Question

A college administrator would like to determine how much time students spend on homework assignments during a typical week. It is known that the standard deviation for the time students spend on homework is 3 hours per week.    A questionnaire is sent to sample of n=100 students and their response indicates a mean of 7.4 hours per week. Now make an interval estimate of population mean so that you are 99% confident that the "true" mean is in your interval. (i.e. compute the 99% confidence interval) - College Statistics Course

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.4

Population standard deviation = = 3

Sample size = n = 100

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2* ( /n)

= 2.576 * ( 3/ 100)

= 0.77

At 99% confidence interval estimate of the population mean is,

- E < < + E

7.4 - 0.77< < 7.4 + 0.77

6.63< <8.17

( 6.63,8.17 )