A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 28 subjects had a mean wake time of 98.1 min and a
standard deviation of 42.1 min. Assume that the 28 sample values appear to be from a normally distributed population and construct a confidence interval estimate of the standard deviation of
the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective?
Find the confidence interval estimate.
min < min
(Round to two decimal places as needed.)

Question

A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 28 subjects had a mean wake time of 98.1 min and a
standard deviation of 42.1 min. Assume that the 28 sample values appear to be from a normally distributed population and construct a confidence interval estimate of the standard deviation of
the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective?
Find the confidence interval estimate.
min < min
(Round to two decimal places as needed.)

Ask by Warren Schultz. in the United States

Mar 25,2025

UpStudy AI · Accepted Answer

Given that the sample standard deviation is $$ s = 42.1 $$ minutes for $$ n = 28 $$ subjects, the confidence interval for the population variance is based on the chi‐square distribution. In particular, a $$ 95\% $$ confidence interval for the variance $$ \sigma^2 $$ is

$$
\left[\frac{(n-1)s^2}{\chi^2_{0.975,\, n-1}},\;\frac{(n-1)s^2}{\chi^2_{0.025,\, n-1}}\right],
$$

and the corresponding interval for the standard deviation $$ \sigma $$ is

$$
\left[\sqrt{\frac{(n-1)s^2}{\chi^2_{0.975,\, n-1}}},\;\sqrt{\frac{(n-1)s^2}{\chi^2_{0.025,\, n-1}}}\right].
$$

For $$ n=28 $$ the degrees of freedom are $$ n-1 = 27 $$. Using a chi-square table (or calculator) we take approximate critical values

$$
\chi^2_{0.975,27} \approx 14.26 \quad \text{and} \quad \chi^2_{0.025,27} \approx 41.34.
$$

Step 1. Compute $$ (n-1)s^2 $$:

$$
(n-1)s^2 = 27 \times (42.1)^2.
$$

First, compute

$$
(42.1)^2 = 42.1 \times 42.1 = 1772.41.
$$

Thus,

$$
27 \times 1772.41 \approx 47855.07.
$$

Step 2. Find the lower bound for $$ \sigma $$:

$$
\sigma_L = \sqrt{\frac{47855.07}{41.34}}.
$$

Calculate the fraction:

$$
\frac{47855.07}{41.34} \approx 1158.55,
$$

so that

$$
\sigma_L \approx \sqrt{1158.55} \approx 34.04 \text{ minutes}.
$$

Step 3. Find the upper bound for $$ \sigma $$:

$$
\sigma_U = \sqrt{\frac{47855.07}{14.26}}.
$$

Compute the fraction:

$$
\frac{47855.07}{14.26} \approx 3356.50,
$$

thus

$$
\sigma_U \approx \sqrt{3356.50} \approx 57.96 \text{ minutes}.
$$

Therefore, the $$ 95\% $$ confidence interval estimate for the standard deviation $$ \sigma $$ is

$$
34.04\text{ min} < \sigma < 57.96\text{ min}.
$$

Regarding the effectiveness of the treatment, note that the confidence interval calculated here concerns the variability (i.e. the standard deviation) of wake times among subjects after treatment. It does not compare the mean wake time with a baseline or control condition. Consequently, this interval does not by itself indicate whether the drug treatment is effective for treating insomnia.
The $$ 95\% $$ confidence interval for the standard deviation of wake times is $$ 34.04 $$ minutes to $$ 57.96 $$ minutes. This interval indicates the range within which the true standard deviation of wake times for the population is likely to fall, but it does not directly show whether the drug treatment is effective for treating insomnia.

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