Question

Carl conducted an experiment to determine if the there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in the sample was 91.1 with a population standard deviation of 0.52 and the mean body temperature for women in the sample was 97.6 with a population standard deviation of 0.45.

Answer 1

Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

Answer 2

Pretty much i just need to know how to calculate the 98% confidence interval and the margin of error for the mean

Answer 3

i dont know about the margin of error, but the 98 confidence interval is the mean \(\pm\)2.33 std deviations.

Answer 4

uhm so the interval would be .52± 2.33? the std deviation is .45

Answer 5

not exactly like that, let me see what it was again, it's a while ago since i had this subject.

Answer 6

do you have any information about the sample size?

Answer 7

mm no i don't I posted all the information it gave me /:

Answer 8

\[91.1-(\frac{ 0.52 }{ \sqrt{n} }*2.33)<M<91.1+(\frac{ 0.52 }{ \sqrt{n} }*2.33)\] where n is the sample size. I beleive that should work.

Answer 9

the bigger the sample size n, the more accurate it becomes.

Answer 10

So with there not being a stated sample size would i not be able to use this formula?

Answer 11

nope, but then again, i don't know how you could possibly give a confidence interval without know the sample size.

Answer 12

mm okay, well do you know how to calculate the margin of error for the mean?

Answer 13

no, i don't know how to calculate the margin of error.

Answer 14

okay, well thank you!

Answer 15

i'm pretty good at probability and statistics but i don't think i ever had that