Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.9 and a standard deviation of 0.65F. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6F as the mean body temperature

Answer 1

SOLUTION

given : sample mean (T) = 98.9, sample standard deviation (s) = 0.65,

sample size (n) = 103

degrees of freedom (df) =n-1 = 103 - 1 = 102

Confidence level (C) = 0.99

a=1-C=1-0.99 = 0.01

Refer t-table or use excel function "=T.INV.2T(0.01,102)" to find the value of t.

to = 2.625

$\dpi{100} \mathbf{formula:}\; confidence\; interval \;for\; the \;population\; mean,\; \mu$

{ } - 1+=

= 98.9 2.625 * 0.65 (103)

= 8.9 +0.168

Lower limit = 98.9 -0.168 = 98.732'F

Upper limit = 98.9 +0.168 = 99.068°F

• 98.732°F <μ < 99.068°F

This suggests that the mean body temperature could be higher than 98.6°F