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Question 525:



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The null hypothesis is that there is no difference in heart-attack rates between the lovastatin group and the inactive pill groups. The alternative hypothesis is that there is a difference.

We will use a 2-proportion test. We observed 57/2,325 with lovastatin and 97/2,081 in the inactive pill version. The two proportions we are calculating are then .0245 and .0466 and the difference is .00221. Can we conclude this difference is greater than chance?

We divide this difference by the square-root of a denominator which accounts for the chance which will provide you with a z-score. The denominator is

1         1
--   +  --     * PQ
n1       n2

Where P = (x1 + x2)/(n1+n2)  Q is 1-P. The x's are just the number of heart-attacks and the n's are the sample size.

P = (57+97)/(2325+2081) = .03495
Q = 1-.03495 = .965
PQ = .03495*.965 = .03373 (the variance)

1/n1 + 1/n2 = 1/2325 + 1/2081 = .000911

So multiply .03373 * .000911 = 0.0000307

Now the square root of this is SQRT(0.0000307) = .00554

So the equation is the observed difference divided by the standard error =  .00221/.00554 = 3.98

That last result of 3.98 is the z-score which is your test-statistic. You now look this value up using the z-score to percentile calculator using the 2-sided area. This gets us a p-value of .000069 or less than .01. Which means the probability the observed difference is due to chance is extremely remote. We would reject the null hypothesis and conclude there is a difference in hear-attack rates.

We are using the normal approximation to the binomial, which greatly simplifies the calculations. It's usually Ok to use the normal approximation to the binomial when n*p and n*q > 5, where p is the proportion getting a heart-attack and q is the proportion not. Using the smallest sample we get, 97/2081 is a p of .0245. That makes .0245*2081 = 51 and q = .975, .975*2081 = 2029, so the assumption of normality, as it applied to using the normal curve to approximate the binomial is fine.

The difference if of .022, while appearing small is a 90% difference (.0466-.0245)/.0245 which is very large and suggest this pill does indeed provide promise for greatly reducing heart-attacks.

Before widely prescribing this drug, researchers would likely need to understand the interaction of this drug with others and if there are other side-effects such as strokes, bleeding etc which would mitigate the benefits.

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