Question

A simple random sample from a population with a normal distribution of 108 body temperatures has x overbarequals98.50degrees Upper F and sequals0.62degrees Upper F. Construct a 90​% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

Statistics and Probability

A simple random sample from a population with a normal distribution of n = 108 body temperatures has = 98.50degrees F and s = 0.62 degrees F.

Since population standard deviation is not known hence T-distribution is applicable for confidence interval calculation.

The confidence interval is calculated as:

μ = ± t(se)

where:

= sample mean
t = t statistic determined by confidence level and degree of freedom, df = n-1
se = standard error = √(s2/n)

= 98.50
t = 1.66 calculated uisng excel formula for t-distribution which is =T.INV.2T(0.10, 107)
se = √(0.62/108) = 0.06

μ = ± t(se)
μ = 98.50 ± 1.66*0.06
μ = 98.50 ± 0.099

So, the 90% confidence level would be:

{98.401, 98.599}