Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005) and its Z score = 2.575.
99% = mean ± 2.575 SEm
SEm = SD/√n
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 for the number of minutes that an adult spends surfing the Internet per day?
1 answer
1 answer