Question

EXerCIse 8.417. Fewer young people are driving. In 1983, 87 percent of 19 year-olds had a driver's license. Twenty-five years later that percentage had dropped to 75 percent (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1,150 nineteen year-olds in 1983 and again in 2008. a. At 95% confidence, what is the marqin of error and the interval estimate of the number of nineteen year old drivers in 1983? Round your answers to four decimal places. Margin of error .0165 Interval estimate .8535 to .8865 b. At 95% confidence, what tis the margin of error and the interval estimate of the number of nineteen year old drivers in 2008? Round your answers to four decimal places Margin of error 0212 Interval estimate 7288 to 7712 c. Is the margin of error the same in parts (a) and (b)? No Why, or why not? The proportions are not the same

Answer 1

a)

sample proportion, = 0.87
sample size, n = 1150
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.87 * (1 - 0.87)/1150) = 0.0099

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(Proportion) = (0-2/34, P), P+=2 / 84,7"))

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0099
ME = 0.0194

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.87 - 1.96 * 0.0099 , 0.87 + 1.96 * 0.0099)
CI = (0.8506 , 0.8894)

b)
sample proportion, = 0.75
sample size, n = 1150
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.75 * (1 - 0.75)/1150) = 0.0128

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(Proportion) = (0-2/34, P), P+=2 / 84,7"))

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0128
ME = 0.0251

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.75 - 1.96 * 0.0128 , 0.75 + 1.96 * 0.0128)
CI = (0.7249 , 0.7751)