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.

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.

Answer 1

There's no sample size so I am lost.

Answer 2

Are you SURE it says 91.1? Maybe 97.1?

Have you considered finding z-score and constructing the desired intervals?

Answer 3

Yes. I copied and pasted. I also have done what you suggested but it did not help me find what I was looking for.

Answer 4

You're saying that you cannot find the z-score for a 98% confidence interval? You don't have a table or a calculator?

Answer 5

The z-score is 2.33.

Answer 6

What I can't find is the margin of error. The equation for it is 2.33(mean/√sample size) but there is no sample size given in the question.

Answer 7

*2.33(standard deviation/√sample size)

Answer 8

Well, generally speaking, lacking sufficient information, we have to go with the wort-case scenario. n = 1.

And, 91.1 is idiotic. Something wrong with this sample. It may be just a misprint.

Answer 9

I completely understand. I contacted my teacher, but she has not responded. I can only take this test once as well.

Answer 10

I guess I can assume that 100 is the sample size and try to work from there.

Answer 11

Thank you for your help.

Answer 12

100 is pretty bold. Good luck.