Chapter 8, Problem R8.4RE

a)

To determine

To construct confidence interval for the population proportion.

Answer to Problem R8.4RE

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

Explanation of Solution

Given:

  $\begin{array}{l}n=880\\ x=171\end{array}$

Confidence level = 0.95

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

The confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

Sample proportion:

  $\widehat{p}=\frac{171}{880}=0.1943$

The confidence level = 0.95

So, level of significance = a=0.05

The z_c=z _a/2 critical value = 1.96 …Using excel formula, =ABS(NORMSINV(0.05/2))

The margin of error is,

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

The confidence interval is,

  $\begin{array}{l}(0.1943-0.0261,0.1943+0.0261)\\ (0.1682,0.2204)\end{array}$

Hence, the 95% confidence interval for population proportion is 0.1682 < p < 0.2204

b)

To determine

To explain the bias in the study.

Answer to Problem R8.4RE

No, we expect more than 171 respondent which are actually drive through red light.

Explanation of Solution

Given:

  $\begin{array}{l}n=880\\ x=171\end{array}$

Confidence level = 0.95

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

The practical approach is that most of the people do not want to admit to drive though a red light and therefore, the proportion is likely higher. That means we are expecting more that 171 respondents have actually drive though a red light.

The margin of error includes sampling error. It does not include any non-sampling error. Therefore, the source of bias not included in the margin of error.