How do I get to the answer?

Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 302 claim to always buckle up. Using a confidence level of 93%, construct a confidence interval for the proportion of the population who claim to always buckle up.

Express the lower limit and upper limit to three decimal places, as needed. Use interval notation and include the parentheses in your answer. For example: (0.54, 0.692)

Answer 1

Confidence Interval for a proportion

phat = sample proportion

n = sample size

z* = critical z value

CI = phat +- z* sqrt(phat*(1-phat)/n)

phat = number that buckle up/total sample size = 302/415 = 0.728

93% CI convers middle of normal curve between 3.5% and 96.5%.

find z* as z-score for 96.5 probability = 1.81

CI = 0.728 +/- 1.81 * sqrt(0.728*(1-0.728)/415)

CI = (0.688,0.768)