# On the “correlation” between a continuous and a categorical variable

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

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Let us get back on the Titanic dataset,

loc_fichier = "http://freakonometrics.free.fr/titanic.RData" download.file(loc_fichier, "titanic.RData") load("titanic.RData") base = base[!is.na(base$Age),]

On consider two variables, the age \(x\) (the continuous one) and the survivor indicator \(y\) (the qualitative one)

X = base$Age Y = base$Survived

It looks like the age might be a valid explanatory variable in the logistic regression,

summary(glm(Survived~Age,data=base,family=binomial)) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -0.05672 0.17358 -0.327 0.7438 Age -0.01096 0.00533 -2.057 0.0397 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 964.52 on 713 degrees of freedom Residual deviance: 960.23 on 712 degrees of freedom AIC: 964.23

The significance test here has a \(p\)-value just below \(4%\). Actually, one can relate it with the value of the deviance (the null deviance and the residual deviance). Recall that\(D=2\big(\log\mathcal{L}(\boldsymbol{y})-\log\mathcal{L}(\widehat{\boldsymbol{\mu}})\big)\)while\(D_0=2\big(\log\mathcal{L}(\boldsymbol{y})-\log\mathcal{L}(\overline{y})\big)\)Under the assumption that \(x\) is worthless, \(D_0-D\) tends to a \(\chi^2\) distribution with 1 degree of freedom. And we can compute the \(p\)-value dof that likelihood ratio test,

1-pchisq(964.52-960.23,1) [1] 0.03833717

(which is consistent with a Gaussian test). But if we consider a nonlinear transformation

summary(glm(Survived~bs(Age),data=base,family=binomial)) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.8648 0.3460 2.500 0.012433 * bs(Age)1 -3.6772 1.0458 -3.516 0.000438 *** bs(Age)2 1.7430 1.1068 1.575 0.115299 bs(Age)3 -3.9251 1.4544 -2.699 0.006961 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 964.52 on 713 degrees of freedom Residual deviance: 948.69 on 710 degrees of freedom

which seems to be “more significant”

1-pchisq(964.52-948.69,3) [1] 0.001228712

So it looks like the variable \(x\) is interesting here.

To visualize the non-null correlation, one can consider the condition distribution of \(x\) given \(y=1\), and compare it with the condition distribution of \(x\) given \(y=0\),

ks.test(X[Y==0],X[Y==1]) Two-sample Kolmogorov-Smirnov test data: X[Y == 0] and X[Y == 1] D = 0.088777, p-value = 0.1324 alternative hypothesis: two-sided

i.e. with a \(p\)-value above \(10%\), the two distributions are not significatly different.

F0 = function(x) mean(X[Y==0]<=x) F1 = function(x) mean(X[Y==1]<=x) vx = seq(0,80,by=.1) vy0 = Vectorize(F0)(vx) vy1 = Vectorize(F1)(vx) plot(vx,vy0,col="red",type="s") lines(vx,vy1,col="blue",type="s")

(we can also look at the density, but it looks like that there is not much to see)

An alternative is discretize variable \(x\) and to use Pearson’s independence test,

k=5 LV = quantile(X,(0:k)/k) LV[1] = 0 Xc = cut(X,LV) table(Xc,Y) Y Xc 0 1 (0,19] 85 79 (19,25] 92 45 (25,31.8] 77 50 (31.8,41] 81 63 (41,80] 89 53 chisq.test(table(Xc,Y)) Pearson's Chi-squared test data: table(Xc, Y) X-squared = 8.6155, df = 4, p-value = 0.07146

The \(p\)-value is here \(7%\), with five categories for the age. And actually, we can compare the \(p\)-value

pvalue = function(k=5){ LV = quantile(X,(0:k)/k) LV[1] = 0 Xc = cut(X,LV) chisq.test(table(Xc,Y))$p.value} vk = 2:20 vp = Vectorize(pvalue)(vk) plot(vk,vp,type="l") abline(h=.05,col="red",lty=2)

which gives a \(p\)-value close to \(5\)%, as soon as we have enough categories. In the slides of the course (STT5100), I claim that actually, the age is an important variable when trying to predict if a passenger survived. Test mentioned here are not as conclusive, nevertheless…

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