Power calculations for population attributable fraction (PAF)
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A collaborator recently asked me to do power calculations for the population attributable fraction (PAF) of a dichotomous exposure, $F$, for an incident outcome, $D$. In a previous iteration, we had focused instead on using standard sample size and power calcuation approaches to get a minimum detectable relative risk (RR) assuming known sample size, outcome rate, exposure prevalence, and 80% power. While there is a paper by Browner and Newman (1989) for sample size and power for the PAF, I think an alternate approach could use the minimum detectable RR that I had already computed and the relationship between the PAF and RR.
Definition of population attributable fraction (PAF)
The population attributable fraction is typically defined by the proportion of disease incidence in a population that is attributable to the exposure of interest, specified by \(PAF = \displaystyle\frac{P(D) - P(D\vert \bar{F})}{P(D)},\) where $P(D)$ is the prevalence of the outcome in the overall population and $P(D\vert \bar{F})$ is the prevalence of the outcome among those that are not exposed. Using Bayes’ rule or the law of total probability, i.e., \(P(D)=P(D\vert F)P(F)+P(D\vert \bar{F})P(\bar{F}),\) the PAF can be expressed in terms of the exposure prevalence $P(F)$ and the relative risk given by $RR=P(D\vert F)/P(D\vert\bar{F})$ as follows [see (1953) and Leviton (1973)]:
\[\begin{align*} PAF &= \displaystyle\frac{P(D) - P(D\vert \bar{F})}{P(D)} \\ &= \displaystyle\frac{P(D\vert F)P(F)+P(D\vert \bar{F})P(\bar{F}) - P(D\vert \bar{F})}{P(D\vert F)P(F)+P(D\vert \bar{F})P(\bar{F})}\\ &= \displaystyle\frac{P(D\vert F)P(F)+P(D\vert \bar{F})(1-P(F)) - P(D\vert \bar{F})}{P(D\vert F)P(F)+P(D\vert \bar{F})(1-P(F))}\\ &= \displaystyle\frac{P(F)[P(D\vert F) - P(D\vert\bar{F})]}{P(\bar{F})[P(D\vert F)-P(D\vert\bar{F})] + P(D\vert\bar{F})}\\ &= \displaystyle\frac{P(F)\left[\displaystyle\frac{P(D\vert F)}{P(D\vert\bar{F})}-1 \right]}{P(F)\left[\displaystyle\frac{P(D\vert F)}{P(D\vert\bar{F})} - 1 \right] + 1}\\ &= \displaystyle\frac{P(F)[RR - 1]}{P(F)[RR - 1] + 1}. \end{align*}\]Thus, if one can calculate the minimum detectable relative risk (RR), then one can calculate the minimum detectable PAF since the PAF is a monotonically increasing function of RR as follows:
\[\begin{align*} PAF &= \displaystyle\frac{P(F)[RR - 1]}{P(F)[RR - 1] + 1} \\ &= \displaystyle\frac{P(F)[RR - 1] + 1 - 1}{P(F)[RR - 1] + 1} \\ &= 1 - \displaystyle\frac{1}{P(F)[RR - 1] + 1} \end{align*}\]So to calculate the minimum detectable PAF, given the sample size, exposure prevalence $P(F)$, overall outcome prevalence (or rate), do the following:
- Use standard software to calculate the minimum detectable relative risk or hazard ratio assuming, say, 80% power.
- Calculate the corresponding PAF using the relationship above. That should be the minimum detectable PAF since there is a monotonic relationship between RR and PAF.
Relevant References:
Browner WS, Newman TB. Sample size and power based on the population attributable fraction. American Journal of Public Health. 1989 Sep;79(9):1289-94.
Crowson CS, Therneau TM, O’Fallon WM. Attributable risk estimation in cohort studies. Technical Report; 2009 Oct 8.
Levin ML. “The occurrence of lung cancer in man.” Acta Unio int contra cancrum 9 (1953): 531-941.
Leviton A. Definitions of attributable risk. American journal of epidemiology. 1973 Sep 1;98(3):231.