Answer in Statistics and Probability for tyza #249071

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):

“p=0.25 \nnn=400 \nnhat{p} = frac{152}{400}=0.38 \nnCI= hat{p}u00b1Z^* times sqrt{ frac{hat{p}(1- hat{p}}{n} }”

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

“CI = 0.38 u00b1 1.645 times sqrt{frac{0.38 times 0.62}{400}} \nn= 0.38u00b1 0.04 \nn= (0.34, 0.42)”