Question

In a sleeping laboratory, 16 volunteers sleep an average of 7.2 hours, with a standard deviation of 1.2 hours. What would be a 90% confidence interval estimate for the mean number of hours that people sleep at the night?

Answer 1

Solution:

Given that,

   = 7.2 ....... Sample mean

s = 1.2 ........Sample standard deviation

n = 16 ....... Sample size

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 16 - 1 = 15

    =    =  _0.05,15 =  1.753

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

=  1.753 * (1.2 / 16 )

=  0.5259

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(7.2   - 0.5259)   <   <  (7.2  + 0.5259)

6.6741  <   <  7.7259

Required interval is (6.6741 , 7.7259)