Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 degrees°F and a standard deviation of 0.56 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 98.6degrees°F as the mean body​ temperature?

a. What is the confidence interval estimate of the population mean mμ​?

b.What does this suggest about the use of 98.6degrees°F as the mean body​ temperature?

Answer 1

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 98.1

S = 0.56

n = 103

df = n – 1 = 102

Confidence level = 99%

Critical t value = 2.6249

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 98.1 ± 2.6249*0.56/sqrt(103)

Confidence interval = 98.1 ± 0.1448

Lower limit = 98.1 - 0.1448 = 97.96

Upper limit = 98.1 + 0.1448 = 98.24

Confidence interval = (97.96, 98.24)

Part b

We cannot use 98.6 degrees °F as the mean body​ temperature, because this value is not lies between above interval.