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Let us get back on the Titanic dataset,

loc_fichier = "http://freakonometrics.free.fr/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…