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?

Answer 1

Answer:

The correct answer is 49.1 minutes to 59.3 minutes

Step-by-step explanation:

Answer 2

We have that the z-score is defined as z=$\frac{X-m}{\sigma}$ where X is the value of the observation, m is the mean of all the observations and sigma is the standard deviation. We also have that the z-score of the ends of the confidence intervals is 2.58 and -2.58. We have then, solving the first equation that :$X-m=z*\sigma \\ and \\ X=z*\sigma+M$
Substituting the values for z, m and sigma, we have that the upper bound of the confidence interval is 90.32 minutes and that the lower bound is 18.08 minutes. Hence, almost all adults (99%) have a total time of watching TV in this interval.