Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 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?

Answer 1

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$.

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.