Question

A random sample of 26 seedless watermelons has a mean weight of 32.71 ounces and a standard deviation of 9.48 ounces. Assuming the weights are normally distributed, construct a 90% confidence interval estimate for the population mean weight of all seedless watermelons.

Answer 1

Solution:

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 = 26 - 1 = 25

    =    =  _0.05,25 = 1.708

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

The margin of error is given by

E =  /2,d.f. * ( / n)

=  1.708 * (9.48 / 26)

= 3.1755

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

( - E ) <   <  ( + E)

(32.71 - 3.1755)   <   <  (32.71 + 3.1755)

29.5345 <   < 35.8855

Required 90% confidence interval is (29.5345 , 35.8855)