Question

A simple random sample from a population with a normal distribution of

107107

body temperatures has

x overbarxequals=98.3098.30degrees Upper F°F

and

sequals=0.680.68degrees Upper F°F.

Construct

aa

9999​%

confidence interval estimate of the standard deviation of body temperature of all healthy humans.

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Answer 1

Chi square critical values at 0.01 level with 106 df = L = 72.251 , R = 147.247

99% confidence interval for is

sqrt [ - t * S / R] < < sqrt [ + t * S /L]

sqrt [ 106 * 0.68² / 147.247] < < sqrt [ 106 * 0.68² / 72.251 ]

0.5770 < < 0.8236

99% CI is ( 0.5770 , 0.8236)