Question

In order to estimate the average time spent on the computer terminals per student at a university, data were collected for a sample of 81 business students over a one week period. Assume the population standard deviation is 1.8 hours. With a 0.95 probability, the margin of error is approximately

Answer 1

The margin of error is approximately 0.39.

Explanation:

Formular for the margin of error is

Margin of error = z* (sd/√n)

Where z* is 95% confidence level = 1.96, sd is 1.8 and n is 81

Margin of error = 1.96 (1.8/√81)

= 1.96(1.8/9)

= 1.96*0.2

= 0.392