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anonymous
 3 years 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.
anonymous
 3 years 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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So have you covered the t distribution yet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354857899830:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354857941165:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does this make sense so far?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354858016253:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, in class do you use a graphic calculator or table?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Both. :/ But I prefer the calculator.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If so, how does that come into play? or am i trialing off??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354858281536:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, so we need to know the total area under that portion of the curve, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So basically, you're looking for the area under the curve right up to one of horizontal lined regions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And the lines are 2 std deviations from the mean, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll return to that later. So, for now, what is the area where the horizontal lines are?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354858667169:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354858861005:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that means the area under the curve up to that point is 10.025 = 0.975

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That will a be the p value that you can enter into the calculator. so invnorm(0.975)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it will be close to 2, but not exactly 2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what result do you get for the z score?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You should get something like 1.96 for the z value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0basically ,that's the reason why 95% of the population is around 2 standard deviations from the mean.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The invnorm() command converts a cumulative probability to a zscore

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's important to calculate the z scores right or these type of problems will be difficult.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you think you can finish from here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can try. :) Thanks so much for the help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use this formula and the values that I just solved for:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354859367648:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Awesome. Only thing really left to do now is plug in and apply. Thanks so much.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let me know what you get....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so dw:1354859570561:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.088+1.96(5/22) 881.96(5/22) right? so 88.44 and 87.55 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02.09 is the margin error
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