A new treatment for strokes is put on trial. There are two equal size groups,one group is given a drug, and one a placebo. The 95% confidence interval for the difference between the two proportion of subjects having strokes was (0.07,0.12). What can we infer:
1) not enough info to infer
2)at most 12% of the people had strokes
3) at least 4% had strokes
I'm not sure on this. I think it might be 1). Can anyone help out?
Thanks
The information indicates that the difference in the proportion of strokes is significant - note that the CI of the difference does not include $0$. However, we cannot infer that at most 12% of subjects had strokes (the two groups could have, say, proportions of $0.30$ and $0.205$, with a $0.095$ difference and CI $0.07-0.12$).
On the other hand, if we assume that the CI is symmetric (e.g., obtained by a normal approximation), we get that the observed difference lays in the center of the interval. In this case, the center corresponds to a $0.095$ proportion difference. Even if we hypothesize that the group with lower proportion of strokes had zero events, we get that the other group had at least a $9.5\%$ proportion of events. Since the problem states that the two groups have equal size, we could conclude that, overall, $\geq 4.75\%$ of patients had a stroke, so that the third possible answer would be correct.
It should be pointed out that, however, the question does not specify the method used to caculate the CI. Some methods that are not based on the normal approximation do not give symmetric CI (for instance, the exact Clopper-Pearson intervals, which are based on the F distribution). If we assume that our CI is asymmetric, we cannot know the observed difference and cannot use the central point of the interval as above. However, it seems highly unlikely that, whatever the method used, the observed difference is $<0.8$ (usually even asymmetric CI are not so unbalanced), and therefore it is "highly probable" that at least $4\%$ of subjects had a stroke.