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A simple random sample from a population with a normal distribution of 97 body temperatures has...

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

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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

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