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).
a. What was the sample mean reduction?
b. What was the sample standard deviation of the reduction amounts?
(a) The sample mean is midway between the endpoints of the confidence interval. So the sample mean is (0.88 + 1.02)/2 = 0.95.
(b) The width of the confidence interval is 
where s is the sample standard deviation. Therefore 
Solving for s yields s = 0.3813.
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