Respuesta :
Given the data temperatures to be;
[tex]98.9,96.6,98.6,99.7,97,97.4,99.4[/tex]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
[tex]\text{(}\mu)=98.2285[/tex]The standard deviation would be derived to be;
[tex]\sigma=1.2230[/tex]The sample size can be gotten from the question to be;
[tex]n=7[/tex]The probability value of a 98% confidence interval is given to be 2.33
We can then derive the answer using the formula below;
[tex]\mu\pm z^x(\frac{\sigma}{\sqrt[]{n}})[/tex]We would substitute into the formula
[tex]\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}[/tex]ANSWER:
[tex](97.152,99.306)[/tex]