Dear Majid,

Pearson's correlation is the wrong test with ordinal data: try using Spearman's correlation instead. It works better when you have more ordinal categories.

Chi square is for nominal data. Chi 2 calculators are on the web. For nominal scale you can use McNemar test for dependent groups and for independent groups Fisher exact probability test. Calculators are on internet.

If your samples are ordinal and dependent use Sign test or Wilcoxon matched pair signed-ranks test. If the groups are independent but on ordinal scale use Mann-Whitney U test

If you have an interval scale, use it then. It is the best scale. Test the difference between groups with t-test for dependent (n-1) och independent (n-2) groups. Give the scores and an interpretation of the scores.

Dear Majid,

Pearson's correlation is the wrong test with ordinal data: try using Spearman's correlation instead. It works better when you have more ordinal categories.

Majid, if I follow what you are saying, you have a quantitative variable that you are carving into some (ordered) categories, and that you then intend to use a Chi-square test to compare two independent groups. Is this right so far? If so, why do you want to carve the quantitative variable into categories? This is almost always a bad idea: It discards information, and reduces power. Why are you not sticking with the original quantitative variable and using a t-test to compare the groups? Thanks for clarifying.

Peter,

this is not a question about correlations

Dear Bruce,

Yes it is right and I agree with you about discarding the information, and reducing the power. But, is this correct that the chi-square test has an ability to comparing the frequencies that ANOVA and T-test do not have it? (This is the main aim of my investigation)

Is there any other way for comparing the frequencies? I should add that I have two independent groups with 60 participants in each one of them.

Hi majid,

instead compare the frequencies, why dont you just compare the mean. That will be more meaningful for your investigation. this method also available in spss package

Majid,

If you want to see if there is an association between an ordinal variable ( bad, average, good...) and a nominal variable (groups) USING FREQUENCIES, there is a test called the Cochran-Armitage test that does this. (see: http://rcompanion.org/handbook/H_09.html ). And there are measures of effect sizes for this type of test. (see: http://rcompanion.org/handbook/H_11.html ).

The advantage of this over a chi-square test of association, is that the the chi-square test knows that the categories are different ("bad" is different than "average" is different than "good"), but it doesn't know that the categories are ordered ("bad" < "average" < "good").

However, echoing Bruce Weaver and Béatrice Ewalds-Kvist: If you have a measured numeric variable to begin with, there is probably no reason to convert it to an ordinal variable. And you will have to decide how to divide the data into categories. How do you decide what the dividing line is between "bad" and "average"? Is this arbitrary? Does this decision affect your results?

I'm following up on Salvatore's point about the Cochran-Armitage (CA) test. Majid listed SPSS as one of the topics. When one uses the SPSS CROSSTABS procedure to carry out a Chi-square test of association, another test, the test of linear-by-linear association (LLA), is also displayed. It is for use when both variables are ordinal, but also works when one is ordinal and the other dichotomous. You can read about it (and see an example) here:

• https://www.uvm.edu/~dhowell/StatPages/More_Stuff/OrdinalChisq/OrdinalChiSq.html

I did not mention this option earlier, because I did not (and still do not) want to encourage needless categorization of a quantitative variable.

By the way, this IBM Technote explains the relationship between the LLA and CA tests: See the second paragraph under Resolving the problem.

• http://www-01.ibm.com/support/docview.wss?uid=swg21480127

HTH.

Salvatore,

Thanks for your suggestion (Cochran-Armitage test).

In fact it (dividing line) was a part of my question.

Bruce,

Thank you for sharing.

I’m not sure but probably it can be a good example of when this approach can help:

https://scholar.google.com/scholar?hl=en&as_sdt=0%2C5&q=The+tension+between+supervision+and+evaluation%3A+Are+tutors+able+to+objectively+evaluate+an+undergraduate+dissertation%3F&btnG=

Majid,

The paper you cite may be a good example of why we recommend against dividing up numeric data into ordinal categories.

Looking at Table 1, they started with 20-point scales (presumably each of 5 4-point Likert items *), and then looked at the categories from this data as low (15).

How do they justify dividing the scales up like this?

I might assume that in the original Likert item both 1 and 2 were labeled "low" so low scores would sum to 5 - 10, and if 3 and 4 were labeled "high", it would sum to 15 - 20 **. Is this most sensible way to divide up these scales? Who knows?

Is this any better than analyzing the numeric values in the scales? I can't imagine it would be. Isn't it better to think about a score of 20 as different than a score of 16, rather than call them the same "high" category?

---------

* It's weird to use a 4-point Likert item when you are asking mentors to evaluate university students. They would have the experience and insight to use, say, 10-point Likert items easily.

** If my theory is correct, there's a problem with them using > rather than >=, etc.

while i fully agree with bruce and others about the problems associated with categorizing continuous variables, there might in some cases be a point in doing so. it's true that you loose some information in the data but, on the other hand, analyzing the residuals in a chi-square-table might give you more insight into how exactly two distributions differ than just knowing that their means are different, which is all you would know from a significant t-test. so, depending on your specific question/problem and provided that you can satisfactorily justify your categories (for example by using the generally acknowledged categories underweight/healthy/overweight/obese in the case of bmi-values), i would not a priori reject your original idea...

There will likely not be any one test that does it all without its pitfall. I quite agree with the fears of Bruce and others in categorising numerical data into categorical, but sometimes you cannot run away from doing that depending on what you are looking for. Everybody in the same class in a college do not have the same mesure of ability, and could be no telling that everyone in sixth form are better than everyone in the fifth form. Yet is justified to be so in the eyes of school administrator. As long as we can justify a generally acceptable knowledge about this categorization (as stated by Thomas) you are ok

The data refered to here by Maji is already an interval data, but to be converted to ordinal data say '0', '1', '2', .. the between two groups might be '1' for presence and '0' for absence. Chi square gives good result for the global significance and logistic regression can probe further the odds in relation with one another. I have used it in most of my studies

if you have the matrix of frequencies (intervals vs Categories) in case you have two groups you can do the ji-squared test (with more than 50 data) or multivariant analysis (if its less than 50) to the absolute value's diference between the cells

for example if the matrix of diferences of frequencies is:

Observed |Bad |Average| Good|

(0,0.1) |0.1 | 0.5 | 0.4 | 1 |

(0.1,0.3) |0.2 | 0.3 | 0.5 | 1 |

(0.3,1) |0.3 | 0.4 | 0.3 | 1|

|0.6 | 1.2 | 1.2 | 6

we get a matrix of estimate frequencies with eij=row(j)*Col(i)/Total

here total is 6 so we inference that:

Estimated |Bad |Average| Good|

(0,0.1) |0.1 | 0.2 | 0.2 | 0.5 |

(0.1,0.3) |0.1 | 0.2 | 0.2 | 0.5 |

(0.3,1) |0.1 | 0.2 | 0.2 | 0.5 |

|0.3 | 0.6 | 0.6 | 3

In case of ji-squared test its just add all the (O(i,j)-E(i,j))^2/E(i,j) then in PQRS find Ji-squared at 0.95 with (3-1)*(3-1) degrees and the sample statistic has to be smaller than the PQRS statistical else theres no independency on at least one variabel