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rjceynar

  • 2 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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  1. zordoloom
    • 2 years ago
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    Statistics right?

  2. rjceynar
    • 2 years ago
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    Right..

  3. zordoloom
    • 2 years ago
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    I can help.

  4. zordoloom
    • 2 years ago
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    So have you covered the t distribution yet?

  5. zordoloom
    • 2 years ago
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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:

  6. zordoloom
    • 2 years ago
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    |dw:1354857899830:dw|

  7. zordoloom
    • 2 years ago
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    |dw:1354857941165:dw|

  8. rjceynar
    • 2 years ago
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    Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?

  9. zordoloom
    • 2 years ago
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    Does this make sense so far?

  10. zordoloom
    • 2 years ago
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    |dw:1354858016253:dw|

  11. zordoloom
    • 2 years ago
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    Now, in class do you use a graphic calculator or table?

  12. rjceynar
    • 2 years ago
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    Both. :/ But I prefer the calculator.

  13. zordoloom
    • 2 years ago
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    ok

  14. zordoloom
    • 2 years ago
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    Following so far?

  15. rjceynar
    • 2 years ago
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    I think so..

  16. rjceynar
    • 2 years ago
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    Because its a 95% confidence interval, wouldnt it include all values withen 2 standard deviations of the mean?

  17. rjceynar
    • 2 years ago
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    If so, how does that come into play? or am i trialing off??

  18. zordoloom
    • 2 years ago
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    |dw:1354858281536:dw|

  19. rjceynar
    • 2 years ago
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    Right, so we need to know the total area under that portion of the curve, right?

  20. zordoloom
    • 2 years ago
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    yes

  21. zordoloom
    • 2 years ago
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    So basically, you're looking for the area under the curve right up to one of horizontal lined regions.

  22. rjceynar
    • 2 years ago
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    And the lines are 2 std deviations from the mean, right?

  23. zordoloom
    • 2 years ago
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    Almost

  24. zordoloom
    • 2 years ago
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    I'll return to that later. So, for now, what is the area where the horizontal lines are?

  25. zordoloom
    • 2 years ago
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    |dw:1354858667169:dw|

  26. zordoloom
    • 2 years ago
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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 .

  27. zordoloom
    • 2 years ago
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    |dw:1354858861005:dw|

  28. zordoloom
    • 2 years ago
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    that means the area under the curve up to that point is 1-0.025 = 0.975

  29. rjceynar
    • 2 years ago
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    Following so far

  30. zordoloom
    • 2 years ago
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    That will a be the p value that you can enter into the calculator. so invnorm(0.975)

  31. zordoloom
    • 2 years ago
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    it will be close to 2, but not exactly 2.

  32. zordoloom
    • 2 years ago
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    what result do you get for the z score?

  33. zordoloom
    • 2 years ago
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    You still there?

  34. zordoloom
    • 2 years ago
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    You should get something like 1.96 for the z value

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

  36. rjceynar
    • 2 years ago
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    .\1.959

  37. rjceynar
    • 2 years ago
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    Ah. Makes sense now.

  38. zordoloom
    • 2 years ago
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    The invnorm() command converts a cumulative probability to a zscore

  39. zordoloom
    • 2 years ago
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    It's important to calculate the z scores right or these type of problems will be difficult.

  40. zordoloom
    • 2 years ago
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    Do you think you can finish from here?

  41. rjceynar
    • 2 years ago
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    I can try. :) Thanks so much for the help.

  42. zordoloom
    • 2 years ago
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    Use this formula and the values that I just solved for:

  43. zordoloom
    • 2 years ago
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    |dw:1354859367648:dw|

  44. zordoloom
    • 2 years ago
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    z=1.96, s=5, n=22

  45. rjceynar
    • 2 years ago
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    Awesome. Only thing really left to do now is plug in and apply. Thanks so much.

  46. zordoloom
    • 2 years ago
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    Let me know what you get....

  47. zordoloom
    • 2 years ago
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    so |dw:1354859570561:dw|

  48. rjceynar
    • 2 years ago
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    88+1.96(5/22) 88-1.96(5/22) right? so 88.44 and 87.55 ??

  49. zordoloom
    • 2 years ago
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    2.09 is the margin error

  50. zordoloom
    • 2 years ago
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    85.9 to 90.1

  51. rjceynar
    • 2 years ago
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    ...woah

  52. zordoloom
    • 2 years ago
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    it's sqrt 22

  53. rjceynar
    • 2 years ago
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    ah.... caught me

  54. rjceynar
    • 2 years ago
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    got it

  55. zordoloom
    • 2 years ago
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    Is that it?

  56. rjceynar
    • 2 years ago
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    thats it. :) Thanks

  57. zordoloom
    • 2 years ago
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    Yep!

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