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  1. #1
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    Help me please

    Hi all,
    please i have only one question and I can finish my qestions exam which isdue to morrow,Please , I beg you to help me.
    Q: The P - value in not the probility that Ho is true given the data . Here is an example: Suppose it is mandated that all newborn babies undergo screening for muscular dystrophy using a heel stick blood test for creatinine phosphokinase (CPK).This test has 100% sensitive (ie., of babies with muscular dystrophy willk test positive) and 99.98% specificity (ie., of babies without muscular dystrophy , 99.98% will test negative and 0.02% will test positive. The incidence of muscular dystrophy at birth is rare -a rough estimate would be ,say, 1 in 5000. Now for a randomly selected newborn , let us test the null hypothesis Ho that the baby does not have muscular dystrophy against the alternative Ho that the baby does have muscular dystrophy. The data cosists of the CPK test result.
    (a) If the decision is to reject Ho when the test is positive , convince yourself (based on the definition of a P-value ) that the value is 0.02%=0.0002.
    (b) Show that given a positive test result , in fact the probability that Ho is true is 50%( Hint :Construct a 2x2 table that shows how many babies (out of, say,100,000) would be fall into each (disease status x test result) category).
    Hence, even when the P-value is near zero , the probability that the null hypothesis is true given the data can be quite high.

    Appreciated your response
    Last edited by keith; 09-29-2005 at 05:43 PM.

  2. #2
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    Write out what you're given:

    A = has disease
    A' = no disease
    B = positive test
    B' = negative test

    sensitivity = P(B|A) = 1
    then P(B'|A) = 0
    specificity = P(B'|A') = 0.9998
    then P(B|A') = 0.0002

    we're also given P(A) = 1/5000 = 0.0002
    then P(A') = 1 - P(A) = 0.9998

    I think the following 2x2 table will help:

    _____________________Has Disease

    Test Result___Pos(B)___2______2_______4

    P(A'|B) = 2/4 = 0.50

    we can also compute P(A'|B) by formula, which is equal to
    P(B|A')*P(A') / P(B)
    = 0.0002*0.9998 / 0.0004
    = 0.5

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