Question

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.

(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?

Short answer

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

Part (b) The likely direction was more than $171$respondents and no, these sources of bias do not include the margin of error.

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Concept

Formula used:$\stackrel{⏜}{p}=\frac{x}{n}$

Part (a) Step 3: Calculation

It is given in the question that,$$\begin{gathered} x=171 \\ n=880 \end{gathered}$$

And the sample proportion is calculated as: $$\begin{gathered} \stackrel{⏜}{p}=\frac{x}{n} \\ =\frac{171}{880} \\ =0.1943 \end{gathered}$$

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 is as: $$\begin{gathered} \stackrel{⏜}{p}-z_{\alpha /2}\times \sqrt{\frac{\stackrel{⏜}{p}\left(1-\stackrel{⏜}{p}\right)}{n}} \\ =0.1943-1.96\times \sqrt{\frac{0.1943(1-0.1943)}{880}} \\ =0.1682 \\ \stackrel{⏜}{p}+z_{\alpha /2}\times \sqrt{\frac{\stackrel{⏜}{p}\left(1-\stackrel{⏜}{p}\right)}{n}} \\ =0.1943+1.96\times \sqrt{\frac{0.1943(1-0.1943)}{880}} \\ =0.2204 \end{gathered}$$

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

Part (b) Step 1: Calculation

It is given in the question that, $$\begin{gathered} x=171 \\ n=880 \end{gathered}$$

And we're $95\%$sure the genuine population share is somewhere between$(0.1682,0.2204)$ As a result, most people will be reluctant to admit to running a red light, hence the proportion is likely to be greater, implying that more than $171$respondents have really done so. These causes of bias are not sampling errors, and as the margin of error only covers sampling errors, they have been excluded from the margin of error.