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

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

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

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

For the confidence level $1鈭�伪=0.95$and find out $z_{伪/2}=z_{0.025}$using table II, we get, $z_{伪/2}=1.96$

Thus, the confidence interval is as: $$\begin{gathered} \stackrel{铃�}{p}鈭�z_{伪/2}脳\sqrt{\frac{\stackrel{铃�}{p}\left(1-\stackrel{铃�}{p}\right)}{n}} \\ =0.1943鈭�1.96脳\sqrt{\frac{0.1943(1鈭�0.1943)}{880}} \\ =0.1682 \\ \stackrel{铃�}{p}+z_{伪/2}脳\sqrt{\frac{\stackrel{铃�}{p}\left(1-\stackrel{铃�}{p}\right)}{n}} \\ =0.1943+1.96脳\sqrt{\frac{0.1943(1鈭�0.1943)}{880}} \\ =0.2204 \end{gathered}$$

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

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.