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I have a data set with two categorical nominal variables (both with 5 categories). I would like to know if (and how) I am able to identify potential correlations between the categories from these two variables.

In other words whether for example the results of category $i$ in variable 1 show a strong correlation with a specific category $j$ in variable 2. Since I have two variables with 5 categories, the total correlation analysis for all the categories would come down to 25 results (at least if it works the way I hope/expect it to work).

I have tried to formulate the problem into concrete questions:

Question 1: Let's say I transfer the categorical variable into 5 different dummy variables per value (category). This same procedure I run for the second variable as well. Then I want to determine the correlation between dummy 1.i and 2.i (for example). Is it statistically correct for me to execute this procedure by means of an ordinary correlation coefficient procedure? Does the correlation coefficient resulting from this procedure provide a proper insight in a correlation among the two dummy variables?

Question 2: If the procedure described in question one is a valid procedure, is there a way to execute this analysis for all categories of 2 (or perhaps more) categorical nominal variables all at once?

The program I am using is SPSS (20).

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  • $\begingroup$ Points made by @Michael Mayer apply to the revised question. $\endgroup$ – Nick Cox Nov 7 '13 at 0:11
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    $\begingroup$ If two variables are not correlated, then you'd have 1/25 in every cell of 5x5 matrix of frequencies. Hence, $\chi^2$ statistics $\sum_{xy}\frac{(O-E)^2}{E}$, where $E=\sum_{xy}O_{xy}/25$ and $O_{xy}$ - observed frequency for any of 5 values of two variables, should be suitable. $\endgroup$ – Aksakal Dec 24 '14 at 5:52
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    $\begingroup$ @Aksakal "Not correlated" is the wrong term here; the variables are nominal, so correlations are not defined. I think you mean independent, but independence doesn't imply equal frequencies either. The cell frequencies under independence depend on the marginal frequencies. $\endgroup$ – Nick Cox Feb 25 '15 at 17:20
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The "focal" association between category $i$ of one nominal variable and category $j$ of the other one is expressed by the frequency residual in the cell $ij$, as we know. If the residual is 0 then it means the frequency is what is expected when the two nominal variables are not associated. The larger the residual the greater is the association due to the overrepresented combination $ij$ in the sample. The large negative residual equivalently says of the underrepresented combination. So, frequency residual is what you want.

Raw residuals are not suitable though, because they depend on the marginal totals and the overall total and the table size: the value is not standardized in any way. But SPSS can display you standardized residuals also called Pearson residuals. St. residual is the residual divided by an estimate of its standard deviation (equal to the sq. root of the expected value). St. residuals of a table have mean 0 and st. dev. 1; therefore, st. residual serves a z-value, like z-value in a distribution of a quantitative variable (actually, it is z in Poisson distribution). St. residuals are comparable between different tables of same size and the same total $N$. Chi-square statistic of a contingency table is the sum of the squared st. residuals in it. Comparing st. residuals in a table and across same-volumed tables helps identify the particular cells that contribute most to chi-square statistic.

SPSS also displays adjusted residuals (= adjusted standardized residuals). Adj. residual is the residual divided by an estimate of its standard error. Interesting that adj. residual is just equal to $\sqrt{N}r_{ij}$, where $N$ is the grand total and $r_{ij}$ is the Pearson correlation (alias Phi correlation) between dummy variables corresponding to the categories $i$ and $j$ of the two nominal variables. This $r$ is exactly what you say you want to compute. Adj. residual is directly related to it.

Unlike st. residual, adj. residual is also standardized wrt to the shape of the marginal distributions in the table (it takes into consideration the expected frequency not only in that cell but also in the cells outside its row and its column) and so you can directly see the strength of the tie between categories $i$ and $j$ - without worrying about whether their marginal totals are big or small relative the other categories'. Adj. residual is also like a z-score, but now it is like z of normal (not Poisson) distribution. If adj. residual is above 2 or below -2 you may conclude it is significant at p<0.05 level$^1$. Adj. residuals are still effected by $N$; $r$'s are not, but you can obtain all the $r$s from adj. residuals, following the above formula, without spending time to produce dummy variables.$^2$

In regard to your second question, about 3-way category ties - this is possible as part of the general loglinear analysis which also displays residuals. However, practical use of 3-way cell residuals is modest: 3(+)-way association measures are not easily standardized and are not easily interpretable.


$^1$ In st. normal curve $1.96 \approx 2$ is the cut-point of 2.5% tail, so 5% if you consider both tails as with 2-sided alternative hypothesis.

$^2$ It follows that the significance of the adjusted residual in cell $ij$ equals the significance of $r_{ij}$. Besides, if there is only 2 columns in the table and you are performing z-test of proportions between $\text {Pr}(i,1)$ and $\text {Pr}(i,2)$, column proportions for row $i$, the p-value of that test equals the significance of both (any) adj. residuals in row $i$ of the 2-column table.

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Directly taken from a document on bivariate statistics with SPSS that lives here:

Chi-square is a useful technique because you can use it to see if there’s a relationship between two ordinal variables, two nominal variables, or between an ordinal and a nominal variable. You look at the assymp. Sig column and if it is less than .05, the relationship between the two variables is statistically significant.

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    $\begingroup$ OK, but three grumbles, one major, two very minor. Chi-square on two ordinal variables ignores the ordering. This isn't the SPSS document, but an elementary introduction by someone else, and they over-simplify, as just mentioned. They didn't copy "Asymp." correctly (example on previous page). The bigger issue for the OP is that correlation is the wrong word here: "association" is the key-word, in terms of measuring, testing and (best of all) modelling the association. $\endgroup$ – Nick Cox Nov 6 '13 at 9:15
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    $\begingroup$ Thanks, I edited the SPSS document bit, it was not my intention to attach any undue authenticity to it . $\endgroup$ – Zhubarb Nov 6 '13 at 9:24

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