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    Finding an upper critical value
    Test 5% level
    Assume X~B(20,0.25) is true
    P(X>=r)=<0.05
    r=7 -> 1-0.8982=0.1018
    r=8 -> 1-0.9591=0.0409 =<0.05

    I forgot S2.
    Why is 9 the CV?
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    (Original post by AlmostNotable)
    Finding an upper critical value
    Test 5% level
    Assume X~B(20,0.25) is true
    P(X>=r)=<0.05
    r=7 -> 1-0.8982=0.1018
    r=8 -> 1-0.9591=0.0409 =<0.05

    I forgot S2.
    Why is 9 the CV?
    Because it is the first value or r s.t P(X>=r) < 0.05 is satisfied.
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    (Original post by Zacken)
    Because it is the first value or r s.t P(X>=r) < 0.05 is satisfied.
    When r=8, P(>=r)=0.0409<0.05
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    (Original post by AlmostNotable)
    When r=8, P(>=r)=0.0409<0.05
    Remember that the tables give you values of P(X=<x), so if you're looking at P(X=<r) then 1 - P(X=<r) = 1 - P(X>r)*** = (1 - P(X >= r+1) as it is a discrete distribution, so if it's greater than r then it's greater or equal to r+1 as there are no values between r and r+1.

    *** Remember that P(X=<r) + P(X>r), where you can read that as probability of something being less than or equal to r, or greater than r, covers all real numbers so the sum = 1.
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    (Original post by AlmostNotable)
    When r=8, P(>=r)=0.0409<0.05
    Urgh, sorry - should have read your question properly.

    P(X >= r) <= 0.05 is what you use to determine your c.v.

    But P(X >= r) = 1 - P(X <= r - 1)

    SO you want

    1- P(X <= r-1) <= 0.05

    Remember that binomial tables are cumulative and the distribution is discrete.

    So r-1 = 8 is the largest value that satisfies the inequality, so r = 9.
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    (Original post by Zacken)
    Urgh, sorry - should have read your question properly.

    P(X >= r) <= 0.05 is what you use to determine your c.v.

    But P(X >= r) = 1 - P(X <= r - 1)

    SO you want

    1- P(X <= r-1) <= 0.05

    Remember that binomial tables are cumulative and the distribution is discrete.

    So r-1 = 8 is the largest value that satisfies the inequality, so r = 9.
    Oh yeah its discrete.
 
 
 
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