# Absolute risk and NNT from logistic regression

I fitted a Logistic regression with a number of variables, and was asked about the absolute risk (AR) and number needed to treat (NNT) for one of the variable with the outcome. I understand how to calculate them in a 2x2 table, but I am wondering if it can be calculated from the logistic regression?

From my understanding, in a Logistic regression we have (say I'm interested in $X_1$):

and since $AR=\pi(\unexposed1)-\pi(\exposed1)$, can we calculate AR like this? \begin{align} \pi(\unexposed) &= \frac{\exp(\beta_0+\beta_2X_2+...)}{(1+\exp(\beta_0+\beta_2X_2+...)} \\[5pt] &\qquad{\rm and} \\[5pt] \pi(\exposed) &= \frac{\exp(\beta_0+\beta_1X_1+\beta_2X_2+...)}{(1+\exp(\beta_0+\beta_1X_1+\beta_2X_2+...)} \end{align} And the NNT is:
$$\frac 1 {\pi(\unexposed)-\pi(\exposed))}$$

But since it varies with different $X_2$, $X_3$, etc., is it possible to get a single number for AR and NNT? Would it be appropriate to calculate the $\pi$'s for all possible $X_n$, and then just take the average?

• Regardless of the method of calculation, please make sure that the study design "allows" you to calculate an NNT and AR. Remember that these are risk-based measures. A case-control study, for example, cannot provide you with estimates of the baseline risk unless fulfill specific assumptions. – user140401 Dec 5 '16 at 2:43
• @Elmer Thank you! That is a very good point. (sorry if this side-tracks the question) May I ask under which specific assumptions/circumstances does a Case-control study "allows" for NNT and AR calculation? I understand that we cannot estimate the incidence from such studies, but in this analysis I know the prevalence from population data, and the proportion of cases in the study is similar to the prevalence. – Inji Dec 5 '16 at 4:37
• Thanks, @lnji. I suggest that you refer to King and Zeng's comprehensive exposition on the estimation of risks and rates from case-control studies: gking.harvard.edu/files/1s.pdf. Good luck! – user140401 Dec 5 '16 at 7:09