Question

a clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. before treatment, 25 subjects had a mean wake time of 104.0 min. after treatment, the 25 subjects had a mean wake time of 81.4 min and a standard deviation of 23.9 min. assume that the 25 sample values appear to be from a normally distributed population and construct a 90% 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 104.0 min before the treatment? does the drug appear to be effective? construct the 90% confidence interval estimate of the mean wake time for a population with the treatment. 73.2 min < μ < 89.6 m (round to one decimal p are different. what does the result su le of 104.0 min before the treatment? does the drug appear to be effective? could be the same. the confidence interval ake time of 104.0 min before the treatment, so the means before and after the treatment this result suggests that the drug treatment an effect.

Explanation:

Step1: Determine the formula

For a confidence interval when the population standard deviation \(\sigma\) is unknown (we use sample standard deviation \(s\) instead), and the sample comes from a normal distribution, the formula is \(\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}}<\mu <\bar{x} + t_{\alpha/2}\frac{s}{\sqrt{n}}\)

We are given \(n = 25\), \(\bar{x}=81.4\), \(s = 23.9\), and the confidence level \(C=0.90\). Then \(\alpha=1 - C=0.10\) and \(\alpha/2=0.05\). The degrees of freedom \(df=n - 1=25-1 = 24\)

Step2: Find the \(t\) - value

Using a \(t\) - distribution table or a calculator, \(t_{\alpha/2,df}=t_{0.05,24}=1.711\)

Step3: Calculate the margin of error \(E\)

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=1.711\times\frac{23.9}{\sqrt{25}}=1.711\times\frac{23.9}{5}=1.711\times4.78 = 8.18\)

Step4: Calculate the confidence interval

The lower limit is \(\bar{x}-E=81.4 - 8.18=73.2\)
The upper limit is \(\bar{x}+E=81.4 + 8.18=89.6\)

Since the mean wake - time before treatment (\(104.0\) min) is not in the confidence interval \(73.2\) min\(<\mu<89.6\) min, the means before and after treatment are different.

Answer:

The confidence interval \(73.2\) min\(<\mu<89.6\) min. The mean wake - time before treatment (\(104.0\) min) is not in the confidence interval. So the means before and after treatment are different. This result suggests that the drug treatment has an effect.