Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.7 and a standard deviation of 16.6. Construct a 99​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol? What is the confidence interval estimate of the population mean mu​?

Answer 1

Solution:

Given that,

   = 3.7 ....... Sample mean

s = 16.6 ........Sample standard deviation

n = 50 ....... Sample size

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 = 50 - 1 = 49  

    =    =  _0.005,49 = 2.680

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

The margin of error is given by

E =  /2,d.f. * ( / n )

=    2.680 * (16.6 / 50 )

=  6.29

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

( - E ) <   <  ( + E)

(3.7 - 6.29)   <   <  (3.7 + 6.29)

-2.59 <   < 9.99

Required interval is (-2.59 , 9.99)