Question

Here are summary statistics for randomly selected weights of newborn​ girls: n=246​, x overbar =30.4​hg, s=6.1hg. Construct a confidence interval estimate of the mean. Use a 98​% confidence level. Are these results very different from the confidence interval 28.9 hg<μ<31.3hg with only16 sample​ values, x overbar=30.1​hg, and s=1.8​hg?

Answer 1

Solution:

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 = 0.01

Also, d.f = n - 1 = 246 - 1 = 245

    =    = t_0.01,245 = 2.342

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

The margin of error is given by

E = t/2,d.f. * (s / n)

= 2.342 * (6.1 / 246)

= 0.91

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

( - E ) <   <  ( + E)

(30.4 - 0.91)   <   <  (30.4 + 0.91)

29.5 <   < 31.1

Now , compare this with 28.9 hg<μ<31.3hg

Two intervals are not different ,  because each confidence interval contains the mean of other confidence interval.