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rjceynar Group Title

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
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  1. zordoloom Group Title
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    Statistics right?

    • one year ago
  2. rjceynar Group Title
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    Right..

    • one year ago
  3. zordoloom Group Title
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    I can help.

    • one year ago
  4. zordoloom Group Title
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    So have you covered the t distribution yet?

    • one year ago
  5. zordoloom Group Title
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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:

    • one year ago
  6. zordoloom Group Title
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    |dw:1354857899830:dw|

    • one year ago
  7. zordoloom Group Title
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    |dw:1354857941165:dw|

    • one year ago
  8. rjceynar Group Title
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    Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?

    • one year ago
  9. zordoloom Group Title
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    Does this make sense so far?

    • one year ago
  10. zordoloom Group Title
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    |dw:1354858016253:dw|

    • one year ago
  11. zordoloom Group Title
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    Now, in class do you use a graphic calculator or table?

    • one year ago
  12. rjceynar Group Title
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    Both. :/ But I prefer the calculator.

    • one year ago
  13. zordoloom Group Title
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    ok

    • one year ago
  14. zordoloom Group Title
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    Following so far?

    • one year ago
  15. rjceynar Group Title
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    I think so..

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

    • one year ago
  17. rjceynar Group Title
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    If so, how does that come into play? or am i trialing off??

    • one year ago
  18. zordoloom Group Title
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    |dw:1354858281536:dw|

    • one year ago
  19. rjceynar Group Title
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    Right, so we need to know the total area under that portion of the curve, right?

    • one year ago
  20. zordoloom Group Title
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    yes

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

    • one year ago
  22. rjceynar Group Title
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    And the lines are 2 std deviations from the mean, right?

    • one year ago
  23. zordoloom Group Title
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    Almost

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

    • one year ago
  25. zordoloom Group Title
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    |dw:1354858667169:dw|

    • one year ago
  26. zordoloom Group Title
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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 .

    • one year ago
  27. zordoloom Group Title
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    |dw:1354858861005:dw|

    • one year ago
  28. zordoloom Group Title
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    that means the area under the curve up to that point is 1-0.025 = 0.975

    • one year ago
  29. rjceynar Group Title
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    Following so far

    • one year ago
  30. zordoloom Group Title
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    That will a be the p value that you can enter into the calculator. so invnorm(0.975)

    • one year ago
  31. zordoloom Group Title
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    it will be close to 2, but not exactly 2.

    • one year ago
  32. zordoloom Group Title
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    what result do you get for the z score?

    • one year ago
  33. zordoloom Group Title
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    You still there?

    • one year ago
  34. zordoloom Group Title
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    You should get something like 1.96 for the z value

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

    • one year ago
  36. rjceynar Group Title
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    .\1.959

    • one year ago
  37. rjceynar Group Title
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    Ah. Makes sense now.

    • one year ago
  38. zordoloom Group Title
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    The invnorm() command converts a cumulative probability to a zscore

    • one year ago
  39. zordoloom Group Title
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    It's important to calculate the z scores right or these type of problems will be difficult.

    • one year ago
  40. zordoloom Group Title
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    Do you think you can finish from here?

    • one year ago
  41. rjceynar Group Title
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    I can try. :) Thanks so much for the help.

    • one year ago
  42. zordoloom Group Title
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    Use this formula and the values that I just solved for:

    • one year ago
  43. zordoloom Group Title
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    |dw:1354859367648:dw|

    • one year ago
  44. zordoloom Group Title
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    z=1.96, s=5, n=22

    • one year ago
  45. rjceynar Group Title
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    Awesome. Only thing really left to do now is plug in and apply. Thanks so much.

    • one year ago
  46. zordoloom Group Title
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    Let me know what you get....

    • one year ago
  47. zordoloom Group Title
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    so |dw:1354859570561:dw|

    • one year ago
  48. rjceynar Group Title
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    88+1.96(5/22) 88-1.96(5/22) right? so 88.44 and 87.55 ??

    • one year ago
  49. zordoloom Group Title
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    2.09 is the margin error

    • one year ago
  50. zordoloom Group Title
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    85.9 to 90.1

    • one year ago
  51. rjceynar Group Title
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    ...woah

    • one year ago
  52. zordoloom Group Title
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    it's sqrt 22

    • one year ago
  53. rjceynar Group Title
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    ah.... caught me

    • one year ago
  54. rjceynar Group Title
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    got it

    • one year ago
  55. zordoloom Group Title
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    Is that it?

    • one year ago
  56. rjceynar Group Title
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    thats it. :) Thanks

    • one year ago
  57. zordoloom Group Title
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    Yep!

    • one year ago
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