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, 28 subjects had a mean wake time of 96.9 min and a standard deviation of 42.3 min. Assume that the 28 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. nothing minless thansigmaless than nothing min ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 96.9

s = 42.3

n = 28

Degrees of freedom = df = n - 1 = 28 - 1 = 27

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,27 = 2.473

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

= 2.473 * (42.3 / 28)

= 19.77

The 98% confidence interval estimate of the population mean is,

- E < < + E

96.9 - 19.77 < < 96.9 + 19.77

77.13 < < 116.67

(77.13,116.67)

The 98% confidence interval is 77.13 to 116.67