Question

41.

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 num-
ber of 19-year-old drivers in 1983?
b. At 95% confidence, what is the margin of error and the interval estimate of the num-
ber of 19-year-old drivers in 2008?
C. Is the margin of error the same in parts (a) and (b)? Why or why not?​

Answer 1

a) ( 0.8509718, 0.8890282)

b) ( 0.7255, 0.7745)

Step-by-step explanation:

(a)

Given that , a = 0.05, Z(0.025) =1.96 (from standard normal table)

So Margin of error = Z × sqrt(p × (1-p)/n) = 1.96 × sqrt(0.87 × (1-0.87) / 1200)

=0.01902816

So 95 % confidence interval is

p+/-E

0.87+/-0.01902816

( 0.8509718, 0.8890282)

(b)

Margin of error = 1.96 × sqrt (0.75 × (1-0.75) / 1200) = 0.0245

So 95% confidence interval is

p+/-E

0.75+/-0.0245

( 0.7255, 0.7745)