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Assume that two drugs were tested. The risk of death for drug 1 is $p_1$ and the risk for drug 2 is $p_2$. We define:

  1. Risk difference (RD) $RD=p_1-p_2$
  2. The number needed to treat (NNT) $NNT=1/|RD|$

If we know the estimated RD as RD* and its standard error as se(RD*), what is the 95% CI for NNT? I can think of two methods for solving this problem. Which one is correct and why?

  1. We first construct the 95% CI for RD, and then obtain 95% CI for NNT by inverting the CI for RD, that is:

    step 1: 95% CI for RD: $RD^* \pm 1.96* se(RD^*)$
    step 2: 95% CI for NNT: $1/(RD^* \pm 1.96* se(RD^*))$
    result: $(7.5, 149)$

  2. We first derive se(NNT*) from se(RD*) by the Delta method, and then calculate the 95% CI for NNT by:

    step 1: $se(NNT^*)= se(RD^*)~|~d(NNT^*)/d(RD^*))$
    step 2: $NNT^* \pm 1.96* se(NNT^*)$
    result: $(1.38, 27.07)$

Obviously the two results are quite different. What is the problem in these two methods?

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This paper can probably help you: rbsd.de/PDF/ci_nnt.pdf (Calculating Confidence Intervals for the Number Needed to Treat - Ralf Bender) –  andrea Dec 30 '12 at 15:29

1 Answer 1

(This isn't the whole answer, I might update later)

NNT is a transform of RD. It's distribution has a different shape. You're calculating CI's based on an assumption of normal distribution (thus the z-score for calculation). If one of those distributions is normal only one is, NNT, or RD, not both (my guess, if one, NNT). In general, with transformed values, if one distribution is normal then you can calculate the CI as you have on that distribution and then transform afterwards. Distributions other than normal need their own CI calc methods. And therefore, at least one of the method you've used is the wrong one for calculating the CI.

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