Question

In a survey, 91 college students were interviewed about how
much

time they spent on school e-mail each week. The sample average was
3.34 hours with a sample standard deviation of 7.728 hours.
Construct a 90% confidence interval for the population mean. Also,
find the margin of error. Round your answers to 3 decimal
places.

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

Answer #1

Solution :

degrees of freedom = n - 1 = 91 - 1 = 90

t/2,df = t0.05,90 = 1.662

Margin of error = E = t/2,df * (s /n)

= 1.662 * ( 7.728 / 91)

Margin of error = E = 1.346

The 90% confidence interval estimate of the population mean is,

± E

= 3.34 ± 1.346

= ( 1.994, 4.686 )

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