Running red lights A random digit dialing telephone survey of$880$drivers asked, "Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?" Of the $880$respondents, $171$admitted that at least one light had been red.${}^{15}$

(a) Construct and interpret a$95\%$confidence interval for the population proportion.

(b) Nonresponse is a practical problem for this survey-only $21.6\%$of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than $171$of the $880$respondents really ran a red light? Why? Are these sources of bias included in the margin of error?

$$\begin{gathered} x=171 \\ n=880 \end{gathered}$$

By dividing the complete number of wins by the sample size, the sample proportion is calculated:

$\hat{p}=\frac{x}{n}=\frac{171}{880}\approx 0.1943$

For confidence level$1-\alpha =0.95$, determine $z_{\alpha /2}=z_{0.025}$using table II (look up$0.025$in the table, the z-score is then the found $z-$score with opposite sign):

$z_{\alpha /2}=1.96$

Then, calculate the margin of error as follow:

$E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=1.96\times \sqrt{\frac{0.1943(1-0.1943)}{880}}\approx 0.0261$

As a result, the confidence interval becomes:

localid="1650283792947" $0.1682=0.1943-0.0261=\hat{p}-E<p<\hat{p}+E=0.1943+0.0261=0.2204$

Need to find the likely direction for the bias.

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 driven 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.