MATH SOLVE

2 months ago

Q:
# A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 15 subjects had a mean wake time of 100.0 min. After treatment, the 15 subjects had a mean wake time of 92.6 min and a standard deviation of 43.7 min. Assume that the 15 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 100.0 min before the treatment? Does the drug appear to be effective?

Accepted Solution

A:

The answers would be that from the standard normal table, the z-critical value at 98% confidence is 2.33

98%C.I. = µ ± z * σ / sqrt of n

= 98.90 ± 2.33 * 42.30 / sqrt of 16

= 98.90 ± 24.64

= 74.26 to 125.54

A 98% confidence interval estimate of the mean wake time for a population with zopiclone treatments is 74.26 to 125.54.

The result does not suggest about the mean wake time of 102.80 minutes before the treatment due to the mean wake time of 102.80 minutes included in the 98% confidence interval which is between 74.26 and 125.54. Zopiclone does not appear effective

98%C.I. = µ ± z * σ / sqrt of n

= 98.90 ± 2.33 * 42.30 / sqrt of 16

= 98.90 ± 24.64

= 74.26 to 125.54

A 98% confidence interval estimate of the mean wake time for a population with zopiclone treatments is 74.26 to 125.54.

The result does not suggest about the mean wake time of 102.80 minutes before the treatment due to the mean wake time of 102.80 minutes included in the 98% confidence interval which is between 74.26 and 125.54. Zopiclone does not appear effective