Chapter 8, Problem 7CRE

(a)

To determine

To construct and interpret a $95\%$ confidence interval for the population proportion.

Answer to Problem 7CRE

We are $95\%$ confident that the true population proportion is between $(0.1682,0.2204)$ .

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}x=171\\ n=880\end{array}$

And the sample proportion is calculated as:

  $\begin{array}{l}\widehat{p}=\frac{x}{n}\\ =\frac{171}{880}\\ =0.1943\end{array}$

For the confidence level $1-\alpha =0.95$ and find out $z_{\alpha /2}=z_{0.025}$ using table II, we get,

  $z_{\alpha /2}=1.96$

Thus, the confidence interval be as:

  $\begin{array}{l}\widehat{p}-z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =0.1943-1.96\times \sqrt{\frac{0.1943(1-0.1943)}{880}}\\ =0.1682\\ \widehat{p}+z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =0.1943+1.96\times \sqrt{\frac{0.1943(1-0.1943)}{880}}\\ =0.2204\end{array}$

Thus, we are $95\%$ confident that the true population proportion is between $(0.1682,0.2204)$ .

(b)

To determine

To find out what is the likely direction of the bias: do you think more or fewer than $171\text{ of the 880}$ respondents really ran a red light and why and explain are these sources of bias included in the margin of error.

Answer to Problem 7CRE

The likely direction was more than $171$ respondents and no, these sources of bias does not include in the margin of error.

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}x=171\\ n=880\end{array}$

And we are $95\%$ confident that the true population proportion is between $(0.1682,0.2204)$ . Thus, most people will not want to admit to driving through a red light and thus the proportion is likely higher, which means that we expect more than $171$ respondents have actually drove through a red light. These sources of bias are not sampling error and the margin of error includes only sampling error, thus these sources of bias have not been included in the margin of error.