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, 16 subjects had a mean wake time of 95.8 min and a standard deviation of 43.2 min. 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 times for a population with the drug treatments. Does the result indicate whether the treatment is​ effective?

Answer 1

Test and CI for One Variance

Method

The chi-square method is only for the normal distribution.
Statistics

N StDev Variance
16   43.5      1892

98% Confidence Interval for estimating of the standard deviation of the wake times for a population with the drug treatments

= (30.5, 73.7)

One-Sample T

N Mean StDev SE Mean      98% CI
16 95.8   43.5     10.9         (67.5, 124.1)

98% Confidence Interval for estimating of the mean wake times for a population with the drug treatments

= (67.5, 124.1)

If the mean wake times for a population without the drug treatments<Lower boundary of 98% CI=67.5 or mean wake times for a population without the drug treatments>Upper boundary of 98% CI for mean=124.1 then we can say that the result indicate the treatment is​ effective. Otherwise it is ineffective. For answering this part we need population mean wake times for a population without the drug treatments.