Chapter 7.2, Problem 41E

(a)

To determine

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

Answer to Problem 41E

0.70

Explanation of Solution

Given:

  $\begin{array}{c}p=70\%=0.70\\ \widehat{p}=67\%=0.67\\ n=1012\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportion is

  ${\mu }_{\widehat{p}}=p=0.70$

The reason is that sample proportion is an unbiased estimator for the population proportion.

(b)

To determine

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

Answer to Problem 41E

0.0144052

Explanation of Solution

Given:

  $\begin{array}{c}p=70\%=0.70\\ \widehat{p}=67\%=0.67\\ n=1012\end{array}$

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

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.70\left(1-0.70\right)}{1012}}\\ =0.0144052\end{array}$

The $10\%$ condition is satisfied, if the sample size of $1012$ is less than $10\%$ of the population size and thus is true since there are more than 10,120 U.S. adults.

(c)

To determine

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

Answer to Problem 41E

The sampling distribution of p is approximately normal

Explanation of Solution

Calculation:

  $np=1012\left(0.70\right)=7-8.4$ And $n\left(1-p\right)=1012\left(0.30\right)=303.6$

The value are both at least 10.

(d)

To determine

To find: the probability of obtaining a sample of 1012 adults in which 67% or fewer say they drink the cereal milk, assuming the milk industry spokesman’s claim is true.

Answer to Problem 41E

0.0188

Yes

Explanation of Solution

Given:

  $\begin{array}{c}p=70\%=0.70\\ \widehat{p}=67\%=0.67\\ n=1012\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Results part (a) and (b):

  $\begin{array}{c}{\mu }_{\widehat{p}}=0.70\\ {\sigma }_{\widehat{p}}=0.0144052\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.67-0.70}{0.0144055}\\ =-2.08\end{array}$

The associating probability using the normal probability table is

  $P\left(\widehat{p}\le 0.67\right)=P\left(z<-2.08\right)=0.0188$

Since the probability is less than $0.05$ , it is unusual to obtain a sample proportion of 67% and therefore it is doubtful that the result of the poll are accurate

(e)

To determine

To find: that this poll gives convincing evidence against the spokesman’s claim.

Answer to Problem 41E

Yes

Explanation of Solution

Given:

  $\begin{array}{c}p=70\%=0.70\\ \widehat{p}=67\%=0.67\\ n=1012\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Results part (a) and (b):

  $\begin{array}{c}{\mu }_{\widehat{p}}=0.70\\ {\sigma }_{\widehat{p}}=0.0144052\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.67-0.70}{0.0144052}\\ =-2.08\end{array}$

The associating probability using the normal probability table is $P\left(Z<-2.08\right)$ is given in the row starting with $-2.08$ and in the column starting with $.08$ of the standard normal probability table in the appendix.

  $\begin{array}{c}P\left(\widehat{p}\le 0.67\right)=P\left(z<-2.08\right)\\ =0.0188\end{array}$

Since the probability is less than $0.05$ , it is unusual to obtain a sample proportion of $67\%$ and thus it is doubtful that the results of the poll are accurate