Question

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver's license. Twenty-five years later that percentage had dropped to 75% (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008 a. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 1983? Round your intermediate answers to four decimal places. Margin of error = (four decimal places) Interval estimate (to four decimal places) = I to b. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 2008? Round your answers to four decimal places. Margin of error = Interval estimate = to c. Is the margin of error the same in parts (a) and (b)? - Select your answer - Why, or why not? The input in the box below will not be graded, but may be reviewed and considered by your instructor. blank

Answer 1

Part a)

Margin of Error = Z(α/2) √ ( (p*q) / n) = 0.0190

p̂ ± Z(α/2) √( (p * q) / n)
0.87 ± Z(0.05/2) √( (0.87 * 0.13) / 1200)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.87 - Z(0.05) √( (0.87 * 0.13) / 1200) = 0.8510
upper Limit = 0.87 + Z(0.05) √( (0.87 * 0.13) / 1200) = 0.8890
95% Confidence interval is ( 0.8510 , 0.8890 )
( 0.8510 < P < 0.8890 )

Part b)

Margin of Error = Z(α/2) √ ( (p*q) / n) = 0.0245

p̂ ± Z(α/2) √( (p * q) / n)
0.75 ± Z(0.05/2) √( (0.75 * 0.25) / 1200)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.75 - Z(0.05) √( (0.75 * 0.25) / 1200) = 0.7255
upper Limit = 0.75 + Z(0.05) √( (0.75 * 0.25) / 1200) = 0.7745
95% Confidence interval is ( 0.7255 , 0.7745 )
( 0.7255 < P < 0.7745 )

C. Is the margin of error the same in parts (a) and (b)?

No, margin of error is not same in part a) and b)

As the percentage OR p̂ had dropped, margin of error increases.