Question

Married with children According to a recent U.S. Bureau of Labor Statistics report, the proportion of married couples with children in which both parents work outside the home is $59\%$.6 You select an SRS of 50 married couples with children and let $\hat{p}$= the sample proportion of couples in which both parents work outside the home.

a. Identify the mean of the sampling distribution of $\hat{p}$

b. Calculate and interpret the standard deviation of the sampling distribution of $\hat{p}$Verify that the 10% condition is met.

c. Describe the shape of the sampling distribution of $\hat{p}$Justify your answer.

Short answer

Part a) The required answer$0.59$

Part b) The required answer $0.0696$

Part c) The required answer is approximately normal.

Step-by-step solution

Part a) Step 1: Given information

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

Part a) Step 2: Calculation

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

${\mu }_{\hat{p}}=p=59\%=0.59$

Part b) Step 1: Given information

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

Part b) Step 2: Calculation

We know,

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

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

${\mu }_{\hat{p}}=p=59\%=0.59$

The sampling distribution's standard deviation is:

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.59(1-0.59)}{50}} \\ =0.0696 \end{gathered}$$

Therefore, 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 $0.59$

Part c) Step 1: Given information

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

Part c) Step 2: Calculation

The sampling distribution of the sample proportions $\hat{p}$is about Normal is the large count's condition is satisfied that is when$np\ge 10$and$n(1-p)\ge 10$

$$\begin{gathered} np=50(0.59)=29.5\ge 10 \\ n(1-p)=50(1-0.59)=20.5\ge 10 \end{gathered}$$

Because the large count's condition is met, the sample proportion's sampling distribution is roughly Normal.