Question

A simple random sample of size n equals 18 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 62 and the sample standard deviation is found to be s equals 20. Construct a 95​% confidence interval about the population mean.

Answer 1

Solution:

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

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 18 - 1 = 17

    =    =  _0.025,17 = 2.110

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.110 * (20 / 18)

= 9.9466

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

( - E ) <   <  ( + E)

(62 - 9.9466)   <   <  (62 + 9.9466)

52.0534 <   < 71.9466

Required 95% confidence interval is (52.0534 , 71.9466)