Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 19 subjects had a mean wake time of 103.0 min. After​ treatment, the 19 subjects had a mean wake time of 76.3 min and a standard deviation of 23.4 min. Assume that the 19 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 103.0 min before the​ treatment? Does the drug appear to be​ effective? Construct the 99​% confidence interval estimate of the mean wake time for a population with the treatment.

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Answer 1

Solution :

degrees of freedom = n - 1 = 19 - 1 = 18

t/2,df = t0.005,18 = 2.878

Margin of error = E = t/2,df * (s /n)

= 2.878 * ( 23.4/ 19)

Margin of error = E = 15.5

The 99% confidence interval estimate of the population mean is

- E < < + E

76.3 - 15.5 < < 76.3 + 15.5

( 60.8 min. < < 91.8 min.)

The confidence interval does not include the mean wake time of 103.0 min.before the treatment so the means before and after the treatment are different this result suggests that the drug treatment has a significant effect