## 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.

• This Question is Open
1. zordoloom

Statistics right?

2. rjceynar

Right..

3. zordoloom

I can help.

4. zordoloom

So have you covered the t distribution yet?

5. zordoloom

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

|dw:1354857899830:dw|

7. zordoloom

|dw:1354857941165:dw|

8. rjceynar

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

9. zordoloom

Does this make sense so far?

10. zordoloom

|dw:1354858016253:dw|

11. zordoloom

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

12. rjceynar

Both. :/ But I prefer the calculator.

13. zordoloom

ok

14. zordoloom

Following so far?

15. rjceynar

I think so..

16. rjceynar

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

17. rjceynar

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

18. zordoloom

|dw:1354858281536:dw|

19. rjceynar

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

20. zordoloom

yes

21. zordoloom

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

22. rjceynar

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

23. zordoloom

Almost

24. zordoloom

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

25. zordoloom

|dw:1354858667169:dw|

26. zordoloom

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

|dw:1354858861005:dw|

28. zordoloom

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

29. rjceynar

Following so far

30. zordoloom

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

31. zordoloom

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

32. zordoloom

what result do you get for the z score?

33. zordoloom

You still there?

34. zordoloom

You should get something like 1.96 for the z value

35. zordoloom

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

36. rjceynar

.\1.959

37. rjceynar

Ah. Makes sense now.

38. zordoloom

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

39. zordoloom

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

40. zordoloom

Do you think you can finish from here?

41. rjceynar

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

42. zordoloom

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

43. zordoloom

|dw:1354859367648:dw|

44. zordoloom

z=1.96, s=5, n=22

45. rjceynar

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

46. zordoloom

Let me know what you get....

47. zordoloom

so |dw:1354859570561:dw|

48. rjceynar

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

49. zordoloom

2.09 is the margin error

50. zordoloom

85.9 to 90.1

51. rjceynar

...woah

52. zordoloom

it's sqrt 22

53. rjceynar

ah.... caught me

54. rjceynar

got it

55. zordoloom

Is that it?

56. rjceynar

thats it. :) Thanks

57. zordoloom

Yep!