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 94.8 min and a standard deviation of 24.9 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?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 94.8

sample standard deviation = s = 24.9

sample size = n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

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

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

= 2.878 * ( 24.9 / 19)

Margin of error = E = 16.4

The 99% confidence interval estimate of the population mean is,

- E < < + E

94.8 - 16.4 < < 94.8 + 16.4

(78.4 min.< < 111.2 min.)

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