Question

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are:98.9 96.6 98.6 99.7 97 97.4 99.4Assume body temperatures of adults are normally distributed. Based on this data, find the 98% confidenceinterval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e.,parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.98% C.I. -

Answer 1

Given the data temperatures to be;

$98.9,96.6,98.6,99.7,97,97.4,99.4$

We would require the following to get the 98% confidence interval of the mean body temperature.

Mean, Standard deviation, sample size, Probability of a confidence interval of 98%.

Using a calculator, we can get the mean to be

$\text{(}\mu)=98.2285$

The standard deviation would be derived to be;

$\sigma=1.2230$

The sample size can be gotten from the question to be;

$n=7$

The probability value of a 98% confidence interval is given to be 2.33

We can then derive the answer using the formula below;

$\mu\pm z^x(\frac{\sigma}{\sqrt[]{n}})$

We would substitute into the formula

$\begin{gathered} \mu\pm z^x(\frac{\sigma}{\sqrt[]{n}}) \\ =98.2285+2.33(\frac{1.2230}{\sqrt[]{7}}) \\ =98.2285\pm1.0770 \\ =(97.152,99.306) \end{gathered}$

ANSWER:

$(97.152,99.306)$