Question

1. a random sample of the birth weight of 186 babies has a mean 0f 3103 g and a standard deviation of 696 g. construct a 90% confidence interval estimate of the mean birth weight of babies.

Answer 1

Solution:

Given that,

n = 186

= 3103   

s = 696  

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

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 186 - 1 = 185

    =    =  _0.05,185 = 1.653

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.653 * (696/ 186)

= 84.3579

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

( - E ) <   <  ( + E)

(3103   - 84.3579)   <   <  (3103  + 84.3579)

3018.6421 <   <  3187.3579

i.e.

(3018.6421 , 3187.3579 )