Chapter 7.2, Problem 42E

(a)

To determine

To calculate: The probability that the proportion in an SRS of 100 students is as small as or smaller than the result of the administration’s sample.

Answer to Problem 42E

The required probability is 0.1446.

Explanation of Solution

Given information:

Percent of students = 67%

Number of students = 15,000

Number of students for SRS = 100

Number of students who support a crackdown = 62

Formula used:

The mean of the sampling distribution of $\widehat{p}$ is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Calculation:

Here,

  $\begin{array}{c}p=67\%\\ =0.67\end{array}$

  $x=62$

  $n=100$

Find the mean of sampling distribution of $\widehat{p}$ by dividing the number of successes by the sample size that is, $\widehat{p}=\frac{x}{n}$

Substitute 62 for x and 100 for n in the above formula $\widehat{p}=\frac{x}{n}$ .

  $\begin{array}{c}\widehat{p}=\frac{62}{100}\\ =0.62\end{array}$

For the mean of the sampling distribution of $\widehat{p}$ , we can use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.67 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.67$

For the standard deviation of the sampling distribution, use the formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.67 for p and 100 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.67\left(1-0.67\right)}{100}}\\ =\sqrt{\frac{0.67\times 0.33}{100}}\\ =\sqrt{\frac{0.2211}{100}}\\ \approx 0.0470213\end{array}$

Now, we find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.62 for $x$ , 0.67 for $\mu$ and 0.0470213 for $\sigma$ in the formula $z=\frac{x-\mu }{\sigma }$ and simplify.

  $\begin{array}{c}z=\frac{0.62-0.67}{0.0470213}\\ \approx -1.06\end{array}$

We can find the probability by using table A as:

  $\begin{array}{c}P\left(\widehat{p}\le 0.62\%\right)=P\left(z<-1.06\right)\\ =0.1446\end{array}$

Hence, the required probability is 0.1446.

(b)

To determine

To determine: The explanation that why the survey does not support the conclusion.

Explanation of Solution

Given information:

Percent of students = 67%

Number of students = 15,000

Number of students for SRS = 100

Number of students who support a crackdown = 62

Formula used:

The mean of the sampling distribution of $\widehat{p}$ is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Explanation:

Here,

  $\begin{array}{c}p=67\%\\ =0.67\end{array}$

  $x=62$

  $n=100$

Find the mean of sampling distribution of $\widehat{p}$ by dividing the number of successes by the sample size that is, $\widehat{p}=\frac{x}{n}$

Substitute 62 for x and 100 for n in the above formula $\widehat{p}=\frac{x}{n}$ .

  $\begin{array}{c}\widehat{p}=\frac{62}{100}\\ =0.62\end{array}$

For the mean of the sampling distribution of $\widehat{p}$ , we can use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.67 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.67$

For the standard deviation of the sampling distribution, use the formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.67 for p and 100 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.67\left(1-0.67\right)}{100}}\\ =\sqrt{\frac{0.67\times 0.33}{100}}\\ =\sqrt{\frac{0.2211}{100}}\\ \approx 0.0470213\end{array}$

Now, we find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.62 for $x$ , 0.67 for $\mu$ and 0.0470213 for $\sigma$ in the formula $z=\frac{x-\mu }{\sigma }$ and simplify.

  $\begin{array}{c}z=\frac{0.62-0.67}{0.0470213}\\ \approx -1.06\end{array}$

We can find the probability by using table A as:

  $\begin{array}{c}P\left(\widehat{p}\le 0.62\%\right)=P\left(z<-1.06\right)\\ =0.1446\end{array}$

Thus, the probability is 0.1446.

The above probability is greater than 0.05.

If the true population proportion is 0.67, then a sample can be obtained with sample proportion of 0.62 because the probability is greater than 0.05.

Hence, it is concluded that the survey does not support the conclusion.