Question

A sample of 114 patients were given a drug to lower cholesterol. A 95% confidence interval for the mean reduction in cholesterol (in mmol/L) was (0.88, 1.02). What was the sample standard deviation of the reduction amounts? The standard deviation was ? mmol/L.

Answer 1

Solution:

95% confidence interval is given

(0.88, 1.02)

Upper limit = 1.02

Lower limit = 0.88

Margin of error = (Upper - Lower)/2 = (1.02 - 0.88)/2 = 0.07

Sample size used = n = 114

Find the sample standard deviation

The interval is constructed using t critical value . (Because the sample SD is used)

c = 95% = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

n = 114

df = n - 1 = 114 - 1 = 113

     =   = 1.981

Now ,

The margin of error is given by

E =  /2,d.f. * ( / n)

0.07 = 1.981 * (s/ 114)

Solving,

s = 0.37728191703  

sample standard deviation of the reduction amounts = 0.37728191703