Significant p-value and relative risk crossing 1 I am trying to convert results I am finding in scientific articles into relative risks so that I can compare studies in a meta analysis (if there is enough homogeneity in the studies). 
I came across a study entitled "Intervention to improve practices of prescribing appropriate medication before an operation; pre-and-post-intervention study". 
In the pre-intervention group, 16 out of 233 patients (so 217 with correct prescription) had an incorrect prescription and post intervention 3 out of 137 patients (134 with correct) had an incorrect prescription. 
In the paper, a p-value of 0.046 was given, using Fisher's exact test. I checked it using a hand calculation and got 0.049. When I put this result into an RR calculator, I get a RR of 0.32 [0.09, 1.07]. Now I am a relative stats rookie but I thought the 95% CI should not cross 1 if my p-value for comparing two proportions is below 0.05.
 A: If you do a two-sided level 0.05 test of hypothesis that the relative risk is different from 1 and get a p-value less than 0.05 then this is equivalent to a two-sided 95% confidence interval that does not contain 1.  So given the p-value of 0.049 you would expect that 1 would fall outside the interval.  What are the possible explanations?
1.  The confidence interval is at a level higher than 95%
2. There is an error in the calculation of the p-value or the confidence interval.
3. The confidence interval is not constructed by inverting the Fisher test (e.g. a chi square approximate test was used for the confidence interval).
I think 3 is very likely to be the reason. The chi square interval is easier to calculate than one using Fisher's test.  Note that the significance level is very close to 0.05 and the upper limit of the confidence interval is just slightjy above 1.0.
A: Fisher test can be inverted to get a confidence interval about the odds ratio, not about the relative risk. Therefore Michael's reason number 3 is necessarily the good one.
