I was struck by dramatic differences in odds ratio for use of force for “compliant” vs. “all” encounters in the NYT article on the recently publicized Harvard study on police use of force (http://www.nytimes.com/2016/07/12/upshot/surprising-new-evidence-shows-bias-in-police-use-of-force-but-not-in-shootings.html). In looking at that data I realized that you could use it to estimate the fraction of stops in which a black person was compliant. I understand that the datasets are different and thus can’t be compared directly, but unless the data are badly flawed, the results are astonishing. Specifically, looking at the 3rd (compliant) and 4th (all) forest plots and selecting the “handcuffed” data (since it is least likely to be reported disparately between officers and civilians) you can derive the following,
(1) Bc*Wt / Wc*Bt = 1.13 (Bc=black compliant, Wt=white total, etc.)
(2) ((Bc+Bu)*Wt) / ((Wc+Wu)*Bt) = 3.17 (Bu = black uncompliant)
Dividing Equation 2 by Equation 1 and a little algebra yields:
Bu/Bc = 1.81 + 2.81*Wu/Wc.
Since Wu/Wc has to be a positive number, that means that black people are uncompliant > 1.81/(1+1.81) = 64% of the time! Unless I’ve made a mistake in my analysis, it seems that if most black parents are really having “the talk” with their kids, other signals overwhelm its efficacy.
Saturday, July 16, 2016
About That Harvard Study on Police Use of Force. . . | Power Line