You may use independent t-test to test for equality between the two cases.

Not quite. You don't want %diff because we don't know how that is distributed. do the regular two sample t test as Prof Etuk suggested and report that result along with the % diff. Best wishes, David Booth

Marta Sandini , I think it would be helpful if you gave a better example of what the data you are measuring looks like. In your example, you say fewer complications, but how is this measured? Like, is that 0, 1, or 2 complications per person? Or is it some kind of continuous measurement or scale? Or is just Yes/No for each person whether there were complications?

My best guess is that you intend the last --- counts for Yes/No for each person. If so, the previous answers are probably not the best approach.

Sal Mangiafico Complications are dichotomous as 0/1 for each person.

If you are looking at the intervention data only †, you can construct a 2 x 2 contingency table of counts with Female/Male on one side and Complication/No-complication on the other. (Image.) You would then apply a chi-square test of association or similar test.

If you need to have a the Treatment/No-treatment factor taken into account, you will need a slightly more complicated approach.

In the attached image, there are 44 female and 56 male doctors. 50% of female doctors had complications and 75% of male doctors had complications. This would be a significant difference in proportions.

R code for this example:

Count = matrix(c(22, 42, 22, 14), nrow=2)

Count

rowSums(Count)

prop.table(Count, margin=1)

chisq.test(Count)

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† If, ... and this makes sense for your situation. How you are collecting this data still isn't entirely clear to me. Like, 1) are you treating each medical procedure as an independent observation with either a complication or no complication, so that your data are essentially: procedure a, yes; procedure b, no; procedure c, no? Or 2) are looking at a proportion of procedures with complications for each doctor, so that your data are Doctor A, 10%; Doctor B, 15%? The chi-square example here assumes the former. It's possible the responses by Ette Etuk and David Eugene Booth assume the latter scenario.

I agree to Salvatore Sir as your data is not measured on continuous scale (it is on nominal scale) and you want to test if the difference in the complications is associated with gender. Chi-square test is good criterion for your scenario. If your data is measured on continuous scale, you may test whether there is any significant difference in the complications generated between male and female. Thank you.

Chi-square test for comparing the change (before vs. after) in proportions of patients with a condition (eg. HTN yes/no) between 2 subgroups