Pregunta
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A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 22 subjects had a mean wake fime of 105.0 min. After treatment, the mean wake time of 101.5 min and a standard deviation of 20.8 min . Assume that the 22 sample values appear to be from a normally distributed population and construct a \( 90 \% \) confidence in mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 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. \( \square \mathrm{min}<\mu<\square \) min (Round to one decimal place as needed.)

Ask by Higgins Weston. in the United States
Mar 15,2025

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The \( 90\% \) confidence interval for the mean wake time after treatment is \( 93.9 \) min to \( 109.1 \) min. Since the pre-treatment mean of \( 105.0 \) min falls within this interval, the drug does not appear to be effective in reducing wake time.

Solución

We are given: - Sample size: \( n=22 \) - Sample mean after treatment: \( \bar{x}=101.5 \) min - Sample standard deviation: \( s=20.8 \) min Since the population is normally distributed and the sample size is small, we use the \( t \)-distribution. The degrees of freedom are: \[ \text{df} = n - 1 = 21. \] For a \( 90\% \) confidence level, the critical value is \( t_{0.95, 21} \) (because there is \( 5\% \) in each tail). A standard \( t \)-table gives approximately: \[ t_{0.95,21} \approx 1.721. \] The standard error (SE) is: \[ \text{SE} = \frac{s}{\sqrt{n}} = \frac{20.8}{\sqrt{22}}. \] Calculating the standard error: \[ \sqrt{22} \approx 4.690,\quad \text{thus}\quad \text{SE} \approx \frac{20.8}{4.690} \approx 4.4. \] The margin of error (ME) is: \[ \text{ME} = t_{0.95,21} \cdot \text{SE} \approx 1.721 \times 4.4 \approx 7.6. \] Thus, the \( 90\% \) confidence interval for the mean wake time after treatment is: \[ 101.5 - 7.6 < \mu < 101.5 + 7.6, \] \[ 93.9 \text{ min} < \mu < 109.1 \text{ min}. \] Since the pre-treatment mean wake time was \( 105.0 \) min, and \( 105.0 \) min falls within the confidence interval \( (93.9, 109.1) \), the data suggest that there is no statistically significant change in mean wake time after treatment. In other words, the drug does not appear to be effective in reducing wake time.

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To construct a 90% confidence interval for the mean wake time after treatment, we first need to find the standard error (SE) and then use the t-distribution because the sample size is small (n=22). 1. **Calculate the Standard Error (SE):** \[ SE = \frac{s}{\sqrt{n}} = \frac{20.8}{\sqrt{22}} \approx 4.43 \text{ min} \] 2. **Find the t-value for a 90% confidence interval** For \( n-1 = 21 \) degrees of freedom and a 90% confidence level, the t-value is approximately \( 1.645 \) (from t-distribution tables). 3. **Construct the confidence interval:** \[ \text{Margin of Error (ME)} = t \times SE = 1.645 \times 4.43 \approx 7.28 \text{ min} \] \[ \text{Confidence Interval} = \bar{x} \pm ME = 101.5 \pm 7.28 \] \[ \text{Lower Limit} = 101.5 - 7.28 \approx 94.2 \text{ min} \] \[ \text{Upper Limit} = 101.5 + 7.28 \approx 108.8 \text{ min} \] So, the 90% confidence interval for the mean wake time after treatment is: \[ 94.2 \text{ min} < \mu < 108.8 \text{ min} \] Now, regarding the initial mean wake time of 105.0 min: the confidence interval includes this value, suggesting that we cannot definitively say whether the drug is effective or not based on this result alone, as it does not provide strong evidence to suggest a significant reduction in wake time. The drug may or may not be effective, as the mean post-treatment wake time falls within the range that was present before the treatment.

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