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

- katieb

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- anonymous

Statistics right?

- anonymous

Right..

- anonymous

I can help.

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## More answers

- anonymous

So have you covered the t distribution yet?

- 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

|dw:1354857899830:dw|

- anonymous

|dw:1354857941165:dw|

- anonymous

Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?

- anonymous

Does this make sense so far?

- anonymous

|dw:1354858016253:dw|

- anonymous

Now, in class do you use a graphic calculator or table?

- anonymous

Both. :/ But I prefer the calculator.

- anonymous

ok

- anonymous

Following so far?

- anonymous

I think so..

- anonymous

Because its a 95% confidence interval, wouldnt it include all values withen 2 standard deviations of the mean?

- anonymous

If so, how does that come into play? or am i trialing off??

- anonymous

|dw:1354858281536:dw|

- anonymous

Right, so we need to know the total area under that portion of the curve, right?

- anonymous

yes

- anonymous

So basically, you're looking for the area under the curve right up to one of horizontal lined regions.

- anonymous

And the lines are 2 std deviations from the mean, right?

- anonymous

Almost

- anonymous

I'll return to that later. So, for now, what is the area where the horizontal lines are?

- anonymous

|dw:1354858667169:dw|

- 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

|dw:1354858861005:dw|

- anonymous

that means the area under the curve up to that point is 1-0.025 = 0.975

- anonymous

Following so far

- anonymous

That will a be the p value that you can enter into the calculator.
so invnorm(0.975)

- anonymous

it will be close to 2, but not exactly 2.

- anonymous

what result do you get for the z score?

- anonymous

You still there?

- anonymous

You should get something like 1.96 for the z value

- anonymous

basically ,that's the reason why 95% of the population is around 2 standard deviations from the mean.

- anonymous

.\1.959

- anonymous

Ah. Makes sense now.

- anonymous

The invnorm() command converts a cumulative probability to a zscore

- anonymous

It's important to calculate the z scores right or these type of problems will be difficult.

- anonymous

Do you think you can finish from here?

- anonymous

I can try. :) Thanks so much for the help.

- anonymous

Use this formula and the values that I just solved for:

- anonymous

|dw:1354859367648:dw|

- anonymous

z=1.96, s=5, n=22

- anonymous

Awesome. Only thing really left to do now is plug in and apply. Thanks so much.

- anonymous

Let me know what you get....

- anonymous

so |dw:1354859570561:dw|

- anonymous

88+1.96(5/22)
88-1.96(5/22)
right? so 88.44 and 87.55 ??

- anonymous

2.09 is the margin error

- anonymous

85.9 to 90.1

- anonymous

...woah

- anonymous

it's sqrt 22

- anonymous

ah.... caught me

- anonymous

got it

- anonymous

Is that it?

- anonymous

thats it. :) Thanks

- anonymous

Yep!

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