Question

Suppose that insurance companies did a survey. They randomly surveyed 450 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) Which distribution should you use for this problem? (Round your answer to four decimal places). B) Construct a 95% confidence interval for the population proportion who claim they always buckle up. C) Calculate the error bound. (Round your answer to four decimal places).

Answer 1

A)

z normal population proportion distribution

p = 330/450 = 0.7333

B)

sample proportion, = 0.7333
sample size, n = 450
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.7333 * (1 - 0.7333)/450) = 0.0208

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.7333 - 1.96 * 0.0208 , 0.7333 + 1.96 * 0.0208)
CI = (0.6925 , 0.7741)

c)

Error Bound =0.7741 - 0.6925 = 0.0816