Chapter 7.2, Problem 42E

(a)

To determine

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

Answer to Problem 42E

0.40

Explanation of Solution

Given:

  $\begin{array}{c}p=40\%=0.40\\ \widehat{p}=44\%=0.44\\ n=1785\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportion $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.40$

Because the sample proportion is an unbiased estimator for the population proportion

(b)

To determine

To Calculate: the standard deviation of the sampling distribution of $\widehat{p}$ .

Answer to Problem 42E

0.0115954

Explanation of Solution

Given:

  $\begin{array}{c}p=40\%=0.40\\ \widehat{p}=44\%=0.44\\ n=1785\end{array}$

Formula used:

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

Calculation:

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.40\left(1-0.40\right)}{1785}}\\ =0.0115954\end{array}$

The 10% condition is satisfied, if the sample size of 1785 is less than 10% of the population size and thus is true since there are more than $17.850$ adults.

(c)

To determine

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

Answer to Problem 42E

The large counts condition is met the reason both n and n(1-p) are greater than or same to 10. The sampling distribution is about normal

Explanation of Solution

Calculation:

  $\begin{array}{l}np=1785\times 0.4=714\ge 10\\ n\left(1-p\right)=1785\times 0.6=1071\ge 10\end{array}$

(d)

To determine

To Explain: about the claim true.

Answer to Problem 42E

0.03%

Explanation of Solution

Given:

  $\begin{array}{c}p=40\%=0.40\\ \widehat{p}=44\%=0.44\\ n=1785\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Results part (a) and (b)

  $\begin{array}{c}{\mu }_{\widehat{p}}=p=0.40\\ {\sigma }_{\widehat{p}}=0.0115954\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.44-0.40}{0.0115954}\\ =3.45\end{array}$

The associating probability using the normal probability table $P\left(Z<3.45\right)$ is given in the row starting with $3.4$ and in the column starting with $.05$ of the standard normal probability table in the appendix.

  $\begin{array}{c}P\left(\widehat{p}>0.60\right)=P\left(z>3.45\right)\\ =1-P\left(Z<3.45\right)\\ =1-0.9997\\ =0.0003\\ =0.03\%\end{array}$

(e)

To determine

To Explain: that this poll gives convincing evidence against the newspaper’s claim.

Answer to Problem 42E

Yes

Explanation of Solution

Given:

  $\begin{array}{c}p=40\%=0.40\\ \widehat{p}=44\%=0.44\\ n=1785\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Results part (a) and (b)

  $\begin{array}{c}{\mu }_{\widehat{p}}=p=0.40\\ {\sigma }_{\widehat{p}}=0.0115954\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.44-0.40}{0.0115954}\\ =3.45\end{array}$

The associating probability using the normal probability table in $P\left(Z<3.45\right)$ is given in the row starting with $3.4$ and in the column starting with $.05$ of the standard normal probability table

  $\begin{array}{c}P\left(\widehat{p}>0.60\right)=P\left(z>3.45\right)\\ =1-P\left(Z<3.45\right)\\ =1-0.9997\\ =0.0003\\ =0.03\%\end{array}$

Since the probability is small, the event of a sample proportion of at least $0.60$ is unlikely to occur by chance and therefore there is convincing evidence against the newspaper’s claim.