Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 14 subjects had a mean wake time of 104.0 min. After​ treatment, the 14 subjects had a mean wake time of 83.9 min and a standard deviation of 22.7 min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 90​% 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 104.0 min before the​ treatment? Does the drug appear to be​ effective?

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 83.9

sample standard deviation = s = 22.7

sample size = n = 14

Degrees of freedom = df = n - 1 = 14 - 1 = 13

At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ =1 - 0.90 =0.10

$\alpha$/2 = 0.05

t$\alpha$/2,df = t_{0.05, 13} = 1.771

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.771 * ( 22.7/ $\sqrt$ 14)

Margin of error = E = 10.7

The 90% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 83.9 ± 10.7

= (73.2, 94.6 )

The confidence interval does not include the mean wake time of 104.0 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