# Use example to explain odds ratio to layperson audience

I found an example of how to express odds-ratios in plain english. Here is the link http://www.pitt.edu/~bertsch/risk.pdf

It says:

How does one express an OR of 0.15 in plain English. Had this been a RR would have said that the intervention reduced the risk by 85%. Because it is an OR we must say that for every 15 persons who experienced the event in the experimental group 100 persons experienced the event in the control group.

I find this a useful explanation for a lay audience. I have two logistic regression models with a categorical independent variable that I need to report odds ratios for.

If the odds ratio for group a (compared to group b) is 1.75, can I say that for every 175 persons in group a that experienced the event, 100 persons in group b did, while controlling for the other variables?

My second model has an odds ratio less than 1 so I would use the quoted example.

Are there any concerns with this approach?

• I'd avoid communicating anything in terms of odds ratios to a lay audience that doesn't consist of gamblers. Average partial effects are much easier to understand & communicate. Commented Aug 29, 2019 at 20:55
• Your interpretation implicitly assumes that the control and treatment groups have equal sizes. Commented Sep 3, 2019 at 17:29
• journalfeed.org/article-a-day/2018/idiots-guide-to-odds-ratios Commented Apr 8, 2020 at 18:16

The nearest I every got to somebody accepting Odds Ratios was this, with the expressed thought process of my listener in italics

• The odds ratio for a $$2\%$$ probability compared to a $$1\%$$ probability is about $$2$$, and so the odds ratio of a $$1\%$$ probability compared to a $$2\%$$ probability is about $$0.5$$.
• OK, though I do not quite see why you say it is "about" rather than "exactly". But I am not worried about that.
• Similarly the odds ratio for a $$99\%$$ probability compared to a $$98\%$$ probability is about $$2$$, and so the odds ratio of a $$98\%$$ probability compared to a $$99\%$$ probability is about $$0.5$$
• That is peculiar, but I can see that you cannot double probabilities of things that are very likely, and I can see that in this case you are going half the way to $$100\%$$ instead of doubling the distance from $$0\%$$. So I will accept that as reasonable in some sense.
• The odds ratio for a $$59\%$$ probability compared to a $$41\%$$ probability is about $$2$$, and so the odds ratio of a $$41\%$$ probability compared to a $$59\%$$ probability is about $$0.5$$
• I have no idea why those particular probabilities give an odds ratio of about $$2$$ rather than something else. But your previous statements were more or less plausible, and I know you are a professional statistician, so I will trust you on this.
• Bigger increases in probabilities lead to larger odds ratios and smaller increases to smaller odds ratios. For example a $$4\%$$ probability compared to a $$1\%$$ probability, and a $$99\%$$ probability compared to a $$96\%$$ probability, and a $$67\%$$ probability compared to a $$33\%$$ probability, all correspond to odds ratios of about $$4$$.
• Those first two examples make a lot of sense given what you said earlier, though the third looks counter-intuitive: one probability is about double the other but you say the odds ratio is about $$4$$ not $$2$$. That is not what I would have guessed. But in a sense I can see the point: a $$67\%$$ probability is what bookmakers might say is "two-to-one-on" while a $$33\%$$ probability is what bookmakers might say is "two-to-one-against" and so they might say one has four times the odds of the other. But I can see I could still get confused and will have to be careful using odds ratios.

Your interpretation is wrong. You are describing an odds of 0.15. There is nothing wrong with the odds ratio as a measure of association. The lay audience usually has no issue interpreting the odds ratio, it's just that they interpret it wrong: they think it's an RR. However, we can say if the OR is statistically significantly different from 1, then we know the RR is different from 1. So it suffices as a measure of association.

You should just point out that the OR != RR. In particular the OR will always exaggerate the effect somewhat. But this can be a good thing since the OR has the distinct quality of expressing associations as exactly $$\infty$$ whereas the RR cannot.