Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.5°F and a standard deviation of 0.61°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°F as the mean body​ temperature? What is the confidence interval estimate of the population mean μ​?

Answer 1

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 98.3ºF and 98.7ºF. 98.6°F is part of the confidence interval, which means that the sample suggests that this is a correct measure.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.99)/(2) = 0.005`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.005 = 0.995`, so Z = 2.575

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575(0.61)/(√(103)) = 0.2`

The lower end of the interval is the sample mean subtracted by M. So it is 98.5 - 0.2 = 98.3ºF.

The upper end of the interval is the sample mean added to M. So it is 98.5 + 0.2 = 98.7ºF.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 98.3ºF and 98.7ºF. 98.6°F is part of the confidence interval, which means that the sample suggests that this is a correct measure.