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, 14 subjects had a mean wake time of 97.6 min and a standard deviation of 41.9 min. Assume that the 14 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

98​% confidence interval estimate of the standard deviation =(28.71, 74.52)

n=14, sample standard deviation=s=41.9

(1-alpha)100% confidence interval for σ={sqrt((n-1)s²/chi-sq( alpha/2 ,n-1)),sqrt((n-1)s²/chi-sq( 1-alpha/2 ,n-1)) }

98% confidence interval for σ={ sqrt(n-1)s²/chi-sq( 0.02/2 ,n-1),sqrt(n-1)s²/chi-sq(1- 02/2 ,n-1) }

={sqrt(13*41.9*41.9/27.69),sqrt(13*41.9*41.9/4.11)}={28.71, 74.52}

chi-sq(1-0.02/2,13)=4.11,                   chi-sq(0.02/2,13)=27.69

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