A simple random sample from a population with a normal distr

A simple random sample from a population with a normal distribution of 102 body temperatures has x bar=98.8 F and s=.66 F. Construct a 98% confidence interval estimate of the standard deviation of body temperatures of all healthy humans. Is it safe to conclude that the population standard deviation is less than .90 F?

Solution

Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.02
^2 right = (1 - Confidence Level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
^2 left = 1 - ^2 right = 1 - 0.01 = 0.99
the two critical values ^2 left, ^2 right at 101 df are 136.971 , 70.901
S.D( S^2 )=0.66
Sample Size(n)=102
Confidence Interval = [ 101 * 0.4356/136.971 < ^2 < 101 * 0.4356/70.901 ]
= [ 43.9956/136.971 < ^2 < 43.9956/70.9007 ]
For Variance = [ 0.3212 , 0.6205 ]       
For S.d = [ Sqrt(0.3212) , Sqrt(0.6205) ] = [ 0.5667, 0.7877]

No, it is n\'t safe to use 0.90 as s.d