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 true proportion of all teachers at the AP institute who would say they have tattoos.

b. Does the interval in part (a) provide convincing evidence that the proportion of all teachers at the institute who would say they have tattoos is different from $0.29$. (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.

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

The given is

$$\begin{gathered} n=98 \\ \hat{p}=23.5\% \end{gathered}$$

The formula is

$E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

For confidence level role="math" localid="1654239338724" $$\begin{gathered} 1-\alpha =0.95, \\ {z}_{\alpha /2}=z_{0.025} \end{gathered}$$

The $z-$score is $1.96$

Then the margin of error is

Substituting in formula E

$$\begin{gathered} =1.96\times \sqrt{\frac{0.235(1-0.235)}{98}} \\ =0.084 \end{gathered}$$

The confidence interval is

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

The $95\%$confidence interval is between $0.151$and $0.319$.

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

From part (a)

$(0.151,0.319)$

The confidence interval comprises $0.29,$indicating that the proportion of all instructors who would say they have tattoos is most likely $0.29$, and that there is no persuasive evidence that the proportion of all teachers at the institute who would say they have tattoos is less than $0.29$.

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

The formula used here is

$Z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

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

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

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

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

The hypothesis is

$$\begin{gathered} H_0:p=0.14 \\ {H}_a:p\ne 0.14 \end{gathered}$$

The test statistic is

$$\begin{gathered} z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \\ =\frac{0.25-0.14}{\sqrt{\frac{0.14(1-0.14)}{100}}} \\ =3.17 \end{gathered}$$

$$\begin{gathered} z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \\ =\frac{0.23-0.14}{\sqrt{\frac{0.14(1-0.14)}{100}}} \\ =2.59 \end{gathered}$$

The P-value is the chance of having the test statistic's value, or a value that is more extreme.

$P=P(Z<-3.17orZ>3.17)=2\times P(Z<-3.17)=2\times 0.0008=0.0016$

$P=P(Z<-2.59orZ>2.59)=2\times P(Z<-2.59)=2\times 0.0048=0.0096$

If the $p-$value is less than the significance level, the null hypothesis must be rejected:

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

As a result, when the responses of the two teachers are taken into account, the conclusion remains unchanged.