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.

Question

anonymous · Answer

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.

anonymous · Answer

please i need help

anonymous · Answer

@mathmale

anonymous · Answer

@uri

anonymous · Answer

@precal

amistre64 · Answer

its all about finding the test statistic

anonymous · Answer

umm
how do you do that @amistre64

amistre64 · Answer

well, theres not much to go on, but the standard error, if thats what we can call it;  would possibly be the sqrt, of the sum of the variances at best.  i dont see any sample sizes given to determine any modificaiton to them

amistre64 · Answer

let me re read it 


calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women.

do we want 2 seperate confidence intervals?

anonymous · Answer

yes i think so

amistre64 · Answer

well, a confidence interval us centered about the mean, and has a zscore adjustment by a standard error, or standard deviation as the case may be

amistre64 · Answer

we are given the means,  can you tell me the zscore related to a 98% confidence interval?  itll have a 1% tail

anonymous · Answer

hold on

anonymous · Answer

is it 2.55?

amistre64 · Answer

i was thinking more like 2.33

amistre64 · Answer

did you try to table it, or do you hav elike a ti83?

amistre64 · Answer

at any rate,  we can then use the setup

mean +-  2.33(sd)  to find the interval with,  and the margin or error is just the 2.33(sd) part

anonymous · Answer

i have a ti 84

anonymous · Answer

but idk how to use it

amistre64 · Answer

ti84, same as a ti83.

hit 2nd, vars, and pick the invnorm function

amistre64 · Answer

then just input your left tail percentage.  in this case we have .9800 in the middle, the tails are .0100

amistre64 · Answer

theres a way to run a CI, but i dont my batteries are dead so i cant review it that well :)

anonymous · Answer

it says area:
          u:
          o:
         paste

anonymous · Answer

but u and o are mu and segma

amistre64 · Answer

for an invnorm function?  i think your on something different

2nd, vars  gets us to the distribution menu .... and invnorm is the inverse normal function that takes a left tail area and gives back a zscore

amistre64 · Answer

http://www.wolframalpha.com/input/?i=invnorm%28.0100%29 heres the wolfs version of it

anonymous · Answer

its Invnorm

amistre64 · Answer

a normal distribution wil have a u of 0, and a o of 1

amistre64 · Answer

z = 2.326 is what we should end up with :)

anonymous · Answer

ok whats next

amistre64 · Answer

use that to determine the interval and the margin of error from a given mean and sd

mean +- 2.326(sd)  gives the interval,  the 2.326(sd) is the margin of error

amistre64 · Answer

i dont see any number of sample size to work with ... so im assuming this should work

anonymous · Answer

so its 95.647~100.326?

anonymous · Answer

doesn;t seem right

amistre64 · Answer

would that be for your woman parts?

amistre64 · Answer

97.6 -2.326(0.45) = 96.55
97.6 +2.326(0.45) = 98.65

anonymous · Answer

oh yeah

amistre64 · Answer

the margin of error is just the 2.326(.45) = 1.05

anonymous · Answer

sorry i thought it would be the 98%

anonymous · Answer

i get it now , for men the margin of error is 2.326(.52), confidence interval is 1.21(91.1) right?

anonymous · Answer

nvm

amistre64 · Answer

margin of error is good

CI is just 91.1 +- (error margin)

anonymous · Answer

you do +-

amistre64 · Answer

mean +, gives the high side of the interval
mean -, gives the low side

amistre64 · Answer

+- is just a way to express that both operation come in to play

anonymous · Answer

mhm i know that

amistre64 · Answer

good :)

anonymous · Answer

whats the second part?

anonymous · Answer

what does CI mean?

amistre64 · Answer

CI, confidence interval

anonymous · Answer

no i meant in the questiuon

anonymous · Answer

what does it mean in this problem

amistre64 · Answer

you tell me your best interpretation and ill correct it if need be

anonymous · Answer

i thikn it means the range of body temperature?

anonymous · Answer

or the chance of it being wrong?

amistre64 · Answer

not quite

we are creating an interval which we are 98% confident will contain the true population mean.

amistre64 · Answer

the population mean is NOT 91.1 .... but we 98% confident that the sample we choose provides us with a way to determine that the population means is someplace within the CI

amistre64 · Answer

spose we are looking for a friends house, we can be 98% sure that it is on such and such a street, but the house we actually choose may or maynot be the place.

amistre64 · Answer

a confidence interval gives us a range of possible options based on the parameter of the sample, since we assume that the mean of the mutliple sample means is normally distributed about the population mean:


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