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    It's for part b) of this question which I'm having problems with.

    Summary Data from a)

    mean = 73.266
    standard deviation = 4.1525

    I'm not really sure how to solve for n, is it a case of trial and error with the values you get from table of 't'-distribution for when q = 0.95 and v = n - 1, or is there a much simpler way to find n?

    I end up with:-

    t(n-1;0.95) < 0.4816(sqrt(n)) (the answer is n=19)

    Thanks for any help!
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    Also I'm slightly stuck with c) and not sure how you form the CI for the variance.
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    If you don't mind me asking, what board is this?


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    (Original post by Black_Materia_)
    Name:  CONFIDENCE INTERVAL.png
Views: 72
Size:  419.6 KB

    It's for part b) of this question which I'm having problems with.

    Summary Data from a)

    mean = 73.266
    standard deviation = 4.1525

    I'm not really sure how to solve for n, is it a case of trial and error with the values you get from table of 't'-distribution for when q = 0.95 and v = n - 1, or is there a much simpler way to find n?

    I end up with:-

    t(n-1;0.95) < 0.4816(sqrt(n)) (the answer is n=19)

    Thanks for any help!
    Are you sure it's 19. I get 21?
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    (Original post by That_Clever_Guy)
    If you don't mind me asking, what board is this?


    Posted from TSR Mobile
    It is part of a Uni seminar sheet but I do remember questions like this when I did OCR Maths S3, so that's why I listed it under "Sixth Form"
 
 
 
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