Question 360460
The following are the values for the before-minus-after values:
d = 12,   0,   1,   -1,   9 ,  -1,   5  , 4

The average m of these values are 29/8 = 3.625.
Now the following corresponding values for {{{(d-m)^2}}} are 
70.14,13.14,6.89,21.39,28.89,21.39,1.89,0.14.
Their sum is 163.87.
The standard deviation of the differences would be {{{s = sqrt(163.87/7) = 4.8384}}}, and the standard error is {{{SE=s/sqrt(8) = 4.8384/sqrt(8) = 1.71}}}.
Now we use a one-tailed matched-pairs t-test.

Null hypothesis: D = 0
Alternative hypothesis: D > 0

{{{t = (m - D)/ SE = (3.625 -0)/1.71}}}, under assumption of the null hypothesis.
{{{t = 2.12}}}
Now {{{P(T < 2.12) = 0.9641}}}, so the p-value is 0.0359.  Since 0.0359>0.01, we don't reject the null hypothesis.  Hence the differences in the sample aren't significant to say that there is a drop in the number of crimes.