A short post to get back – for my nonlife insurance course – on the interpretation of the output of a regression when there is a categorical covariate. Consider the following dataset

> db = read.table("http://freakonometrics.free.fr/db.txt",header=TRUE,sep=";") > attach(db) > tail(db) Y X1 X2 X3 995 1 4.801836 20.82947 A 996 1 9.867854 24.39920 C 997 1 5.390730 21.25119 D 998 1 6.556160 20.79811 D 999 1 4.710276 21.15373 A 1000 1 6.631786 19.38083 A

Let us run a logistic regression on that dataset

> reg = glm(Y~X1+X2+X3,family=binomial,data=db) > summary(reg) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -4.45885 1.04646 -4.261 2.04e-05 *** X1 0.51664 0.11178 4.622 3.80e-06 *** X2 0.21008 0.07247 2.899 0.003745 ** X3B 1.74496 0.49952 3.493 0.000477 *** X3C -0.03470 0.35691 -0.097 0.922543 X3D 0.08004 0.34916 0.229 0.818672 X3E 2.21966 0.56475 3.930 8.48e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 552.64 on 999 degrees of freedom Residual deviance: 397.69 on 993 degrees of freedom AIC: 411.69 Number of Fisher Scoring iterations: 7

Here, the reference is modality . Which means that for someone with characteristics , we predict the following probability

where denotes the cumulative distribution function of the logistic distribution

For someone with characteristics , we predict the following probability

For someone with characteristics , we predict the following probability

(etc.) Here, if we accept (against ), it means that modality cannot be considerd as different from .

A natural idea can be to change the reference modality, and to look at the -values. If we consider the following loop, we get

> M = matrix(NA,5,5) > rownames(M)=colnames(M)=LETTERS[1:5] > for(k in 1:5){ + db$X3 = relevel(X3,LETTERS[k]) + reg = glm(Y~X1+X2+X3,family=binomial,data=db) + M[levels(db$X3)[-1],k] = summary(reg)$coefficients[4:7,4] + } > M A B C D E A NA 0.0004771853 9.225428e-01 0.8186723647 8.482647e-05 B 4.771853e-04 NA 4.841204e-04 0.0009474491 4.743636e-01 C 9.225428e-01 0.0004841204 NA 0.7506242347 9.194193e-05 D 8.186724e-01 0.0009474491 7.506242e-01 NA 1.730589e-04 E 8.482647e-05 0.4743636442 9.194193e-05 0.0001730589 NA

and if we simply want to know if the -value exceeds – or not – 5%, we get the following,

> M.TF = M>.05 > M.TF A B C D E A NA FALSE TRUE TRUE FALSE B FALSE NA FALSE FALSE TRUE C TRUE FALSE NA TRUE FALSE D TRUE FALSE TRUE NA FALSE E FALSE TRUE FALSE FALSE NA

The first column is obtained when is the reference, and then, we see which parameter should be considered as null. The interpretation is the following:

- and are not different from
- is not different from
- and are not different from
- and are not different from
- is not different from

Note that we only have, here, some kind of intuition. So, let us run a more formal test. Let us consider the following regression (we remove the intercept to get a model easier to understand)

> library(car) > db$X3=relevel(X3,"A") > reg=glm(Y~0+X1+X2+X3,family=binomial,data=db) > summary(reg) Coefficients: Estimate Std. Error z value Pr(>|z|) X1 0.51664 0.11178 4.622 3.80e-06 *** X2 0.21008 0.07247 2.899 0.00374 ** X3A -4.45885 1.04646 -4.261 2.04e-05 *** X3E -2.23919 1.06666 -2.099 0.03580 * X3D -4.37881 1.04887 -4.175 2.98e-05 *** X3C -4.49355 1.06266 -4.229 2.35e-05 *** X3B -2.71389 1.07274 -2.530 0.01141 *

--- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 1386.29 on 1000 degrees of freedom Residual deviance: 397.69 on 993 degrees of freedom AIC: 411.69 Number of Fisher Scoring iterations: 7

It is possible to use Fisher test to test if some coefficients are equal, or not (more generally if some linear constraints are satisfied)

> linearHypothesis(reg,c("X3A=X3C","X3A=X3D","X3B=X3E")) Linear hypothesis test Hypothesis: X3A - X3C = 0 X3A - X3D = 0 - X3E + X3B = 0 Model 1: restricted model Model 2: Y ~ 0 + X1 + X2 + X3 Res.Df Df Chisq Pr(>Chisq) 1 996 2 993 3 0.6191 0.892

Here, we clearly accept the assumption that the first three factors are equal, as well as the last two. What is the next step? Well, if we believe that there are mainly two categories, and , let us create that factor,

> X3bis=rep(NA,length(X3)) > X3bis[X3%in%c("A","C","D")]="ACD" > X3bis[X3%in%c("B","E")]="BE" > db$X3bis=as.factor(X3bis) > reg=glm(Y~X1+X2+X3bis,family=binomial,data=db) > summary(reg) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -4.39439 1.02791 -4.275 1.91e-05 *** X1 0.51378 0.11138 4.613 3.97e-06 *** X2 0.20807 0.07234 2.876 0.00402 ** X3bisBE 1.94905 0.36852 5.289 1.23e-07 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 552.64 on 999 degrees of freedom Residual deviance: 398.31 on 996 degrees of freedom AIC: 406.31 Number of Fisher Scoring iterations: 7

Here, all the categories are significant. So we do have a proper model.

OpenEdition suggests that you cite this post as follows:

Arthur Charpentier (September 26, 2013). Logistic regression and categorical covariates. *Freakonometrics*. Retrieved September 14, 2024 from https://doi.org/10.58079/ourf

Hi, thanks for this tutorial.

I have some questions, Can we proceed to delete X3C based on their p-values in the glm output ?

Hi,

I assume that I have to fully qualify the db$X3 variable in your for loop because I have not attached the dataset; otherwise your code does not work for me. (i.e., “relevel(X3,LETTERS[k])” vs “relevel(db$X3,LETTERS[k])”.

indeed ! sorry… actually (so might have guessed) I have generated that dataset. So I did create the variables, created the dataset, and then change the initial code to hide the generating code. But I always forgot that I have to attach the dataset ! I will fix that ! Thanks !

oks Arthur. Thank you. I think your next post it’s very interesting.

You can check this link of plot in ROC curve, too:

http://www.r-bloggers.com/an-example-of-roc-curves-plotting-with-rocr/

Keep in touch,

Martí

Hello,

I think in this example we have a problem of underdispersion when we split: residual deviance/df = 0.40=phi

“Residual deviance: 397.69 on 993 degrees of freedom “.

On the other hand, I think it will be interesting explain how we can treat collinearity in categorical variables in logistic regression. Maybe, in the next post?

Thank you for your contributions and examples.

Martí

thanks Martí for the post,

I did not plan to discuss undispersion, but you’re right… I should. About “collinearity in categorical variables”, I am clearly not an expert. Even for “collinearity in continuous variables”, I usually see an interpretation problem. But I’d be glad to discuss those issues. Or at least to give my point of view.

About next posts, I plan to discuss ROC curve construction for my students, actually.

Keep in touch