4 * 1.96 = 7.84 / 1 = 7.84 ^2 = 61.4656 =62
=62

14)
A simple random sample of 400 individuals provides 100 Yes responses.
a. What is the point estimate of the proportion of the population that would provide Yes responses? (to
2 decimals)
Using…
n=400 x=100
100 / 400 = .25
=.25
b. What is your estimate of the standard error of ,
ρ σ
p
p
(to 4 decimals)?
Using…
= .25 n=400
ρ
.25 * .75 = .1875 / 400 = .00046875
√.00046875 = .02165
=.0217
c. Compute the 95% confidence interval for the population proportion.

Enter your answer (to 4 decimals) using parentheses and a comma, in the form (n1,n2).
Point estimate
± margin of error
1.96 * .0217 = .042532
.25 ± .042532 = .207468 , .292532
=(.2075,.2925)
15) In a survey, the planning value for the population proportion isp* = .35. How large a sample should be taken to provide a 95% confidence interval with a margin of error of .05?

17)
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?
Margin of error = (four decimal places)
87% in 1983
=.019
Interval estimate (to three decimal places) = to
=.851 to .889
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 =
=.0245
Interval estimate = to

=.7255 to .7745
c.
Is the margin of error the same in parts (a) and (b)?
=no

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