Question

Find the 90% confidence interval for population mean if a sample returned the values

68, 104, 128, 122, 60, 64

Answer 1

Solution:

We are given a data of sample size n = 6

68, 104, 128, 122, 60, 64

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (68 + 104 + .......+ 64)/6

= 91

Now ,

s=   

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 30.718072856219

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 = 6 - 1 = 5

    =    =  _0.05,5 = 2.015

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 2.015 * (30.718072856219 / 6)

= 25.2693104707

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

( - E ) <   <  ( + E)

(91 - 25.2693104707)   <   <  (91 + 25.2693104707)

65.7306895293 <   <  116.269310471

Required 90% confidence interval is (65.7307 , 116.2693)