Of the 98 teachers who responded, $23.5\%$said that they had one or more tattoos.

(a) Construct and interpret a $95\%$confidence interval for the actual proportion of teachers at the AP institute who would say they had tattoos.

(b) Does the interval in part (a) provide convincing evidence that the proportion of teachers at the institute with tattoos is not $0.14$(the value cited in the Harris Poll report)? Justify your answer.

(c) Two of the selected teachers refused to respond to the survey. If both of these teachers had responded, could your answer to part (b) have changed? Justify your answer

Given in the question that, of the $98$teachers who responded, $23.5\%$said that they had one or more tattoos.

According to the information, we observed that

$n=98$

$\hat{p}=23.5\%=0.235$

For confidence level, $1-\alpha =0.95$, determine, $z_{\alpha /2}=z_{0.025}$

$z_{\alpha /2}=1.96$

The margin of error is

localid="1650685513171" $$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =1.96\times \sqrt{\frac{0.235(1-0.235)}{98}} \\ \approx 0.084 \end{gathered}$$

The confidence interval is

localid="1650685516522" $$\begin{gathered} 0.151=0.235-0.084 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.235+0.084 \\ =0.319 \end{gathered}$$

Of the $98$teachers who responded, $23.5\%$said that they had one or more tattoos.

We have to find the hypothesis

$H_0:p=0.14$

$H_0:p\ne 0.14$

Determine the value of the test-statistic:

localid="1650685524452" $$\begin{gathered} z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \\ =\frac{0.235-0.14}{\sqrt{\frac{0.14(1-0.14)}{98}}} \\ \approx 2.71 \end{gathered}$$

The localid="1650685529133" $P$-value is the chance of getting the test statistic's result, or a number that is more severe. Calculate the localid="1650685533236" $P$-value in table localid="1650685536989" $A$as follows:

localid="1650685554330" $$\begin{gathered} P=P(Z<-2.71orZ>2.71) \\ =2\times P(Z<-2.71) \\ =2\times 0.0034 \\ =0.0068 \end{gathered}$$

Reject the null hypothesis if the P-value is less than the significance level:

$P<0.05\Rightarrow Reject{H}_0$

Of the $98$teachers who responded, $23.5\%$said that they had one or more tattoos.

From the given values

The sample proportion is calculated by dividing the number of successes by the sample size:

If both professors indicated they had one or more tattoos, the following would happen:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{23.5\%\times 98+2}{100} \\ \approx \frac{25}{100} \\ =0.25 \end{gathered}$$

If both teachers indicated they didn't have any tattoos, then:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{23.5\%\times 98+0}{100} \\ \approx \frac{23}{100} \\ =0.23 \end{gathered}$$

Let's determine the hypothesis

$H_0:p=0.14$

$H_0:p\ne 0.14$

Compute the test statistic's value:

localid="1650685575009" $$\begin{gathered} P=P(Z<-3.17orZ>3.17) \\ =2\times P(Z<-3.17) \\ =2\times 0.0008 \\ =0.0016 \end{gathered}$$

localid="1650685581138" $$\begin{gathered} P=P(Z<-2.59orZ>2.59) \\ =2\times P(Z<-2.59) \\ =2\times 0.0048 \\ =0.0096 \end{gathered}$$

Reject the null hypothesis if thelocalid="1650685612366" $P$-value is less than the significance level.

$P<0.05\Rightarrow Reject{H}_0$