In order to estimate the average time spent per student on the computer terminals at a local 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 .95 probability, the margin of error is approximately .

With a .95 probability, the margin of error is = 0.392

Step-by-step explanation:

Given -

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.

Sample size ( n ) =81

Standard deviation = 1.8 hours

= 1 -.95 =.05

= = 1.96           (Using Z table)

With a .95 probability, the margin of error is

Margin of error  =  

                          =

                           =

Margin of error   = 0.392

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