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}

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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.)

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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.
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.

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