Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 2121 subjects had a mean wake time of 101.0101.0 min. After​ treatment, the 2121 subjects had a mean wake time of 77.977.9 min and a standard deviation of 22.322.3 min. Assume that the 2121 sample values appear to be from a normally distributed population and construct a 95​% 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 101.0 min before the​ treatment? Does the drug appear to be​ effective?

Answer 1

sample std dev ,    s =    22.3000
Sample Size ,   n =    21
Sample Mean,    x̅ =   77.9000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   20          
't value='   tα/2=   2.0860   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   22.3000   / √   21   =   4.866259
margin of error , E=t*SE =   2.0860   *   4.86626   =   10.150838
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    77.90   -   10.150838   =   67.749162
Interval Upper Limit = x̅ + E =    77.90   -   10.150838   =   88.050838
95%   confidence interval is (   67.75   < µ <   88.05   )

As we can see the confidence interval does not contain the 101 and value less than 101. It means drug was effective in treating insomnia.

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