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

Confidence level$=0.95$

We know, the formula of

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

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

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 $=1.96鈥�$Using the excel formula $=ABS(NORMSINV(0.05/2\left)\right)$

So,

The margin of error is,

$$\begin{gathered} E=1.96脳\sqrt{\frac{0.1943脳(1-0.1943)}{880}} \\ 鈬�E=0.0261 \end{gathered}$$

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 $0.1682<p<0.2204$

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

Confidence level $=0.95$

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

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.