Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9 F and a standard deviation of 0.67 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 98.6F as the mean body​ temperature?

What is the confidence interval estimate of the population mean ​?

Answer 1

Given
X̅ = 98.9 F ....... Sample Mean body temperature
n = 103      ....... Sample Size
s = 0.67 F       ....... Sample Standard Deviation

Since the population standard deviation is unknown, we use the t-distribution
Degrees of Freedom = df = n - 1 = 103 - 1 = 102

For 99% Confidence interval

α = 0.01,      α/2 = 0.005
From t tables of Excel function T.INV.2T (α, degrees of freedom) we find the t value
t = T.INV.2T (0.01, 102) = 2.625
We take the positive value of t

Confidence interval is given by

= (98.7267, 99.0733)
= (98.73, 99.07)         ... Rounded to 2 decimals

99% Confidence interval is

98.6 ^oF is less than the lower limit 98.73 ^oF of the confidence interval

Hence, 98.6 ^oF is not a standard mean body temperature considering 99% confidence level and the current sample of 103 body temperatures

99% confidence interval estimate of the population mean lies between the interval

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