I need to compare the means between total number of violent incidents pre/post intervention. Can I use paired T-Test?

Question

I have the total number of incidents occurred within a given setting pre (e.g. 100) and post (e.g. 60) intervention and need to find out if the reduction is statistically significant. Would a paired T-Test be appropriate? For some reason I have my doubts...Anyone can help?

Answer 1

Unfortunately, a paired t-test cannot work with only a single pair.

One idea is to try a two-sample test of proportions. Say you have two four-month chunks of data (120 days pre with 40 incidents and 120 post-change with 33).

The output will look something this:

. prtesti 120 33 120 40, level(95) count

Two-sample test of proportions                     x: Number of obs =      120
                                                   y: Number of obs =      120
------------------------------------------------------------------------------
             |       Mean   Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
           x |       .275    .040761                      .1951099    .3548901
           y |   .3333333   .0430331                      .2489899    .4176768
-------------+----------------------------------------------------------------
        diff |  -.0583333   .0592732                     -.1745066      .05784
             |  under H0:   .0593927    -0.98   0.326
------------------------------------------------------------------------------
        diff = prop(x) - prop(y)                                  z =  -0.9822
    H0: diff = 0

    Ha: diff < 0                 Ha: diff != 0                 Ha: diff > 0
 Pr(Z < z) = 0.1630         Pr(|Z| > |z|) = 0.3260          Pr(Z > z) = 0.8370

Here the null hypothesis H0 is that the incident rate is the same in the two 120 day periods. This is the claim you presumably want to argue against. The output reports results for three alternative hypotheses. I think the appropriate alternative is that the incidence rate declined, so Ha is diff < 0 is the test to focus on. The p-value for that one is 0.1630. The interpretation of that number is the following: if there was no difference in the incidence rate in the two periods, you would still see 30 or fewer incidents 16% of the time by chance alone. This means that the data is fairly consistent with there being no effect. It does not mean that there is no effect, just that you can't distinguish it with the data you have. In other words, you cannot reject the null hypothesis. If you want some more intuition about hypothesis testing, there is a nice example here.

With more data, something more sophisticated could be done, with fewer assumptions, but this is probably the best route for the data you have.