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A random sample of 90 UT business students revealed that on average they spent 8 hours...

A random sample of 90 UT business students revealed that on average they spent 8 hours per week on Facebook. The population standard deviation is assumed to be 2 hours per week.

If we want to develop a 95% confidence interval for the average time spent per week on Facebook by all UT business students, the margin of error of this interval is

The 95% confidence interval for the average time spent per week on Facebook by all UT business students is

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Answer #1

At 95% confidence interval the criical value is z0.025 = 1.96

Margin of error = z0.025 *

                        = 1.96 * 2/

                        = 0.4132

The 95% confidence interval for population mean is

+/- ME

= 8 +/- 0.4132

= 7.5868, 8.4132

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