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 105.0 min. After​ treatment, the 19 subjects had a mean wake time of 99.4 min and a standard deviation of 21.9 min. Assume that the 19 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 105.0 min before the​ treatment? Does the drug appear to be​ effective?

PART 2:

What does the result suggest about the mean wake time of 105.0 min before the​ treatment? Does the drug appear to be​ effective? The confidence interval_____ includes

does not include the mean wake time of 105.0 min before the​ treatment, so the means before and after the treatment _______ are different. could be the same. This result suggests that the drug treatment ________. does not have has a significant effect.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 99.4 min.

sample standard deviation = s = 21.9 min.

sample size = n = 19

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

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,18 = 2.101

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

= 2.101 * ( 21.9 / 19)

Margin of error = E = 10.6

The 95% confidence interval estimate of the population mean is,

   - E < < + E

99.4 - 10.6 < < 99.4 + 10.6

( 88.8 min. < < 110.0 min.)

The confidence interval include the mean wake time of 105.0 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