Question

A random sample of 20 students at a university showed an average age of 21.4 years and a sample standard deviation of 2.6 years. The 98% confidence interval for the true average age of all students in the university is?

Enter in the upper limit of your confidence interval

Answer 1

Solution :

Given that,

  n = 20

   = 21.4

s = 2.6  

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

Our aim is to construct 98% confidence interval.   

c = 0.98

= 1 - c = 1 - 0.98 = 0.02

  /2 = 0.02 2 = 0.01

Also, d.f = n - 1 = 20 - 1 = 19

    =    =  _0.01,19 = 2.539

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.539* ( 2.6/ 20 )

= 1.476

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

( - E ) <   <  ( + E)

( 21.4 - 1.476 )   <   <  ( 21.4 + 1.476 )

19.924 <   < 22.876

Required.98% confidence interval is ( 19.924 , 22.876 )

The upper limit of your confidence interval is 22.876