Question

A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 20 hours. The population standard deviation is 2 hours. What is the 95% confidence interval for the mean?
19.88−20.12 hours
19.76−10.24 hours
19.88−20 hours
20−20.12 hours

Answer 1

The z-value that corresponds to a two-tailed 95% confidence interval is z = +/- 1.96. Then the bounds of the confidence interval can be determined as:
Lower bound = mean - z*SD/sqrt(n) = 20 - 1.96*2/sqrt(100) = 20 - 0.12 = 19.88 hours
Upper bound = mean + z*SD/sqrt(n) = 20 + 1.96*2/sqrt(100) = 20 + 0.12 = 20.12 hours
So the first choice is the correct answer: 19.88-20.12 hours