Question

HELP PLEASE!!! I WILL GIVE A MEDAL!!!

Carlisle conducted an experiment to determine if the there is a difference in mean body temperature for men and women. He found that the mean body temperature for men in the sample was 97.9 with a population standard deviation of 0.57 and the mean body temperature for women in the sample was 98.6 with a population standard deviation of 0.55.
Assuming the population of body temperatures for men and women is normally distributed, calculate the 99% confidence interval and the margin of error for the mean body temperature for both men and women.

Answer 1

I don't understand how to find the confidence interval with only the mean and standard deviation

Answer 2

The question asks for the confidence interval for the men not for the individual values, thus
standard error of the mean = s.d./sqrt(n)
is the correct measure of variability and we assume means are distributed normally..

Answer 3

For men, the pdf is
f(x)=0.699899e−1.53894(x−97.9)2

For a=1.47, we have
∫a+97.997.9−a0.699899e−1.53894(x−97.9)2dx=0.99009

Which means that 99% of men have their body temperature between 96.43 and 99.37

Answer 4

I understand that, but how will I get "n" which is the sample size?

Answer 5

I didn't learn that

Answer 6

oh well now you did! hope i helped!