Question

A data set includes 109 body temperatures of healthy adult humans having a mean of 98.3degrees°F and a standard deviation of 0.54 degrees°F. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6 degrees°F as the mean body​ temperature?

Answer 1

A data set includes n= 109 body temperatures of healthy adult humans having a mean of =98.3degrees°F and a standard deviation of s=0.54 degrees°F.

Since the population standard deviation is unknown and sample size is greater than 30 so, we can assume the sample is from a normal distribution but we will use t-distribution to calculate the confidence interval which is calculated as:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level and by degree of freedom n-1 using excel tool. df=n-1
s_M = standard error = √(s²/n)

M = 98.3 df=109-1=108
t = 2.62 at 99% confidence interval and degree of freedom, df=n-1=109-1=108 by excel formula =T.INV.2T(0.01,108)
s_M = √(0.54²/109) = 0.05

μ = M ± t(s_M)
μ = 98.3 ± 2.62*0.05
μ = 98.3 ± 0.136

The confidence interval is:

CI [98.164, 98.436].

b) If we use 98.6 degrees°F against the calculated confidence interval as we can see that the confidence interval does not include the 98.6 degrees°F so, we cannot use the value as the mean body temeperature