Question

If n=23, ¯xx¯(x-bar)=48, and s=3, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

< μμ <

Answer 1

Solution:

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

Our aim is to construct 98% confidence interval.   

c = 98% = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.01

Also, d.f = n - 1 = 23 - 1 = 22

    =    =  _{0.01 , 22} = 2.508

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 2.508 * (3 / 23)

= 1.56886

= 1.6

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

( - E ) <   <  ( + E)

(48 - 1.6)   <   <  (48 + 1.6)

46.4 <   < 49.6