Hi,

from your sentence " Ideally, I'd like to be able to say something like "B level individuals re-offend at a significantly higher rate than C level individuals" I conclude that "re-offence" is your response variable, and "program level" is your predictor. Read the following text only when this is correct so far.

ANOVA requires a metric (continuous, at leat interval-scaled) response with symmetric errors. You do not have a metric response, so ANOVA cannot apply here.

The Chi² test compares freuency distribution. In fact you do have frequencies here, and you can regard the distribution of "re-offence" depending on the "program level" (or the other way around - this is a symmetric problem). But what will the test tell you at the end? - It will tell you how likely you should expect to get such (or more "extreme") data under the assumption that there is no correlation between "re-offence" and "program level". If this is unlikely enough, you might decide to believe that the assumption of "no correlation" might not be appropriate (-> "reject" it and claim a correlation).

Going back to your question: would this help? I don't think so. Thus, although some Chi² testing might be validly performed with such data, it won't help you answering your question. So I'd say: neither ANOVA nor Chi² should be used.

What then? Let's again have a look at your response: it is binary, dichotomous, yes/no. You want to fugure out the differences in the proportions (of, say, "yes"-answers) between differen "program levels". A typical way to tackle such a problem directly is to calculate and compare the odds between the groups. The odds ratio between two "program levels" answers your question.

In a very simple case, if you have only 2 different "program levels" (so you'd have a 2x2 table) the whole problem reduces to Fisher's exact test, where the (one and only) odds ratio is estimated and you get a confidence interval for it and a p-value (for the hypothesis that the "true" odds ratio was 1).

Now your predictor has not only 2 but 4 levels. This is dealt in generalized linear models of the binary family, the so-called "logistic models". Look for "logistic regression". These models model the (log) odds ratios depending on any combination of any kind of predictor (in your case a single 5-level categorical predictor), and they give you all the (log) odds ratios (with confidence intervals and p-values). I cannot tell you where to this exactly in SPSS (i don't have it), but I am sure that you will find it when you search for "logistic regression (models)". Maybe some SPSS user can help you there.

Hi,

from your sentence " Ideally, I'd like to be able to say something like "B level individuals re-offend at a significantly higher rate than C level individuals" I conclude that "re-offence" is your response variable, and "program level" is your predictor. Read the following text only when this is correct so far.

ANOVA requires a metric (continuous, at leat interval-scaled) response with symmetric errors. You do not have a metric response, so ANOVA cannot apply here.

The Chi² test compares freuency distribution. In fact you do have frequencies here, and you can regard the distribution of "re-offence" depending on the "program level" (or the other way around - this is a symmetric problem). But what will the test tell you at the end? - It will tell you how likely you should expect to get such (or more "extreme") data under the assumption that there is no correlation between "re-offence" and "program level". If this is unlikely enough, you might decide to believe that the assumption of "no correlation" might not be appropriate (-> "reject" it and claim a correlation).

Going back to your question: would this help? I don't think so. Thus, although some Chi² testing might be validly performed with such data, it won't help you answering your question. So I'd say: neither ANOVA nor Chi² should be used.

What then? Let's again have a look at your response: it is binary, dichotomous, yes/no. You want to fugure out the differences in the proportions (of, say, "yes"-answers) between differen "program levels". A typical way to tackle such a problem directly is to calculate and compare the odds between the groups. The odds ratio between two "program levels" answers your question.

In a very simple case, if you have only 2 different "program levels" (so you'd have a 2x2 table) the whole problem reduces to Fisher's exact test, where the (one and only) odds ratio is estimated and you get a confidence interval for it and a p-value (for the hypothesis that the "true" odds ratio was 1).

Now your predictor has not only 2 but 4 levels. This is dealt in generalized linear models of the binary family, the so-called "logistic models". Look for "logistic regression". These models model the (log) odds ratios depending on any combination of any kind of predictor (in your case a single 5-level categorical predictor), and they give you all the (log) odds ratios (with confidence intervals and p-values). I cannot tell you where to this exactly in SPSS (i don't have it), but I am sure that you will find it when you search for "logistic regression (models)". Maybe some SPSS user can help you there.

Dear Courtney,

Although I am not doing Statistics for many years, Chi-Square is a test og goodness of fit, whereas ANOVA is a technique that you apply when you would like to compare say, types of wheat, which has better production. Then the is the factor of soil. Better soil, better production. Then you have to make an analysis of variance, to a part that is due to factor soil and another part which is due to better production of Wheat.

In order to solve your problem, there is a nice "CookBook" to look:

Introductory Mathematical Statistics By Erwin Kreyszig. There you will find your answer.

Hi,

completely agree with Jochen Wilhelm.

In SPSS you will find it in "analyze" > "regression" > "(binary) logistic".

As a starting point you may have a look at http://www.ats.ucla.edu/stat/spss/output/logistic.htm where you can find an example output of spss with interpretation, and how to find and interpret the odd ratios.

(Small annotation to Jochen's answer: in my humble opinion, I think it's 4-level categorial predictor, not to leave Courtney even more confused ;)

Hope this helps, and good luck with your thesis

Volker

Oh yes, Volker, you are right: It's a 4-level predictor (A,B,C,D)! I am not very confortable with these small numbers ;) I promis that I will practice...

Thanks you for pointing out this mistake and sorry to Cortney for the confusion this might have caused.

Thank you all, so much. These answers have helped immensely (and saved me from having to deal with another problem I was having!). I really appreciate everyone lending me their time and aid!

Dear Jochen and colleagues

I Agree with you on the logistic regression, but I have a reserve about the Fisher Exact Test in this case, because It assumes that there is a sample from one population like occurs in a test of independence, As pointed out by Fisher, this leads under a null hypothesis of independence to a hypergeometric distribution of the numbers in the cells of the table (http://en.wikipedia.org/wiki/Fisher_exact_test).

But a problem like this there are as minimun 2 populations, then de hypergeometric would not be adecuate and the Exact Test would not be Exact at all.

Thank you Guillermo for pointing this out. However, I do not really understand it. Isn't the test about the null that both samples come from the same population? If not, at least from two populations with identical freuqncy distributions? If the latter, how should the test "see" the difference to the former case?

Dear Jochen

We have to consider several things. The Chi Square Test is an approximation of the exacts test involved. There are 3 different tests under the same name: "Chi Square Test".

1) Test of independence: The Null hypotesis is that the categorical variables from a sample of one population of individuals are independent vs that they were somehow asociated. Then Pij= Pi. P.j where Pij is the probability of cell i, j th and Pi. is the marginal probability of level i of the row cathegorical variable and P.j is the marginal probability of level i of the column cathegorical variable.

2) Test of Homogeneity of proportions. The Null hypotesis is that the probability of a event is the same in each of m populations. The population are recognized by a categorical variable and the response event for another. For example if each row of the contingence table correspond to one population:

H0) Pi.=Pi'. for i,i'= 1, 2, ...m

3) Test of Goodness of fi. The Null hypotsis is that the probability of each cell is correctly specified by a tested model. Pij= postulated by the model (Pij)

The Fisher Exact Test is exact when used in 1) and I think that it may be adapted for 3). In 2) we have m populations, from each of them we get an independent sample. Then the categorical variable that defines the population is not a response variable but an explanatory variable and the distribution required by the Fisher test (Hypergeometric) is not valid. For each of the m samples we will have its own hypergeometric distribution and the combined sample would be a product of hypergeometrics.

The example of the teenagers (http://en.wikipedia.org/wiki/Fisher%27s_exact_test) is confused. There you have a sample of teenagers that might be divided by sex into male and female on the one hand, and those that are and are not currently dieting on the other. That is OK for Fisher Test if the sex is a response variable and you did a simple random sample without reposition of a population of teenagers (e.g. High School pupils) without discriminating a priori by sex nor by diet. But if you got an stratified random sample where you get a simple random sample of male teenagers and other simple random sample of female teenagers, then you were in the situation 2. Besides the hypotesis tested would be that the sex is independent of the dietarian habits, instead that the proportion of dieting individuals is higher among the women than among the men that corresponds better to the situation 2.

This hypotesis may be tested with the Fisher Exact Test but it should be written

H0) P(diet |female) =P(diet|male) because in the case of independence they would be the same that its marginal probability..

Pearson Chi-Square is suitable to test if there is a significant correlation between a "Program level" and individual re-offended. coding variables not effect on the computational results.

The results indicate the existence of a relationship between the two variables with highly significant (p

Given all the limitations, she should still go with regression analysis.

Dear Jagdish

If you are talking of LSE Regression Analysis, NO, if you are talking about Logistic Regression Analysis, YES.

I hope this help.