A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval (z-score = 2.58) for the number of minutes that an adult spends surfing the Internet per day?

2 answers

99% = mean ± Z SEm

SEm = SD/√n
50min