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 point 2 minutes, with a standard deviation of 14 point 0 minutes. What is the 99 percent confidence interval open paren z - score is equal to 2 point 5 8 close paren for the number of minutes that an adult spends surfing the Internet per day? Remember, the margin of error, ME , can be determined using the formula ME is equal to the fraction with numerator z times s and denominator square root of n