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.37

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) The $95\%$confidence interval for the population proportion is $0.1682<p<0.2204$

Part b) No, we expect more than $171$respondents which is actually drive through the red lights.

Step-by-step solution

Part a) Step 1: Given information

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

Confidence level$=0.95$

Part a) Step 2: Explanation

We know, the formula of

Sample proportion: $\hat{p}=\frac{x}{n}$

The margin of error: $E=Z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

The confidence interval: $(\hat{p}-E,\hat{p}+E)$

Sample proportion: $\hat{p}=\frac{171}{880}=0.1943$

The confidence level is $0.95$

The level of significance is $=\mathrm{a}=0.05$

The $z_c=z_{a/2}$critical value $=1.96\dots$Using the excel formula $=ABS(NORMSINV(0.05/2\left)\right)$

So,

The margin of error is,

$$\begin{gathered} E=1.96\times \sqrt{\frac{0.1943\times (1-0.1943)}{880}} \\ \Rightarrow E=0.0261 \end{gathered}$$

And the confidence interval is,

$$\begin{gathered} (0.1943-0.0261,0.1943+0.0261) \\ =(0.1682,0.2204) \end{gathered}$$

Therefore the $95\%$confidence interval for a population proportion is $0.1682<p<0.2204$

Part b) Step 1: Given information

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

Confidence level $=0.95$

The $95\%$confidence interval for population proportion is $0.1682<p<0.2204$

Part b) Step 2: Explanation

The practical approach is that most people do not want to admit to running a red light, so the proportion is likely to be higher. That means we expect more than $171$respondents to have driven through a red light.

The margin of error takes into account sampling error. It contains no non-sampling errors. As a result, the source of bias is not accounted for in the margin of error.