Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 20 subjects had a mean wake time of 105.0 min. After​ treatment, the 20 subjects had a mean wake time of 98.5 min and a standard deviation of 21.52 min. Assume that the 20 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? Construct the 95​% confidence interval estimate of the mean wake time for a population with the treatment.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 98.5

sample standard deviation = s = 21.52

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_{0.025, 19 = 2.093}

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

= 2.093 * ( 21.52 / 20)

Margin of error = E = 10.1

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

  ± E

= 98.5 ± 10.1

= ( 88.4, 108.6 ) 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