Your calculation of the inverse of 05% turned out to be 0.005? How can this be correct to calculate the NNT?
The confidence internal for the hazard ratio il is huge for such a large sample size and it could easily be a 0.98 HR, or a 2% protection
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Your calculation of the inverse of 05% turned out to be 0.005? How can this be correct to calculate the NNT?
The confidence internal for the hazard ratio il is huge for such a large sample size and it could easily be a 0.98 HR, or a 2% protection