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

@PlsHaveMercyLilB @SolomonZelman

Answer 3

@jagr2713

Answer 4

@ganeshie8

Answer 5

@IMStuck do you think you can help

Answer 6

I am trying really hard...

Answer 7

Same number of men as there are women?

Answer 8

I would assume so, judging by the fact that that info isn't listed.

The confidence interval will have the form
\[\left((\bar{x}_1-\bar{x}_2)-Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}},~(\bar{x}_1-\bar{x}_2)+Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}}\right)\]
where \(\bar{x}_1-\bar{x}_2\) is the estimated mean difference, \(Z_{\alpha/2}\) is the critical \(z\) value for a \((1-\alpha)100\%\) confidence level, \(\sigma_1\) and \(\sigma_2\) are the corresponding standard deviations, and \(n_1\) and \(n_2\) are the respective sample sizes.

You have all the info necessary to construct the interval except for the sample sizes.

Answer 9

if have to do all that?
@SithsAndGiggles

Answer 10

Well yes, I'm not planning on doing it for you. Do you have the sample size(s) given?

Answer 11

no

Answer 12

That's strange. You need the info to proceed with the interval :(

Answer 13

everything i listed at the top is given

Answer 14

@abb0t @aaronq @aum @AriPotta can yall help me

Answer 15

@campbell_st

Answer 16

@AriPotta u think u can help

Answer 17

maybe. i'm thinking

Answer 18

please help me i am so lost... i really dont know how to do this!

Answer 19

@AriPotta

Answer 20

Well, I've done what I can... Is this problem taken from a textbook, written by a professor, etc?

Answer 21

no

Answer 22

can anybody tell me what the confidence intervals
@campbell_st @ash2326 @ash2326 @abb0t @AriPotta

Answer 23

@AriPotta

Answer 24

sorry, i can't help. i keep losing my internet connection

Answer 25

awwwww

Answer 26

@Prestianne help

Answer 27

@Kainui

Answer 28

@abb0t help please

Answer 29

hey, what's wrong??

Answer 30

i was looking for some answers to ur first question and i found something, does this help?? Make sure you read the whole thing, its not that long when u read it carefully. go to that website:
http://openstudy.com/study#/updates/5272cc39e4b077aaee848854

Answer 31

Looks like those guys ran into the same problem - no sample size is given.

As for the margin of error - it's the \(Z_{\alpha/2}\sqrt{\cdots}\) term. Not having the sample size is the issue, though. You simply can't answer this question without it.