Question

An experiment that investigated the effectiveness of a nicotine lozenge for subjects who wanted to quit smoking. Before the treatments began, subjects answered background questions, including how many cigarettes they smoked per day. Among the 81 subjects in the study, the average was 22.0 cigarettes per day, and the standard deviation was 10.8 cigarettes per day.

Give the margin-of-error for a 99% confidence interval for the average number of cigarettes.

Answer 1

Solution :

Given that,

Population standard deviation =    = 10.8
Sample size = n =81

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   ( Using z table )

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

= 2.576* ( 10.8/ 81)

=  3.0912