anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Statistics right?
anonymous
  • anonymous
Right..
anonymous
  • anonymous
I can help.

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

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