Question

A simple random sample from a population with a normal distribution of 99 body temperatures has x overbarequals98.90degrees Upper F and sequals0.62degrees Upper F. Construct a 95​% 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.

Answer 1

Solution:

Confidence interval for population standard deviation is given as below:

Sqrt[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ < sqrt[(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

We are given

Confidence level = 95%

Sample size = n = 99

Degrees of freedom = n – 1 = 98

Sample standard deviation = S = 0.62

χ²_{α/2, n – 1} = 127.2821

χ²_{1 -α/2, n– 1} = 72.5009

(By using chi square table)

Sqrt[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ < sqrt[(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

Sqrt[(99 – 1)* 0.62^2 / 127.2821] < σ < sqrt[(99 – 1)* 0.62^2/ 72.5009]

Sqrt(0.2960) < σ < sqrt(0.5196)

0.5440 < σ <   0.7208

Lower limit = 0.5440

Upper limit = 0.7208