Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73degreesF. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body​ temperature?

Answer 1

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI =  $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

or

Lower limit of CI =  $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

or

Upper limit of CI =  $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F