Answer to Question #249071 in Statistics and Probability for tyza

Question

The Gauteng traffic department records show that 25% of all drivers wear seatbelts. In a random sample of 400 cars stopped at a roadblock in Gauteng, 152 of the drivers were wearing seatbelts. A 90% confidence interval for the proportion in the population who wear seatbelts is given by (round final answer to two decimal places):

Answer 1

$p=0.25 \\ n=400 \\ \hat{p} = \frac{152}{400}=0.38 \\ CI= \hat{p}±Z^* \times \sqrt{ \frac{\hat{p}(1- \hat{p}}{n} }$

$Z^*= 1.645$ (for 90 % confidence interval)

$CI = 0.38 ± 1.645 \times \sqrt{\frac{0.38 \times 0.62}{400}} \\ = 0.38± 0.04 \\ = (0.34, 0.42)$