Question

Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

a) x = ? b) n = ? c) p' = ? d)

Which distribution should you use for this problem? (Round your answer to four decimal places.)

e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.

i. state the confidence interval

ii. sketch graph

iii. calculate error bound

Answer 1

Page No Dain L a n=430, x= 330 6 = x= 330 - 0.7676 n 430 we we ar normal distribution for this problem because op! 76.75 > 10 95% CI for population proportion & + Zx 12 x Bert i 1 0.7674 + Z0.05/2 X 10.7676(1-0.7676) 430 0.7676 ± 1.96 X 0.0203 0.7674 + 0.0397 (0.7674 - 0.0397, 0.7676 + 0.0397) (0.7277 0.8071) 0.7277 < p < 0.8071 A gsto (I *AAN 0.7277 0 0. 8071 error bound & upper booond lower bound 2 0.8071 0.7277 2 Error bound = 0.0794