Question

a sample size n =44 has sample mean =56.9 and Sample standard deviation s =9.1. a. construct a 98% confidence interval for the population mean meu b. if the sample size were n =30 would the confidence interval be narrower or wider? please show work to explain

Answer 1

Given that,

   = 56.9 ....... Sample mean

s = 9.1 ........Sample standard deviation

n = 44 ....... Sample size

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 = 0.022 = 0.01

Also, d.f = n - 1 = 44 - 1 = 43  

    =    =  _0.01,43 =  2.416

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

The margin of error is given by

E =  /2,d.f. * ( / n)

=  2.416 * (9.1 / 44 )

=  3.3145

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

( - E ) <   <  ( + E)

(56.9 - 3.3145)   <   <  (56.9 + 3.3145)

53.585 <   <  60.215

Required 98% interval is (53.585 , 60.215)

if the sample size were n =30 instead of 44 , then  the confidence interval would be wider.

Because , as sample size decreases , then the margin of error E increases , so that the interval becomes wider..