Question

Current Attempt in Progress Suppose you want to estimate the proportion of cars that are sport utility vehicles (SUVs) being driven in Kansas City, Missouri, at rush hour by standing on the corner of 1-70 and 1-470 and counting SUVs. You believe the figure is no higher than 0.40. If you want the error of the confidence interval to be no greater than 03, how many cars should you randomly sample? Use a 90% level of confidence. Appendix A Statistical Tables (Round your answer up to the nearest integer) Sample

Answer 1

Given,

Proportion of cars that are SUVs, p= 0.40

Error of confidence interval, margin of error, E = 0.03

For 90% confidence interval, Z = 1.645

We know for margin of error for computing population proportion is given by

$E=Z \times \sqrt{\frac{p(1-p)}{n}}$

$\Rightarrow n=\left ( \frac{Z}{E} \right )^{2} \times p(1-p)$

So, number of cars should you randomly sample ,

$\Rightarrow n=\left ( \frac{1.645}{0.03} \right )^{2} \times 0.4(1-0.4)$

            $=721.6067\approx 722$

Answer : n = 722