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Math Help - Continuity correction and probability

  1. #1
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    Continuity correction and probability

    The standard medication for a certain ailment is known to give relief to 60% of sufferers. A new drug has been developed which is claimed to be better than the standard medication. A random sample of 50 patients is given the new drug and 35 of them obtain relief. What is the P-value for the test of H0: the new drug is the same as the standard versus H1: the new drug is better than the standard is, using a normal approximation with continuity correction?

    I know that you need to add 0.5 to the X for the continuity correction part but how do you actually solve this without knowing the SD?

    Thanks a lot
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  2. #2
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    Re: Continuity correction and probability

    Hey sazzy.

    You can use what is known as a Wald test which uses the sample to get an estimate of the standard deviation which is commonly referred to as the standard error.

    Given your estimate of p (call it p_hat), the estimate of the standard deviation is given by SQRT(p_hat*(1 - p_hat)/n) where n is the number of observations in your total sample.

    You can derive this using the MLE estimator for p_hat and the invariance principle considering estimators of a function of an MLE estimate (in this case your function f is f(x) = x*(1-x) and x is your MLE estimate for your proportion parameter p).
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  3. #3
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Continuity correction and probability

    p=0.60
    n=50
    \mu _0=n*p = 30

    \sigma =\sqrt{n p (1-p)} = 3.4641

    P\left[\left.\mu >\mu _0\right|\mu ,\sigma \right]=0.074457 > 0.05\text{   }H_0: \text{Accept}

    New drug is same as old one (p>0.05)
    Last edited by MaxJasper; October 19th 2012 at 12:12 AM.
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