HELP with STATISTICS PROBLEM

Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 398 drivers and find that 281 claim to always buckle up. Construct a 90% confidence interval for the population proportion that claim to always buckle up. Round your answer to three decimal places and use interval notation, for example, (0.689,0.787)

Answer 1

Let's assume the population follows Binomial distribution. For n=398 and x=281, we have a proportion of interest of p=x/n=0.706. The equation for the Binomial confidence interval is given by the Wald interval:

( p - z_1-∝/2 √p(1-p)/n , p + z_1-∝/2 √p(1-p)/n ) where z_1-∝/2 is the z value for the desired level of confidence. In this case, z value for 90% CI is 1.64, therefore the confidence interval is (0.66, 0.74).