Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 22 subjects had a mean wake time of 105.0 min. After​ treatment, the 22 subjects had a mean wake time of 95.7 min and a standard deviation of 22.2 min. Assume that the 22 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 105.0 min before the​ treatment? Does the drug appear to be​ effective?

Answer 1

Answer:

Step-by-step explanation:

We want to determine a 99% confidence interval estimate of the mean wake time for a population with drug treatments.

Number of sample, n = 22

Mean, u = 95.7 min

Standard deviation, s = 22.2 min

For a confidence level of 90%, the corresponding z value is 2.58.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

95.7 ± 2.58 × 22.2/√22

= 95.7 ± 12.21

The lower end of the confidence interval is 95.7 - 12.21 = 83.49 min

The upper end of the confidence interval is 95.7 + 12.21 = 107.91 min

The result suggests that there is no significant difference between the mean wake time before and after treatment. Therefore, the drug does not appear to be effective.