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

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).

Re: Continuity correction and probability

p=0.60

n=50

$\displaystyle \mu _0=n*p$ = 30

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

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

New drug is same as old one (p>0.05)