Confidence Interval for Standard Deviation:
sqrt( (n-1)*s^2/ chi-square alpha/2, n-1 DOF) < sigma < sqrt( (n-1)*s^2/ chi-square 1-alpha/2, n-1 DOF)
where n = 17, s = 41.6, alpha = 0.01, DOF = degrees of freedom
Noname K.
asked 04/11/22A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug,
17
subjects had a mean wake time of
90.8
min and a standard deviation of
41.6
min. Assume that the
17
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?
Confidence Interval for Standard Deviation:
sqrt( (n-1)*s^2/ chi-square alpha/2, n-1 DOF) < sigma < sqrt( (n-1)*s^2/ chi-square 1-alpha/2, n-1 DOF)
where n = 17, s = 41.6, alpha = 0.01, DOF = degrees of freedom
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