The headline screams “You’re 45% more likely to be murdered in de Blasio’s Manhattan”.
The evidence? Sixteen people have been killed so far this year in Manhattan, against only eleven over the same period last year.
Does this evidence indicate you are more likely to be murdered, as the headline says? To find out, I tested whether a constant murder rate could explain the results. The probability of getting murdered over the same period last year may be approximately 11/Manhattan’s population = 11/1,630,000 = 0.0000674 = .00674%.
Is it likely that with the same murder rate this period this year, one would get a number as high as 16 murders? Yes.
This can be seen by calculating the 95% confidence interval for 11/1,630,000, which according to 3 different statistical methods, spans 5 to 20. That is, even with a constant murder rate, due to statistical fluctuations, the murders over this period could easily have been as low as 5 or as high as 20. Just like if one flips a coin 10 times, one may get 3 heads the first time and 6 the next, without the chance of a head changing.
Doing this more properly means comparing the two rates directly. I did this using three different methods, all of which found no significant difference.
The article also reports that the number of shooting incidents is higher this year, 50 instead of 31. Using the three different statistical methods again, this was (barely) significantly different. So here the journalist has a point. But this should be taken with a big grain of salt. Journalists are always looking for “news”, and if they repeatedly look at how many people have been murdered/shot, eventually they are guaranteed to find an apparent difference, because all possible statistical fluctuations will happen eventually.
The statistics and the code are here.
I only did all this and wrote this post because Hal Pashler saw someone tweet the NYPost piece. Hal knew I had previously looked into the statistics of proportions and asked whether the headline was justified. I invite others to disagree with my calculations if they have a better way of doing it. I don’t think different methods will give a very different result, however.