Question

A sample of 15 cows were randomly selection from the cow population in Texas and given a special feed supplement for two months. The sample mean weight gain with the supplement for these 15 cows was 55 pounds. The sample standard deviation of weight gained over two months is 12 pounds. Weight gain for cows is known to be normally distributed. Calculate a 99% confidence interval for mean weight gain for a two month period with this supplement for all cows in the population.

Answer 1

Solution:

Given that,

n = 15

   = 55

s = 12

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 = 15 - 1 = 14  

    =    =  _0.005,14 = 2.977

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

The margin of error is given by

E =  /2,d.f. * ( / n )

=  2.977 * (12/ 15)

= 9.2239

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

( - E ) <   <  ( + E)

(55 - 9.2239)   <   <  (55 + 9.2239)

45.7761 <   < 64.2239

Required.99% confidence interval is (45.7761 , 64.2239)