Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 387 drivers and find that 287 claim to always buckle up. Construct a 91 % confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5] ___________

Answer 1

Sample proportion = 287 / 387 = 0.742

= 1 - 0.91 = 0.09

Z/2 = Z0.09/2 = Z0.045 = 1.6954

91% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.742 - 1.6954 * sqrt [ 0.742 * ( 1 - 0.742) / 387 ] < p < 0.742 + 1.6954 * sqrt [ 0.742 * ( 1 - 0.742) / 387 ]

0.704 < p < 0.780

91% CI is ( 0.704 , 0.780 )