**Freakonometrics » R-english**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

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=";") > 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.

**leave a comment**for the author, please follow the link and comment on their blog:

**Freakonometrics » R-english**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.