Question

In a sample of 50 households, the mean number of hours spent on social networking sites during the month of January was 45 hours. In a much larger study, the standard deviation was determined to be 8 hours. Assume the population standard deviation is the same. What is the 95% confidence interval for the mean hours devoted to social networking in January?

Answer 1

95% Confidence Interval = 45 + or - 1.96 x 8/sqrt(50) = 45 + or - 1.96 x 1.131 = 45 + or - 2.217 = 42.783 or 47.217