# u-shape for logistic regression?

I'm stuck and have real problems what to interpret from my current results. Maybe you can help me out? Thanks!

Lets say... I'm investigating on the influence of health factors on dying.

Dependent variable: death yes/no after 10 years time

Independent variables: ml wine per day, cigarettes per day, gramm of fruits or vegetables per day, minutes of excercises per day etc. ...

I'm doing a logistic regression, since I have a binary dependent variable:

model.binomial <- glm(dv_death ~
wine +
cigarettes +
fruits +
excercise,
data = complete_dataset, family = binomial(link = logit))


I have a questions and I might just lost sight, but..:

If I put in the model all variables (wine, cigarettes, fruits and excercise), all of them are significant. If I only use the independent variable "wine", it is not significant (same goes for all other variables: I have to admit, I also have a correlation between wine + cigarettes of 0.55, but VIFs and Eigenscores are alright). However... when I look at the wine and death data specifically by using:

ggplot(complete_dataset, aes(x=complete_dataset$$wine, y=complete_dataset$$death))+ geom_point(size=2, alpha=0.4)+
stat_smooth(method="loess", colour="blue", size=1.5)+
xlab("Wine")+
ylab("Death (yes = 1)")+
theme_bw()


... I get this kind of plot:

For me this seems to be a u-shape correlation: Too little wine and too much wine reduces your probability of dying, so either be an alcoholic or do not every take a sip...

However, the variable is not significant. Can I test for a u shape in a logistic regression? Or am I on the completely wrong track?

(Don't worry - this is a made up example so pour yourself a drink..)

Update due to the comments:

I added an independent variable squared wine to the model.

Full model without winesquared: wine is not significant.

Full model with wine + winesquared: both are significant - wine (p<0.001), wine squared (p<0.01)

Single model without winesquared: wine is not significant

Single model winesquared only: winesquared is not significant

"Single" model with wine and winesquared: both are significant - both at p<0.1

Update thanks to @Roland: GAM Model:

model.binomial.gam <- mgcv::gam(dv_death ~
s(wine) +
cigarettes +
fruits +
excercise,
data = complete_dataset, family = binomial(link = logit), select = TRUE)
summary(model.binomial.gam)
Estimate Std. Error z value       Pr(>|z|)
(Intercept)                -0.9217701  0.3225723  -2.858       0.004269 **
cigarettes                 -8.0936235  3.5047369  -2.309       0.020925 *
fruits                      0.3063182  0.0838298   3.654       0.000258 ***
excercise                   0.1126536  0.0273186   4.124 0.000037284368 ***

Approximate significance of smooth terms
edf Ref.df Chi.sq p-value
s(wine) 2.478      9  16.55 0.00014 ***

• It may help you to read this: How to use boxplots to find the point where values are more likely to come from different conditions? Commented Dec 5, 2018 at 17:40
• Is your death indicator reverse coded (is 1 actually alive and 0 dead?). Usually, the best survival is for intermediate alcohol consumption. Commented Dec 5, 2018 at 18:02
• Have you checked whether each independent variable is linear in the log of odds of death? Note that you're assuming that when you run the logistic regression, and by assuming that wine is linear in the log of odds of death you're already assuming a certain non-linear relationship between wine and death. Start by understanding what relationship between wine and death you've assumed to hold and whether that assumption is satisfied. Commented Dec 5, 2018 at 18:12
• I would fit a GAM to account for possible (and likely) nonlinear relationships on the log-odds scale. Using package mgcv would also combine this with shrinkage and take care of variable selection. Commented Dec 6, 2018 at 7:15
• @ColorStatistics I'm not too sure if I understand you correctly, thank you and sorry. I have reasoning and evidence for both aussumptions (wine & death have a linear vs. a non-linear relationship). With a binary dv I thought I can only run a logistic model. Commented Dec 6, 2018 at 8:57