Question

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 _________.

Answer 1

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

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
`(\sigma )` = 1.8 hours

`\alpha` = 1 -.95 =.05

`Z_{(\alpha )/(2)}` =
`Z_{(.05 )/(2)}` = 1.96 (Using Z table)

With a .95 probability, the margin of error is

Margin of error =
`Z_{(\alpha )/(2)}(\sigma )/(√(n))`

=
`Z_{(.05 )/(2)} (1.8 )/(√(81))`

=
`1.96* 0.2`

Margin of error = 0.392