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, 29 subjects had a mean wake time of 90.4 min and a standard deviation of 42.5 min. Assume that the 29 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?

Find the confidence interval estimate.

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Answer 1

Solution :

Given that,

sample standard deviation = s = 42.5 min.

sample size = n = 29

confidence interval = 98%

Degrees of freedom = df = n - 1 = 29 - 1 = 28

At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

t/2,df = t0.01,28 = 2.47

Margin of error = E = t/2,df * (s /n)

= 2.47 * (42.5 / 29)

Margin of error = E = 19.49

The 95% confidence interval estimate of the population mean is,

- E < < + E

90.4 - 19.49 < < 90.4 + 19.49

70.91 min. < < 109.89 min.

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