Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 13 subjects had a mean wake time of 105.0 min. After​ treatment, the 13 subjects had a mean wake time of 77.6 min and a standard deviation of 23.2min. Assume that the 13 sample values appear to be from a normally distributed population and construct a 99​% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 min before the​ treatment? Does the drug appear to be​ effective?

Answer 1

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
77.6 ± Z (0.01/2 ) * 23.6/√(13)
Lower Limit = 77.6 - Z(0.01/2) 23.6/√(13)
Lower Limit = 60.74
Upper Limit = 77.6 + Z(0.01/2) 23.6/√(13)
Upper Limit = 94.46
99% Confidence interval is ( 60.74 , 94.46 )

Since 105 does not contained in confidence interval and all values in confidence interval

are less than 105, we have sufficient evidence to suuport the claim that the drug appear to be​ effective.