Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment,

2323

subjects had a mean wake time of

100.0100.0

min. After​ treatment, the

2323

subjects had a mean wake time of

74.674.6

min and a standard deviation of

20.920.9

min. Assume that the

2323

sample values appear to be from a normally distributed population and construct a

9999​%

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

100.0100.0

min before the​ treatment? Does the drug appear to be​ effective?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 74.6

sample standard deviation = s = 20.9

sample size = n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,22 = 2.819

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

= 2.819 * (20.9 / 23)

Margin of error = E = 12.3

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

- E < < + E

74.6 - 12.3 < < 74.6 + 12.3

( 62.3 min. < < 86.9 min.)

The confidence interval does not include  the mean wake time of 100.0 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