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Thread: 99% CI for failure proportions

  1. #1
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    99% CI for failure proportions




    Hi,

    Is this working correct for a 99% CI for the difference between two proportions:

    p(A) = 11 failures out of 200
    p(B) = 6 failures out of 200


    p(A) - p(B) +/- 2.575 x [(p(A) x (1-p(A))/n) + (p(B) x (1-p(B))/n)^0.5]

    0.025 +/- 2.575 x 0.020134
    = -0.02684, 0.076845

    My textbook gives the formula p +/- (z critical value)[p(1-p)/n]^0.5
    I've just put the second term in for p(B) but I'm not sure if this is the right thing to do.

    Cheers

  2. #2
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    The formula that the text has given is for the CI for one sample, where as for the difference of proportions the formula that you have given is perfect.
    i.e (p1 - p2) +/- Z(critical) * sqrt[sq(SE1) + sq(SE2)].
    where:
    SE1 = p1(1-p1)/n and
    SE2 = p2(1-p2)/n

    Cheers
    Mahi..
    Last edited by Mahi; 09-15-2008 at 04:37 AM. Reason: In the previous mail I have taken the difference of sq(SE1) and sq(SE2), which should be actually addition.

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