Question

Here are summary statistics for randomly selected weights of recent​ girls: n equals 168​, x = 27.2 ​hg, s = 7.5 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 25.4 hg < u < 29.4 hg with only 19 sample​ values, x= 27.4 ​hg, and s =n3.1 ​hg?

Answer 1

Solution:

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

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 168 - 1 = 167     

    =    =  _0.005,167 = 2.606

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.606 * (7.5 / 167)

= 1.5

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

( - E ) <   <  ( + E)

(27.2 - 1.5)   <   <  (27.2 + 1.5)

25.7 <   < 28.7

Required 99% confidence interval is 25.7 hg <   < 28.7 hg

Now ,

Compare this with given interval 25.4 hg < u < 29.4 hg  with only 19 sample​ values, x= 27.4 ​hg, and s =3.1 ​hg?

Results are very different.