Questions: 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 102.0 min. After treatment, the 18 subjects had a mean wake time of 79.2 min and a standard deviation of 21.3 min. Assume that the 18 sample values appear to be from a normally distributed population and construct a 95% 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 102.0 min before the treatment? Does the drug appear to be effective? Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment.

Solution Steps

In a clinical trial to test the effectiveness of a drug for treating insomnia in older subjects, the following data was collected:

• Mean wake time before treatment: \(\bar{x}_{\text{before}} = 102.0\) min
• Mean wake time after treatment: \(\bar{x}_{\text{after}} = 79.2\) min
• Standard deviation after treatment: \(s = 21.3\) min
• Sample size: \(n = 18\)

To construct a \(95\%\) confidence interval for the mean wake time after treatment, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 79.2\) min (sample mean after treatment)
• \(t\) is the critical value from the t-distribution for \(n-1\) degrees of freedom at the \(95\%\) confidence level.
• \(s = 21.3\) min (sample standard deviation)
• \(n = 18\) (sample size)

The critical value \(t\) for \(n-1 = 17\) degrees of freedom at the \(95\%\) confidence level is approximately \(2.1\).

Substituting the values into the formula:

\[ 79.2 \pm 2.1 \cdot \frac{21.3}{\sqrt{18}} \]

Calculating the margin of error:

\[ \frac{21.3}{\sqrt{18}} \approx 5.03 \]

Thus, the margin of error is:

\[ 2.1 \cdot 5.03 \approx 10.6 \]

Now, we can calculate the confidence interval:

\[ (79.2 - 10.6, 79.2 + 10.6) = (68.6, 89.8) \]

The \(95\%\) confidence interval for the mean wake time after treatment is:

\[ 68.6 \text{ min} < \mu < 89.8 \text{ min} \]

Since the entire confidence interval is below the mean wake time before treatment (\(102.0\) min), we conclude that the drug appears to be effective in reducing the mean wake time.

Final Answer