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The had 750 total trump voters. If you assume then that 600 support Trump say then there are about 300 presented with black and 300 with white. Given 50% support we can calculate approximate 95% CI on 50% which will be about +/-6%. Given that we are only comparing 51 to 59 then we have no difference. Things would be worse for the oppose trump group. I'm certain that if the possible assumptions for proportion supporting Trump were tried they would all say the same thing, no effect.

I'm rather unimpressed that they haven't performed a statistical test. Whatever some people may say about p-values they can be rather useful in cases like this, although odds tail and 95% might work as well.

Even if this wasn't flawed, there is always the file drawer problem. There are lots of researchers doing these sorts of analyses, some must produce interesting but incorrect results.

Fabio Machado

There's no proper control group in this set-up either. Where is the group that wasn't prompted at all? I would also like to see what the result were for non-white Trump supporters.


Ken: Good point. I thought they might have just enough to claim p < 0.05 but that is not the case. They would have needed about 600 on each side to make it. Surprised that as researchers, they skimmed over statistical significance. (On a further check, they have 750 voters which were further split into Trump supporters versus non-supporters so the sample size on the non-Trump supporters side must be even smaller.)

Fabio: Excellent catch too. They argue that there were "almost no" non-white Trump supporters, which is not true. They would have to over-sample though to get that estimate. I also like to run tests including the null control group - but often it is tricky to design. In this case, I would either show a neutral image without a person or not show an image at all. Naysayers will then argue that I would have introduced another potential confounder in either case.

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