Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 17 subjects had a mean wake time of 100.0 min. After​ treatment, the 17 subjects had a mean wake time of 90.3 min and a standard deviation of 21.8 min. Assume that the 17 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?

Answer 1

Explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 21.8

n = number of samples = 17

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 17 - 1 = 16

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.921

Margin of error = 2.921 × 21.8/√17

= 15.44

The confidence interval for the mean wake time for a population with drug treatments is

90.3 ± 15.44

The upper limit is 90.3 + 15.44 = 105.74 mins

The lower limit is 90.3 - 15.44 = 74.86 mins

The result suggests that the mean wake time might have really reduced since the values barely fall above 100 minutes as in before treatment with a high degree of confidence. Therefore, the drug is effective.