I am conducting research investigating two different groups of women: one group of Short-Stature Women (SSW) and another with Non-Short Women (NSW). We have the hypothesis that SSW has an inaccurate auto-perception of their current body size (CBS). We assessed CBS with a figure rating scale, consisting of 9 different silhouettes, ranging from 1 to 9 (which would assess something like from "malnourished" to "very obese"). We know that CBS is highly determined by the person's Body Mass Index (BMI). Hence, we would like to know if the BMI is a significantly better "predictor" of CBS in NSW compared to SSW.

My first approach was to run two Spearman rho rank correlation test (one for each group (SSW and NSW), between BMI and CBS) and then to compare both coefficients using the Fisher's Z-test. Nevertheless, I am pretty sure that I can treat CBS as an ordinal variable, right? So, it seems to me like an ordinal regression (using CBS as DV and BMI as IV) would be more suitable to my data. If this is true, my problem is that I do not know how to compare coefficients for ordinal regression. Would it be comparing Nagelkerke-R²? How can I do this?

To summarize I have 2 main questions: a) Is Spearman rho rank correlation with comparison of the coefficients by Fisher's Z-test adequate in this case? and b) If ordinal regression is more adequate, how can I compare the coefficients of the regression between my two groups?


1 Answer 1


Yes, CBS sounds ordinal. If you're only interested in comparing bivariate relationships, comparison of Spearman's $\rho$s seems fair enough to me. However, an ordinal regression model would allow you to estimate the independent relationships of BMI and stature while controlling the effects of each other predictor. You could also test whether these factors moderate each other by including an interaction term. For more on that, see "How to test whether a regression coefficient is moderated by a grouping variable?" and "What is the correct way to test for significant differences between coefficients?"

The method described in the latter question is correct for your purposes in multiple regression (though it wasn't in that OP's case). If you want to calculate the probability that your two predictors' slope coefficients would differ by at least as much as they do in your sample if (1) you were to collect another equivalent sample from the exact same population, and if (2) your predictors are actually equally related to CBS, then you can do this with a z-test: $$Z = \frac{b_1 - b_2}{\sqrt{SE_{b_1}^2 + SE_{b_2}^2}}$$

You can convert the resultant z statistic to a probability using pnorm in or with other methods described here: "How to deal with Z-score greater than 3?"

By the way, if you have continuous height data, you probably ought to reconsider dichotomizing stature into short and non-short. This wastes information that could improve your regression model and correlation estimates; you would probably get smaller standard errors by entering both BMI and height data as originally measured (if you have it) as predictors in multiple ordinal regression of CBS.

  • $\begingroup$ Thanks Nick. First, I do have continuous height data, but the design of the research is to strictly compare SSW (<150cm defined by World Health Organization) and NSW (>160 cm). So I have nobody between 150-160 cm. Hence, my interest is really comparing findings between groups. I was somehow aware that I could include a dummy variable in the model to compare regression coefficients, but only for a linear regression. Actually, my SPSS output for ordinal regression using BMI as IV and CBS as DV does not show something like a "coefficient" of the linear regression, except for the pseudo-R². $\endgroup$ Commented Jun 10, 2014 at 3:02
  • $\begingroup$ @NassibBueno: that might be an unfortunate consequence of using SPSS ;) you should be able to write the syntax manually to produce slope coefficients for your predictors if necessary; there are such components of the ordinal regression model. Maybe it's time to switch to R? I'm getting slope coefficients out by default using polr from the MASS package... $\endgroup$ Commented Jun 10, 2014 at 5:43
  • $\begingroup$ You are right, I began to use SPSS due to it "deceiving" user-friendly interface, and now I feel its lack of extra resources. Nevertheless, I am able to use Stata, and I will find out how to calculate coefficients for the ordinal regression there and then I will apply the methods described by you to compare coefficients. If I fail in last case, I am at least safe to use the Spearman correlation with Fisher Z-test approach, right? Thank you again, Nick. $\endgroup$ Commented Jun 11, 2014 at 2:05
  • $\begingroup$ Yeah, it's not an unsafe method AFAIK; it's just rather limited. You can do a lot more with ordinal regression than with comparison of bivariate correlations. $\endgroup$ Commented Jun 11, 2014 at 2:46
  • $\begingroup$ Nick, I think I did it. I run the ologit command in Stata (and realized that SPSS do show me a coefficient for the ordinal regression, but he calls it "estimate"). The I found both coefficients and SE, for SSW (0.1754 ; 0.06262) and NSW (0.2639, 0.0528). I applied your equation and found a Z-value of -13.4, which means a a really small p-value, right? (I could not calculate the exact p). Hence I conclude that BMI predicts CBS in significantly different ways between SSW and NSW (P < .01), right? the problem is that the CI95% I receive are: SSW [0.0526 to 0.2981] and NSW: [0.1603 to 0.3675]. $\endgroup$ Commented Jun 11, 2014 at 2:59

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