Question

A simple random sample from a population with a normal distribution of 108 body temperatures has x overbarxequals=98.30 degrees Upper F°F and sequals=0.61 degrees Upper F°F. Construct aa 98% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

Solution:

Given a = 1-0.98 = 0.02, the critical value is X^2 with 0.01 and df = 108-1 = 107 is 75.93 (from chisquare table).
The critical value is X^2 with 0.99 and df = 97 is 143.94.

So 98% confidence interval is

(√(n-1)*s^2/143.94, √(n-1)*s^2/75.93)
--> (√97*0.61^2/143.94, √97*0.61^2/75.93)
--> (0.0254, 0.0483)