Question

a clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. before treatment, 18 subjects had a mean wake time of 104.0 min. after treatment, the 18 subjects had a mean wake time of 83.1 min and a standard deviation of 22.7 min. assume that the 18 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 \(n = 18\), the degrees of freedom \(df=n - 1=18-1 = 17\). 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 \(t_{\frac{\alpha}{2}}\) with \(df = 17\) is \(t_{0.05,17}=1.740\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) when the population standard deviation \(\sigma\) is unknown is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 22.7\) (sample standard deviation) and \(n = 18\).

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Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) (when \(\sigma\) is unknown) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=83.1\) (sample mean).

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Answer:

\(73.8\) min\(<\mu<92.4\) min

Since the \(90\%\) confidence interval \((73.8,92.4)\) does not contain the value \(104.0\) (the mean wake - time before treatment), it suggests that the mean wake - time after treatment is significantly different from the mean wake - time before treatment. So, the drug appears to be effective.