Chapter 12.2, Problem 52E

(a)

To determine

To construct: and define a confidence interval of 95 percent for the real proportion of AP Institute teachers who will say they have tattoos.

Answer to Problem 52E

(0.151, 0.319)

Explanation of Solution

Given:

  $\begin{array}{l}n=98\\ \widehat{p}=23.5\%=0.235\end{array}$

Formula used:

The margin of error is

  $E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

Calculation:

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

  $z_{\alpha /2}=1.96$

The margin of error is

  $\begin{array}{c}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =1.96\times \sqrt{\frac{0.235\left(1-0.235\right)}{98}}\\ =0.084\end{array}$

The confidence interval

  $0.151=0.235-0.084=\widehat{p}-E<p<\widehat{p}+E=0.235+0.084=0.319$

There is confidence in 95 percent that the true proportion of the population is between 0.151 and 0.319

(b)

To determine

To Explain: that the interval in part (a) provides strong proof that the proportion of teachers with tattoos at the institute is not 0.14, supports the answer.

Answer to Problem 52E

Yes

Explanation of Solution

Given:

  $\begin{array}{c}n=98\\ \widehat{p}=23.5\%=0.235\end{array}$

Formula used:

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

Calculation:

  $\begin{array}{c}{H}_0:p=0.14\\ H_1:p\ne 0.14\end{array}$

The test-statistic:

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.235-0.14}{\sqrt{\frac{0.14\left(1-0.14\right)}{98}}}\\ =2.71\end{array}$

The P-value is

  $\begin{array}{c}P=P\left(Z<-2.71\text{ or Z > 2}\text{.71}\right)\\ =2\times P\left(Z<-2.71\right)\\ =2\times 0.0034\\ =0.0068\end{array}$

Reject the null hypothesis if the P-value is lower than the significance level: $P<0.05\Rightarrow \text{Reject }{H}_0$

There is ample proof to support the claim.

(c)

To determine

To Explain: that the two teachers chosen failed to respond to the survey. If both of these teachers had answered, the response may have changed to Part (b), explain the response.

Answer to Problem 52E

No

Explanation of Solution

Given:

  $\begin{array}{c}n=98\\ \widehat{p}=23.5\%=0.235\end{array}$

Formula used:

  $\begin{array}{c}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

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

Although both teachers said they had no tattoos at all:

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

The hypothesis testing

  $\begin{array}{c}{H}_0:p=0.14\\ H_1:p\ne 0.14\end{array}$

The test-statistic:

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

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

The P-value is

  $\begin{array}{c}P=P\left(Z<-3.17\text{ or Z > 3}\text{.17}\right)\\ =2\times P\left(Z<-3.17\right)\\ =2\times 0.0008\\ =0.0016\\ P=P\left(Z<-2.59\text{ or Z > 2}\text{.59}\right)\\ =2\times P\left(Z<-2.59\right)\\ =2\times 0.0048\\ =0.0096\end{array}$

If the P-value is less than the level of significance, the null hypothesis is rejected: $P<0.05\Rightarrow \text{Reject }{H}_0$

And it is found that when the two teachers' answer is used the inference does not change.