Chapter 7, Problem R7.4RE

(a)

To determine

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

Answer to Problem R7.4RE

0.15

Explanation of Solution

Given:

  $\begin{array}{l}n=1540\\ p=15\%=0.15\end{array}$

The mean of the sampling distribution of $\widehat{p}$ is equal to the population proportion:

  ${\mu }_{\widehat{p}}=p=0.15$

(b)

To determine

To Calculate: and interpret the standard deviation of the sampling distribution of $\widehat{p}$ and check that the 10% condition is met.

Answer to Problem R7.4RE

0.0091

10% condition is met

Explanation of Solution

Given:

  $\begin{array}{l}n=1540\\ p=15\%=0.15\end{array}$

Formula used:

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

Calculation:

The standard deviation of the sampling distribution of $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.15\left(1-0.15\right)}{1540}}\\ =0.0091\end{array}$

The proportion of adults who jog in a random sample of 1540 adults varies on average by 0.0091 from the mean proportion of 0.15.

The 10% condition is met, because the 1540 adults is less than 10% of the population of all adults.

(c)

To determine

To find: the sampling distribution of $\widehat{p}$ approximately normal.

Answer to Problem R7.4RE

Approximate normal

Explanation of Solution

Given:

  $\begin{array}{l}n=1540\\ p=15\%=0.15\end{array}$

Calculation:

The sampling distribution of $\widehat{p}$ is approximately normal if $np$ and $n\left(1-p\right)$ are both at least 10.

  $\begin{array}{c}np=1540\times 0.15=231\\ n\left(1-p\right)=1540\times \left(1-0.15\right)=1309\end{array}$

Since both are greater than 10 the distribution is approximately normal.

(d)

To determine

To find: the probability that between 13% and 17% of people jog in a random sample of 1540 adults.

Answer to Problem R7.4RE

0.9722

Explanation of Solution

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The sampling distribution of $\widehat{p}$ is approximately normal with mean 0.15 and standard deviation 0.0091.

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.13-0.15}{0.0091}\\ =-2.20\end{array}$

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.17-0.15}{0.0091}\\ =2.20\end{array}$

Finding the corresponding probability using table

  $\begin{array}{c}P\left(0.13\le \widehat{p}\le 0.17\right)=P\left(-2.20<z<2.20\right)-P\left(z<-2.20\right)\\ =0.9861-0.0139\\ =0.9722\end{array}$