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We think the example used in this paper seems to contain an error:
“We use data from a randomised trial comparing intensive versus standard insulin treatment in patients with diabetes mellitus and acute myocardial infarction.8 From figure 1 in the paper, the control group mortality rates at 2 and 4 years were 0.33 and 0.49 respectively. The reported hazard ratio was h = 0.72 with 95% confidence interval 0.55 to 0.92. The number needed to treat at 2 years is thus estimated as 1/(0.33^0.72–0.33) = 8.32. The 95% confidence interval for the number needed to treat is obtained from equation 1 setting h to 0.55 and then 0.92, giving 4.7 to 32.7.”
However the formula for the NNT is given as 1/[Sc(t)^h-Sc(t)] where h is the hazard ratio and Sc(t) is the survival probability at time t in the control arm. This formula is derived from 1/[hazard in treatment-hazard in control] = 1/[[1-Sc(t)^h]-[1-Sc(t)]]. However, in the example quoted above rather than use the survival probability the example has used the mortality probability. The survival probabilities in the example are 1-0.33=0.67. The correct NNT should be 1/[0.67^0.72-0.67]=12.58 (with CI 7.56 to 45.84). Equivalent to 1/[1-0.67^h-0.33].