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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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zordoloomBest ResponseYou've already chosen the best response.1
So have you covered the t distribution yet?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
I'm willing to work through it to show you how to do this problem. If you haven't covered the tdistribution. Then Just use this formula:
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354857899830:dw
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354857941165:dw
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Does this make sense so far?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354858016253:dw
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Now, in class do you use a graphic calculator or table?
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
Both. :/ But I prefer the calculator.
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
Because its a 95% confidence interval, wouldnt it include all values withen 2 standard deviations of the mean?
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
If so, how does that come into play? or am i trialing off??
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354858281536:dw
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
Right, so we need to know the total area under that portion of the curve, right?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
So basically, you're looking for the area under the curve right up to one of horizontal lined regions.
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
And the lines are 2 std deviations from the mean, right?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
I'll return to that later. So, for now, what is the area where the horizontal lines are?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354858667169:dw
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
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 .
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354858861005:dw
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
that means the area under the curve up to that point is 10.025 = 0.975
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
That will a be the p value that you can enter into the calculator. so invnorm(0.975)
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
it will be close to 2, but not exactly 2.
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
what result do you get for the z score?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
You should get something like 1.96 for the z value
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
basically ,that's the reason why 95% of the population is around 2 standard deviations from the mean.
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
The invnorm() command converts a cumulative probability to a zscore
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
It's important to calculate the z scores right or these type of problems will be difficult.
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Do you think you can finish from here?
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
I can try. :) Thanks so much for the help.
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Use this formula and the values that I just solved for:
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
dw:1354859367648:dw
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
Awesome. Only thing really left to do now is plug in and apply. Thanks so much.
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
Let me know what you get....
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
so dw:1354859570561:dw
 one year ago

rjceynarBest ResponseYou've already chosen the best response.0
88+1.96(5/22) 881.96(5/22) right? so 88.44 and 87.55 ??
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
2.09 is the margin error
 one year ago
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