Question

In Exercises 9–16, assume that each sample is a simplerandom sample obtained from a population with a normal distribution.

Insomnia Treatment A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. After treatment with zopiclone, 16 subjects had a mean wake time of 98.9 min and a standard deviation of 42.3 min (based on data from “Cognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults,” by Sivertsen et al.,Journal of the American Medical Association,Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population and construct a 98% confidence interval estimate of the standard deviation of the wake timesfor a population with zopiclone treatments. Does the result indicate whether the treatment is effective?

Short answer

The 90% confidence interval estimate of the standard deviation of the wake times for a population with zopiclone treatments is $29.6\min <\sigma <71.6\min$.

And, the confidence interval does not indicate the effectiveness of the treatment.

Step-by-step solution

Given information

The sample number of values is $n=16$.

The mean wake time is 98.9 min.

The sample standard deviation is $s=42.3$min.

The level of confidence is 98%.

Compute the critical values and confidence interval

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =16-1 \\ =15 \end{gathered}$$

The level of confidence is 98%, which implies that the level of significance is 0.02.

Using the Chi-square table, the critical values at 0.02 level of significance and at 15 degrees of freedom arerole="math" localid="1648109418028" ${\chi }_L^2=5.2293$ and ${\chi }_R^2=30.5779$.

The 98% confidence interval estimate of the standard deviation of the wake times for a population with zopiclone treatments is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(16-1\right){42.3}^2}{30.5779}}<\sigma <\sqrt{\frac{\left(16-1\right){42.3}^2}{5.2293}} \\ 29.6<\sigma <71.6 \end{gathered}$$

Therefore, the 90% confidence interval estimate of the standard deviation of the wake times for a population with zopiclone treatments is role="math" localid="1648109480702" $29.6min<\sigma <71.6min$.

And, the estimated confidence interval of the standard deviation of the wake times for a population with zopiclone treatments does not indicate whether the treatment is effective.