Question

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

Answer #1

At 95% confidence interval the criical value is
z_{0.025} = 1.96

Margin of error = z_{0.025} *

= 1.96 * 2/

= 0.4132

The 95% confidence interval for population mean is

+/- ME

= 8 +/- 0.4132

= 7.5868, 8.4132

The mean number of hours of study time per week for a sample of
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Группа выборов ответов
(1.994,4.686) margin of error=1.346
(1.994,4.686) margin of error=2.692
(1.731,4.949) margin of error=1.609
(1.731,4.949) margin of error=3.218
(1.994,4.686) margin of error=3.340

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Group of answer choices
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Question 31
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