Question

A simple random sample from a population with a normal distribution of 97 body temperatures has x overbarequals99.10degrees Upper F and sequals0.68degrees Upper F. Construct an 80​% confidence interval estimate of the standard deviation of body temperature of all healthy humans. LOADING... Click the icon to view the table of​ Chi-Square critical values. nothingdegrees Upper Fless thansigmaless than nothingdegrees Upper F

Answer 1

df = n - 1 = 97 - 1 = 96

Chi square critical values at 0.20 degrees of freedom with 96 df = $\chi^{2}$ L = 78.725 ,$\chi^{2}$U = 114.131

80% confidence interval for  $\sigma$ is

Sqrt ( (n-1) S² / $\chi^{2}$ U) < $\sigma$ < Sqrt ( (n-1) S² / $\chi^{2}$ L)

Sqrt [ 96 * 0.68² / 114.131 ) < $\sigma$ < Sqrt [ 96 * 0.68² / 78.725 )

0.6237 < $\sigma$ < 0.7509