I am comparing two groups of sex offenders, those who are satisfied with sex offender treatment programs versus though who are not, splitting them by satisfaction or not, and then seeing if they recidivate. I am pretty sure I would use a two-groups t test to compare how many recidivated in the satisfied group versus the non-satisfied. Is this right?
Hypothesis: Those who are satisfied with the sex offender treatment program are less likely to recidivate than those who are not satisfied.
How do you come to think of a t-test?!
You could test the hypothesis that the proportion of recidivates is equal in both groups. The ideal test for this purpose is Fisher's exact test. The estimate of interest is odds-ratio (the ratio of the odds being a recidivate when one is not satisfied and the odds being a recidivate when one is satisfied - or the other way around).
How do you come to think of a t-test?!
You could test the hypothesis that the proportion of recidivates is equal in both groups. The ideal test for this purpose is Fisher's exact test. The estimate of interest is odds-ratio (the ratio of the odds being a recidivate when one is not satisfied and the odds being a recidivate when one is satisfied - or the other way around).
Hello August. I have some questions.
Hi, August,
If you are measuring recidivism as a nominal variable, then you should use Chi square to test. This assumes, of course, a small sample size.
If your research is being conducted in the US and you track recidivism for 2-3 years, I would be very surprised if you found significant results. Recidivism rates are so high overall that it nearly produces a ceiling effect.
I suggest that you carefully consider how to measure recidivism. Technical violations for non-compliance with parole conditions is the most common trigger for re-incarceration. You may want to focus on a new crime rather than a new arrest or technical violation.
Peter, excellent point about measurement of recidivism.
But I don't understand why you say that using a Chi-square test "assumes, of course, a small sample size." If the sample size is too small, some of the expected counts will be low, and the Chi-square approximation will break down. For 2x2 tables with low expected counts, the N-1 Chi-square or the Fisher-Irwin test (aka., Fisher's exact test) are preferred over Pearson's Chi-square. See Ian Campbell's website and Statistics in Medicine article (link below) for the details.
HTH.
http://www.iancampbell.co.uk/twobytwo/twobytwo.htm
Thank you all for your answers and recommendations! I ended up using the chi-square test and it worked out greatly for the research I was performing.
Dear August
The Chi-square test will be suitable for your data because your study deals with a frequency of individual in each category.
Good Luck
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