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Part 1
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 1818 subjects had a mean wake time of 103.0103.0 min. After​ treatment, the 1818 subjects had a mean wake time of 81.381.3 min and a standard deviation of 21.621.6 min. Assume that the 1818 sample values appear to be from a normally distributed population and construct a 9595​% 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 103.0103.0 min before the​ treatment? Does the drug appear to be​ effective?
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Part 1
Construct the 9595​% confidence interval estimate of the mean wake time for a population with the treatment.
enter your response here minless than<muμless than<enter your response here min
​(Round to one decimal place as​ needed.)

We are asked to construct a 95% confidence interval for the mean wake time after the treatment and assess whether the drug treatment appears effective in reducing wake time.

### Step 1: Given information
- Sample size \( n = 18 \)
- Sample mean after treatment \( \bar{x} = 81.3 \) minutes
- Sample standard deviation \( s = 21.6 \) minutes
- Confidence level \( 95\% \)

### Step 2: Calculate the margin of error (E)

The formula for the margin of error for a confidence interval of the mean is:

\[
E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]

where:
- \( t_{\alpha/2} \) is the critical value of the t-distribution for \( \alpha = 0.05 \) (for a 95% confidence level) with \( df = n - 1 = 18 - 1 = 17 \).
- \( s \) is the sample standard deviation.
- \( n \) is the sample size.

Using a t-table or calculator, the critical value \( t_{0.025, 17} \) for a 95% confidence level with 17 degrees of freedom is approximately 2.110.

Now, calculate the margin of error:

\[
E = 2.110 \times \frac{21.6}{\sqrt{18}} = 2.110 \times \frac{21.6}{4.2426} \approx 10.75
\]

### Step 3: Confidence interval

The confidence interval is given by:

\[
\bar{x} - E < \mu < \bar{x} + E
\]

Substituting the values:

\[
81.3 - 10.75 < \mu < 81.3 + 10.75
\]

\[
70.5 < \mu < 92.1
\]

### Step 4: Interpretation

The 95% confidence interval for the mean wake time after treatment is between **70.5 minutes** and **92.1 minutes**.

### Step 5: Effectiveness of the drug

The mean wake time before the treatment was 103.0 minutes. Since the entire confidence interval (70.5 min to 92.1 min) lies below the pre-treatment mean of 103.0 minutes, we can conclude that the drug treatment **does appear to be effective** in reducing wake time in older subjects.

---

Would you like any further details or have any questions? Here are some related questions to deepen your understanding:

1. How is the t-distribution different from the normal distribution in this context?
2. Why is the t-distribution used instead of the z-distribution here?
3. What would the confidence interval look like if we had a larger sample size?
4. How does the confidence level affect the width of the confidence interval?
5. Can we conclude causality from this confidence interval analysis alone?

**Tip:** A larger sample size reduces the margin of error, leading to a narrower confidence interval.

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 103.0 min. After treatment, the 18 subjects had a mean wake time of 81.3 min and a standard deviation of 21.6 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 103.0 min before the treatment? Does the drug appear to be effective?

Construct a 95% confidence interval for the mean wake time after drug treatment in older subjects with a sample size of 18. The analysis includes calculating the margin of error and assessing the effectiveness of the drug in reducing wake time. Suitable for undergraduate statistics students.

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