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, 11 subjects had a mean wake time of 94.9 min and a standard deviation of 41.4 min. Assume that the 11 sample values appear to be from a normally distributed population and construct a 90​% 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 <σ< ____min

Answer 1

We need to construct the 90% confidence interval for the population variance. We have been provided with the following information about the sample variance ( s = 41.4 ) and sample size:

s^2	1713.961713.96
n	1111

The critical values for α=0.1 and df = 10 degrees of freedom are

χL2 ​= χ^2(1−α/2, n−1​ ) = 3.9403

χU2 ​ = (χ^2α/2, n−1 ) ​= 18.307.

The corresponding confidence interval is computed as shown below:

CI(Variance)​ = ​(n−1)s2/​χ^2(1−α/2, n−1​ ) ,​(n−1)s2​/(χ^2α/2, n−1 )

CI(variance) = 10*1713*.9617/ 3.9403 , 10*1713.9617/ 18.307

CI(variance) = (936.2301,4349.822)​

Now that we have the limits for the confidence interval, the limits for the 90% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

CI(Standard Deviation) = sqrt(936.2301​,4349.822​) = (30.5979,65.9532)

Therefore, based on the data provided, the 90% confidence interval for the population variance is 936.2301 < σ2 < 4349.822, and the 90% confidence interval for the population standard deviation is 30.5979 < σ < 65.9532.