Question

A simple random sample from a population with a normal distribution of 99 body temperatures has x overbarequals98.20degrees Upper F and sequals0.64degrees Upper F. Construct a 90​% 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 ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

c = 0.90

s = 0.64

n = 99

At 90% confidence level the is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

_/2,df = _0.05,98 = 122.11

and

_{1- /2,df} = _0.95,98 = 76.16

²_L = ²_/2,df = 122.11

²_R = ²_{1 - /2,df} = 76.16

The 90% confidence interval for is,

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

0.64 ( 99 - 1 ) / 122.11 < < 0.64( 99 - 1 ) / 76.16

0.57 < < 0.73

( 0.57 , 0.73)