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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 _________.

User Smitelli
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Answer:

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

User Adam Chetnik
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