6.13 Meta-Analysis 89the natural logarithm function. Then L has a chi - square distribution
with 2 df.
Now suppose we have k independent tests, and we let L k = − 2 ln( V ),
where V is the product of the k independent uniforms. So L k = − 2 ln( U 1 ,
U 2... U k ) = − 2 ln( U 1 ) − 2 ln( U 2 ) −... − 2 ln( U k ). L k is the sum of k
independent chi - square random variable with 2 df , and hence it is
known to have a chi - square distribution with df equal to the sum of the
df for the chi - square random variables being summed. So L k is chi -
square with 2 k df. V is the probability that all k null hypotheses are
true, which, under the independence assumption, is the product of the
individual p - values. Because L k is a simple transformation of V with a
known chi - square distribution, it is more convenient to work with L k
rather than V.
We fi rst illustrate this with a consulting application that I provided
to a medical device company. The company conducted a clinical trial
in the United States and some countries in Europe. The device was a
cutting balloon catheter used for angioplasty. The manufacturer believed
that the restenosis rate would be lower for the cutting balloon compared
with conventional balloon angioplasty. Historically, the conventional
approach had a disappointing 40% restenosis rate. Since the manufac-
turer expected the new method would have about a 25% rate, which
would clearly be a clinically signifi cant improvement, they used these
assumptions to determine the necessary sample size.
Initially, the plan was to get FDA approval, which only required a
study in the United States. But recruitment was going slower than they
had hoped. So they chose to expand the trial to several sites in European
countries. Unfortunately, the results were not consistent across the
various countries. See Table 6.3.
We see that country E (which is the United States) had the lowest
rate and it is below the anticipated 25%. Ironically had the company
waited until the required number patients were treated in the United
States, they would have had a successful trial. But even though coun-
tries C and D also have rate signifi cantly lower than 40%, countries A
and B do not, raising the question as to why. Using country as a main
effect, an ANOVA clearly shows a signifi cant difference between coun-
tries. The most likely explanation is difference in the techniques of the
physicians in the European countries, where they may have had less
experience with the cutting balloon catheter, or differences in the sever-
ity of the disease across the various countries.