Charles Scripter wrote:

Here we fit the NSW “before” region to a slope + constant background,

while the “after” region is fit only to a slope (chance resulted in

this slope passing through, or very close to zero, eliminating the

need for a constant, 4th parameter). The origin is located at year 1900.Data for years 1919 and 1920 were rejected as “anomalous” and the

“before” region was extended back to 1907 to offset this loss of data.

It is amazing what you can show if you pick the right subset of the

data. In particular, if you want to “find” a decreasing trend, just

follow Charles’ example and pick your start year (1907) at a local

maximum, and your stop year at a local minimum (1918). Works every

time.

Charles’ model:

Rate = (AYear+B), if Year <=20 = (CYear), if Year >20 where A = -0.124825 B = 4.00198 C = 0.0516219

This additional parameter, and use of slopes rather than constants,

results in a fit which is much more pleasing to the eye

The general practice in statistics is to avoid subjective measures like

“more pleasing to the eye”. The objective measure favoured is the

chi-square test. The chi-square statistic is

sum(((o[i]-e[i])/sd)^2), where o[i] is the ith observation, e[i] is the

ith value predicted by the model and sd is the standard deviation.

Since we just want to compare two models, we can multiply through by

sd^2 and just work out sum((o[i]-e[i])^2) for each model. The model for

which this value is lower has an objectively better fit.

The results (using years 1907-1937):

Charles’s model: 4.96

My step model: 4.43

Charles model has a worse fit, in spite of the fact that his model has

more parameters and in spite of the fact that the start year was chosen

to favour his model.

This analysis, with it’s positive slope (parameter C) in the post-law

region, apparently contradicts the claimed effect of “decreasing

homicide rates” correlated with (or “caused” by) 1921 law.

Since you forced the fitted line for the post-law region to go through 0

in 1900, it is impossible for the slope to be anything but positive.

If, on the other hand, you allow the line to have an arbitrary

intercept, you will discover that the slope is not significantly

different from 0.