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For a group of 22 college football players, the mean heart rate after a morning workout session was 88 beats per minute with a standard deviation of 5. Find the 95% confidence interval of the true mean for all college football players after a workout session.

Mathematics
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Statistics right?
Right..
I can help.

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Other answers:

So have you covered the t distribution yet?
I'm willing to work through it to show you how to do this problem. If you haven't covered the t-distribution. Then Just use this formula:
|dw:1354857899830:dw|
|dw:1354857941165:dw|
Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?
Does this make sense so far?
|dw:1354858016253:dw|
Now, in class do you use a graphic calculator or table?
Both. :/ But I prefer the calculator.
ok
Following so far?
I think so..
Because its a 95% confidence interval, wouldnt it include all values withen 2 standard deviations of the mean?
If so, how does that come into play? or am i trialing off??
|dw:1354858281536:dw|
Right, so we need to know the total area under that portion of the curve, right?
yes
So basically, you're looking for the area under the curve right up to one of horizontal lined regions.
And the lines are 2 std deviations from the mean, right?
Almost
I'll return to that later. So, for now, what is the area where the horizontal lines are?
|dw:1354858667169:dw|
So the horizontal lines add up to .05. So x+x=.05 That means x=.025. That is the area under one of the horizontal striped .
|dw:1354858861005:dw|
that means the area under the curve up to that point is 1-0.025 = 0.975
Following so far
That will a be the p value that you can enter into the calculator. so invnorm(0.975)
it will be close to 2, but not exactly 2.
what result do you get for the z score?
You still there?
You should get something like 1.96 for the z value
basically ,that's the reason why 95% of the population is around 2 standard deviations from the mean.
.\1.959
Ah. Makes sense now.
The invnorm() command converts a cumulative probability to a zscore
It's important to calculate the z scores right or these type of problems will be difficult.
Do you think you can finish from here?
I can try. :) Thanks so much for the help.
Use this formula and the values that I just solved for:
|dw:1354859367648:dw|
z=1.96, s=5, n=22
Awesome. Only thing really left to do now is plug in and apply. Thanks so much.
Let me know what you get....
so |dw:1354859570561:dw|
88+1.96(5/22) 88-1.96(5/22) right? so 88.44 and 87.55 ??
2.09 is the margin error
85.9 to 90.1
...woah
it's sqrt 22
ah.... caught me
got it
Is that it?
thats it. :) Thanks
Yep!

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