Question

Carl conducted an experiment to determine if there is a difference in the mean body temperature between men and women. He found that the mean body temperature for a sample of 100 men was 91.1 with a population standard deviation of 0.52 and the mean body temperature for a sample of 100 women 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

Using complete sentences, explain what these confidence intervals mean in the context of the problem.

Answer 2

@prettygirl876

Answer 3

what do you know about the setup?

Answer 4

sorry can't help

Answer 5

Im looking up what I know as far as a formula

Answer 6

thats a good step to take :)

Answer 7

This was in a most recent algebra 2 lesson, but I didnt really understand fully.
The formulas in my lesson are for finding the z-score and the margin of error, I am just unsure of where to go from here. Help would be greatly appreciated!

Answer 8

a confidence interval, its the range of values that we are some degree of confidence that the true population parameter should exit within.

i am 95% confident that my age is somewhere between 35 and 87 ... thats a confidence interval.

Answer 9

how do we define the interval?

we use the sample parameter as a basis, assuming that it is a normal value to obtain and not something that is unusual.

we then spread a blanket out, the spread is defined by the margin of error, the larger the margin of error the more confident we are that we have our true parameter within it.

Answer 10

Okay, so a 98% confidence interval has a z-score of 2.33

Answer 11

good, so we want be within 2.33 standard errors about our sample mean right?

Answer 12

Yes, so do I now plug this into the formula for the margin of error?

Answer 13

whats your margin of error formula?

Answer 14

Which is given to me as (z-score)*(standard deviation)/(sample size)

Answer 15

sqrt(sample size)

Answer 16

Oh right, my bad
So it would be:
2.33*0.52/100

Answer 17

100^2

Answer 18

Or square root of 100

Answer 19

recall that variance = (standard deviation)^2

\[\sqrt{\frac{\sigma^2}{n}}=\frac{\sigma}{\sqrt{n
}}\]

Answer 20

5.2/10 = .52 for our margin of error

Answer 21

Thats different from the formula im given, are we looking for the margin of error for the population mean or portion?

Answer 22

you posted as your formula:

standard deviation
----------------
sample size

this is in error, i was just demonstrating why it was in error

sample size is sqrted.

5.2/sqrt(100) = 5.2/10 = .52 as our standard error

z(.52) is our margin with respect to that error ... our margin of error

mean +- margin of error is our spread for our level of confidence

Answer 23

Okay I see, wouldnt the standard deviation appear as 0.52/sqrt(100) = 0.52/10 = 0.052?

Answer 24

Sorry for all of the questions, I want to make sure I understand :)

Answer 25

most of your hypothesis tests and such are modifications or simplifications of the same basic formula that can be traced back to your z score formula

and yes, my memory hates me, the population standard deviation of 0.52

so .52/10 is appropriate

Answer 26

Okay, just checking. So then we would multiply 0.052 by the z-score of 2.33=0.12116

Answer 27

doing fine so far

Answer 28

This would be our margin of error, and then we apply this to the mean? To get the range 90.97-91.22

Answer 29

So for the men, we are 98% confident that the body temp. will be between 90.97-91.22

Answer 30

thats our interval for males.

your wording is not setting right for me.

We are 98% confident that the actual population mean of the temperature of men can be found within our endpoints. Something to that effect

Answer 31

but its fine to me regardless, im just not grading it

Answer 32

Okay thank you so much, ill follow the same steps for the women. I appreciate your time!

Answer 33

youre welcome