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. min < μ < min (round to one decimal place as needed.)

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 25\), the degrees of freedom \(df=n - 1=25-1 = 24\). For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\). Using the \(t\) - distribution table or a calculator, the critical value \(t_{\frac{\alpha}{2},df}=t_{0.05,24}=1.711\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) for a \(t\) - distribution is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 23.9\) (sample standard deviation) and \(n = 25\).

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Step3: Calculate the confidence interval

The sample mean \(\bar{x}=81.4\). The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\).

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Answer:

\(73.2\) min\(<\mu<89.6\) min