Question

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours. With a 0.95 probability, the margin of error is approximately

0.26

1.96

0.21

1.64

Answer 1

Solution :

Given that,

Population standard deviation =    =1.2

Sample size n =81

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96* ( 1.2 / 81 )

E= 0.26

Margin of error = E = 0.26