Question

A certain drug was tested for it's effectiveness with insomnia. In a random sample of 18 adults treated with the drug for insomnia, the standard deviation of awake time was 42.3 minutes. Assume the sample comes from a population that is normally distributed. Find a 98% confidence interval for the population standard deviation of awake times for all adults who are treated with the drug. Sketch a curve and shade the appropriate region with notation. Show all work using proper notation.

Answer 1

Solution :

Given that,

c = 0.98

s = 42.3

n = 18

At 98% confidence level the is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

_/2,df = _0.01,17 = 33.41

and

_{1- /2,df} = _0.99,17 = 6.41

The 98% confidence interval for is,

s (n-1) / _/2,df < < s (n-1) / _{1- /2,df}

42.3( 18 - 1 ) / 33.41< < 42.3( 18- 1 ) / 6.41

30.17< < 68.90

( 30.17, 68.90)