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