4 196 784 1 784 2 14 A simple random sample of 400 individuals

4 196 784 1 784 2 14 a simple random sample of 400

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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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