Question

Insomnia Treatment A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone, 16 subjects had a mean wake time of 102.8 min. After treatment with zopiclone, the 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 mean wake time for a population with zopiclone treatments. What does the result suggest about the mean wake time of 102.8 min before the treatment? Does zopiclone appear to be effective?

Short answer

The 98% confidence intervalof the meanwake time for a population after treatment with the drug zopiclone is\(71.4{\rm{ }}\min < \mu < 126.4{\rm{ }}\min \).

The treatment with the drug zopiclone does not appear to be effective since the mean before the treatment is included in the interval.

Step-by-step solution

Given information

To test the effectiveness of the drug zopiclone for treating insomnia, critical trial was conducted on 16 (n) subjects. The mean wake time of subjects before treatment is 102.8 min; and after treatment,it is 98.2 min\(\left( {\bar x} \right)\) with standard deviation of 42.3 min (s).

Check the requirements

It is assumed that values appear to be normally distributed.

Further, assume that the samples are selected randomly with unknown population standard deviation.

Thus, the t-distribution would be used here.

Describe the formula for confidence interval

The formula for\(\left( {1 - \alpha } \right)\% \)confidence interval is\(\bar x - E < \mu < \bar x + E\).

Here, E is margin of error which is given by,\(E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\)

Where,\({t_{\frac{\alpha }{2}}}\)is the critical value with level of significance\(\alpha \).

Here,\({\rm{\bar x}}\)represents the sample mean of wake time of subjects after treatment with zopiclone and\({\rm{\mu }}\)represents the population mean of wake time of subjects after treatment with zopiclone.

Calculate the critical value

98% level of confidence implies\(\alpha = 0.02\).

The degree of freedom is,

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 16 - 1\\{\rm{ }} = 15\end{array}\)

In the t-distribution table, find the value corresponding to the row value of degree of freedom 15 and column value of area in one tail 0.01 is 2.602 which is critical value\({{\rm{t}}_{{\rm{0}}{\rm{.01}}}}\).

Therefore, the critical value\({{\rm{t}}_{0.01}}\)is 2.602.

Calculate margin of error

Margin of error is given by,

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\\ = 2.602 \times \frac{{42.3}}{{\sqrt {16} }}\\ = 27.5212\end{array}\)

Therefore, the margin of error is 27.52.

Construct the confidence interval

The 98% confidence interval for mean wake time after the treatment is,

\(\begin{array}{l}\bar x - E < \mu < \bar x + E = 98.9 - 27.52 < \mu < 98.9 + 27.52\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 71.38 < \mu < 126.42\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \approx 71.4 < \mu < 126.4\end{array}\)

Therefore, 98% confidence interval is \(71.4{\rm{ }}\min < \mu < 126.4{\rm{ }}\min \).

Interpret the result

The 98% confidence intervalof the meanwake time for a population after treatment with the drug zopiclone is\(71.4{\rm{ }}\min {\rm{ }} < {\rm{ }}\mu {\rm{ }} < {\rm{ }}126.4{\rm{ }}\min \).

Here,the confidence interval includes mean wake time of 102.8 minbefore the treatment which means the mean wake time after the treatment with zopiclone is not different. Therefore, the treatment with the drug zopiclone does not have significant effect.