How to plot a logistic regression

[Cath]

Hi,
I did a logistic Regression and trying to plot it now. However, I do not find a way to plot it nicely. I'm not very familiar with R so I'm first trying to do it without interaction effects and just one variable.

Here is my data:

> dput(head(Wahl2013))
structure(list(Wahlbeteiligung = structure(c(1L, 1L, 1L, 1L, 
1L, 1L), .Label = c("ja, habe gewaehlt", "nein, habe nicht gewaehlt"
), class = "factor"), Geschlecht = structure(c(2L, 2L, 1L, 1L, 
1L, 1L), .Label = c("0", "1"), class = "factor"), Gebj = c(4L, 
1L, 1L, 1L, 2L, 2L), Zweitstimme = c(1, 0, 0, 0, 0, 0), eig.Pst.Klima = c(4, 
6, 7, 5, 5, 4), Salienz.Klima = c(1, 4, 4, 3, 2, 3), Bildung = c(3L, 
3L, 1L, 3L, 2L, 1L), Atomenergie = c(3, 3, 3, 5, 3, 1)), na.action = structure(c(`2` = 2L, 
`11` = 11L, `22` = 22L, `29` = 29L, `50` = 50L, `58` = 58L, `72` = 72L, 
`76` = 76L, `77` = 77L, `85` = 85L, `96` = 96L, `108` = 108L, 
`112` = 112L, `119` = 119L, `120` = 120L, `124` = 124L, `125` = 125L, 
`130` = 130L, `142` = 142L, `143` = 143L, `151` = 151L, `160` = 160L, 
`175` = 175L, `183` = 183L, `190` = 190L, `196` = 196L, `219` = 219L, 
`229` = 229L, `234` = 234L, `238` = 238L, `243` = 243L, `248` = 248L, 
`261` = 261L, `269` = 269L, `277` = 277L, `279` = 279L, `285` = 285L, 
`286` = 286L, `287` = 287L, `311` = 311L, `313` = 313L, `319` = 319L, 
`324` = 324L, `331` = 331L, `334` = 334L, `347` = 347L, `348` = 348L, 
`351` = 351L, `352` = 352L, `359` = 359L, `368` = 368L, `373` = 373L, 
`374` = 374L, `380` = 380L, `385` = 385L, `391` = 391L, `398` = 398L, 
`410` = 410L, `412` = 412L, `422` = 422L, `423` = 423L, `434` = 434L, 
`435` = 435L, `442` = 442L, `449` = 449L, `453` = 453L, `462` = 462L, 
`463` = 463L, `466` = 466L, `473` = 473L, `483` = 483L, `484` = 484L, 
`534` = 534L, `546` = 546L, `547` = 547L, `554` = 554L, `561` = 561L, 
`568` = 568L, `573` = 573L, `583` = 583L, `596` = 596L, `612` = 612L, 
`618` = 618L, `619` = 619L, `625` = 625L, `638` = 638L, `645` = 645L, 
`677` = 677L, `692` = 692L, `726` = 726L, `734` = 734L, `738` = 738L, 
`741` = 741L, `751` = 751L, `759` = 759L, `767` = 767L, `768` = 768L, 
`770` = 770L, `774` = 774L, `784` = 784L, `792` = 792L, `793` = 793L, 
`800` = 800L, `805` = 805L, `821` = 821L, `834` = 834L, `857` = 857L, 
`867` = 867L, `869` = 869L, `877` = 877L, `895` = 895L, `896` = 896L, 
`898` = 898L, `912` = 912L, `918` = 918L, `925` = 925L, `928` = 928L, 
`931` = 931L, `939` = 939L, `946` = 946L, `949` = 949L, `956` = 956L, 
`1001` = 1001L, `1009` = 1009L, `1016` = 1016L, `1018` = 1018L, 
`1019` = 1019L, `1031` = 1031L, `1032` = 1032L, `1054` = 1054L, 
`1058` = 1058L, `1062` = 1062L, `1063` = 1063L, `1068` = 1068L, 
`1089` = 1089L, `1090` = 1090L, `1101` = 1101L, `1102` = 1102L, 
`1121` = 1121L, `1154` = 1154L, `1156` = 1156L, `1162` = 1162L, 
`1170` = 1170L, `1174` = 1174L, `1181` = 1181L, `1182` = 1182L, 
`1183` = 1183L, `1191` = 1191L, `1196` = 1196L, `1201` = 1201L, 
`1215` = 1215L, `1233` = 1233L, `1267` = 1267L, `1270` = 1270L, 
`1294` = 1294L, `1297` = 1297L, `1305` = 1305L, `1315` = 1315L, 
`1330` = 1330L, `1335` = 1335L, `1338` = 1338L, `1340` = 1340L, 
`1345` = 1345L, `1352` = 1352L, `1370` = 1370L, `1373` = 1373L, 
`1379` = 1379L, `1380` = 1380L, `1400` = 1400L, `1410` = 1410L, 
`1427` = 1427L, `1438` = 1438L, `1439` = 1439L, `1440` = 1440L, 
`1444` = 1444L, `1447` = 1447L, `1468` = 1468L, `1469` = 1469L, 
`1473` = 1473L, `1485` = 1485L, `1486` = 1486L, `1490` = 1490L, 
`1498` = 1498L, `1499` = 1499L, `1500` = 1500L, `1503` = 1503L, 
`1506` = 1506L, `1516` = 1516L, `1520` = 1520L, `1522` = 1522L, 
`1536` = 1536L, `1545` = 1545L, `1557` = 1557L, `1565` = 1565L, 
`1569` = 1569L, `1576` = 1576L, `1577` = 1577L, `1579` = 1579L, 
`1586` = 1586L, `1596` = 1596L), class = "omit"), row.names = c(1L, 
3L, 4L, 5L, 6L, 7L), class = "data.frame")

model3 <- glm(Zweitstimme ~ Atomenergie, data = Wahl2013)
> summary(model3)

Call:
glm(formula = Zweitstimme ~ Atomenergie, data = Wahl2013)

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.17189  -0.12128  -0.07067  -0.02007   1.03054  

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.081152   0.022555  -3.598 0.000332 ***
Atomenergie  0.050609   0.006139   8.244 3.81e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 0.08119716)

    Null deviance: 118.71  on 1395  degrees of freedom
Residual deviance: 113.19  on 1394  degrees of freedom
AIC: 460.49

Number of Fisher Scoring iterations: 2

> ggplot(Wahl2013, aes(x=Atomenergie, y=Zweitstimme)) + 
+   geom_point(shape=1, position=position_jitter(width=.05,height=.05)) + 
+   stat_smooth(method="glm", method.args=list(family="binomial"), se=FALSE)

This Looks like this:

I do not find a way, to make it look better. Thanks for help!

[FJCC]

To do a logistic regression, I believe you have to set the family argument of glm() to binomial().

model3 <- glm(Zweitstimme ~ Atomenergie, data = Wahl2013,
                family = binomial(link = "logit"))

To plot the result, you can try calculating the observed probability of Zweitstimme = 1 at each Atomenergie against the probability calculated from the logistic fit.

[technocrat]

If the outcome Zweitstimme is not binary 1/0, as appears from the plot, a logistic regression with family = binary does not work. Values of Zweitstimme are clustered around 1 and 0, but exceed it. To perform a logistic regression it would be necessary to round those values to 1 or 0.

And then what results does not plot like an ordinary least squares regression, but similarly to

See my post

[Matthias]

I am not sure if this is a suitable case for logistic regression, because the x-axis values (Atomenergie) aren't numerical but more categorical (don't like, somehow like, like...) encoded as numeric. It probably works better with something like age or so.
Also here the ratio between 0 and 1 (elected this party vs. did not elected this party?) is more relevant, also to show this in the plot.
Here I would suggest to calculate the percentage and show just this, either for the positive case or for both cases with stacked bars. Then the trend should be visible, maybe even a linear regression could do the job (although the same rule applies, technically it shouldn't be used for categorical values)