Chapter 12.2, Problem 57E

(a)

To determine

To construct: and interpret the 95% confidence interval.

Answer to Problem 57E

(0.151, 0.319)

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

For confidence level $1-\alpha =0.95$ , $z_{\alpha /2}=z_{0.025}$ using table II look up 0.025 in the table

The z-score is

  $z_{\alpha /2}=1.96$

The margin of error is

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

The confidence intervals are

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

There are 95% confidence that the true population proportion is between 0.151 and 0.319.

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

(b)

To determine

To Explain: that the part (a) is providing the convincing evidence or not.

Answer to Problem 57E

No

Explanation of Solution

From the part (a):

(0.151, 0.319)

It is observed that that the confidence interval contains 0.29, which means that it is likely that the proportion of all teachers who would say they have tattoos is 0.29 and therefore there is no convincing proof that the proportion of all teachers at the institute who would say they have tattoos is different from 0.29.

(c)

To determine

To Explain: that two of the chosen teachers refused to response to the survey, if both the teachers responses, could the answer to part (b) have changed or not, justify.

Answer to Problem 57E

No

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

If both teachers said that they had one or more tattoos:

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

If both teachers said that they did not have one or more tattoos: $\widehat{p}=\frac{x}{n}=\frac{23.5\%\times 98+0}{100}=\frac{23}{100}=0.23$

The hypothesis is:

  $\begin{array}{l}{H}_0:p=0.14\\ H_a:p\ne 0.14\end{array}$

The test-statistic:

  $\begin{array}{l}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.25-0.14}{\sqrt{\frac{0.14\left(1-0.14\right)}{100}}}=3.17\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.23-0.14}{\sqrt{\frac{0.14\left(1-0.14\right)}{100}}}=2.59\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme.

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

If the P-value is smaller than the significance level, so reject the null hypothesis:

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

Therefore it is observed that the conclusion does not change when the response of the two teachers is included.