Statistical test to assess counts between years

[Jessica Berrett]

Hi all, is there a simple test to assess if there is a statistical difference between counts for years?

[Nick Cox]

You could try fitting a regression to test linear trend. That seems unlikely to be very interesting or useful.

[Leonardo Guizzetti]

Are the data averaged/summed across individual annual counts or are the 3-year intervals just the reporting frequency?

Do you just want to know if there are differences in counts between years? If so, a simple eyeball test is pretty clear. There is sharp drop after 2008. If you don't care to know where such differences are, a chi-squared test is adequate, but as Nick said, it's fairly uninteresting nor useful. Using the linear regression approach would almost certainly yield a negative slope, but again, that's self-evident so fairly pointless to test.

[George Ford]

Fisher's exact test?

help cc

look at cci, if you know the exposure counts.

[Carlo Lazzaro]

Jessica:
as an aside to previous helful advice, you may want to consider (with the usual caveats about overdispersion) -poisson-:

 poisson Counts i.Year

Iteration 0:  Log likelihood = -14.261213  
Iteration 1:  Log likelihood = -14.261213  

Poisson regression                                      Number of obs =      5
                                                        LR chi2(4)    =   6.01
                                                        Prob > chi2   = 0.1987
Log likelihood = -14.261213                             Pseudo R2     = 0.1739

------------------------------------------------------------------------------
      Counts | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
        Year |
          2  |  -.0521858   .1865645    -0.28   0.780    -.4178454    .3134739
          3  |  -.3163373   .2005118    -1.58   0.115    -.7093332    .0766585
          4  |  -.3398678   .2018878    -1.68   0.092    -.7355606     .055825
          5  |  -.3639654   .2033209    -1.79   0.073    -.7624671    .0345363
             |
       _cons |   4.077537   .1301889    31.32   0.000     3.822372    4.332703
------------------------------------------------------------------------------

. testparm i.Year

 ( 1)  [Counts]2.Year = 0
 ( 2)  [Counts]3.Year = 0
 ( 3)  [Counts]4.Year = 0
 ( 4)  [Counts]5.Year = 0

           chi2(  4) =    6.06
         Prob > chi2 =    0.1944

.

[George Ford]

Add robust and that should handle the overdispersion problem. Those results look plausible.