Question

A recent University of Chicago research project surprised many young adults when it was reported that older Americans (ages 50-75) continue to have active sex lives. Specifically, it was found that respondents are having sex 2.4 times per week, on average, and deviated from that mean by 0.87. The study was based on interviews with a random sample of 500 older Americans. What is the margin of error (remember, this is the value that we add to and subtract from the sample mean in order to build the interval - NOT the interval itself!) at the 99% level of confidence? (Round to hundredths.)

Answer 1

n = sample size = 500

Mean =

Sample standard deviation = s = 0.87

Here population standard deviation is not known but sample size n is large (n = 500 > 30). so we use z interval

Confidence level = c = 0.99

z critical value for (1+c)/2 = (1 + 0.99)/2 = 0.995

zc = 2.58 (From statistical table of z values)

Margin of error (E) :

E = 2.58 * 0.038908

E = 0.100 (Round to 3 decimal)

Margin of error = 0.100

99% Confidence interval is

99% Confidence interval is (2.3, 2.5)