

The standard deviation of the sampling distribution is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.59\left(1-0.59\right)}{50}}\\ =0.0696\end{array}$

The proportion of couples in which both parents work outside the home among 50 married couples varies on average by $0.0696$ from the mean of $0.59$

(c)

To determine

To Calculate:

Answer to Problem 34E

Approximately normal

Explanation of Solution

Given:

  $\begin{array}{c}p=\text{59\% =0}\text{.59}\\ n\text{=50}\end{array}$

The sampling distribution of the sample proportions $\widehat{p}$ is about Normal is the large counts condition is satisfied that is when $np\ge 10$ and $n\left(1-p\right)\ge 10.$

  $\begin{array}{c}np=50\left(0.59\right)=29.5\ge 10\\ n\left(1-p\right)=50\left(1-0.59\right)=20.5\ge 10.\end{array}$

Since the large counts condition is satisfied, the sampling distribution of the sample proportion is approximately Normal