Question

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?
Remember, the margin of error, ME, can be determined using the formula mc019-1.jpg.
49.1 minutes to 59.3 minutes
50.3 minutes to 58.1 minutes
54.2 minutes to 58.1 minutes
54.2 minutes to 59.3 minutes

Answer 1

Margin of error = z * δ/√n 

sample population = 50
mean = 54.2 minutes
standard deviation = 14.0 minutes
z-score = 2.58

2.58 * (14/√50) = 2.58 * 14/7.07 = 2.58 * 1.98 = 5.1084 or 5.11

54.2 + 5.11 = 59.31 minutes
54.2 - 5.11 = 49.09 minutes

Answer is: 49.1 minutes to 59.3 minutes

Answer 2

The correct answer is 49.1 minutes to 59.3 minutes!