[R-sig-eco] Logistic regression with 2 categorical predictors
ONKELINX, Thierry
Thierry.ONKELINX at inbo.be
Mon Oct 20 17:04:51 CEST 2014
Dear Andrew,
anova() and summary() test different hypotheses. anova() tests is at least one level is different from the others. summary() tests if the coefficient is different from zero.
Multiple comparison of different interaction levels is probably the most relevant in this case. The easiest way is to make a new variable.
snapper2$inter <- with(snapper2, interaction(age, test))
model <- glm(cbind(prefer,avoid) ~ 0 + inter, data=snapper2, family=binomial)
library(multcomp)
mc <- glht(model, mcp(inter = "Tukey))
summary(mc)
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
Thierry.Onkelinx at inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data.
~ John Tukey
-----Oorspronkelijk bericht-----
Van: r-sig-ecology-bounces at r-project.org [mailto:r-sig-ecology-bounces at r-project.org] Namens Andrew Halford
Verzonden: maandag 20 oktober 2014 16:06
Aan: r-sig-ecology at r-project.org
Onderwerp: [R-sig-eco] Logistic regression with 2 categorical predictors
Hi Listers,
I am trying to run a logistic regression to look at the effects of experiment type and age on the behavior of fish in a choice chamber experiment.
I am using the glm approach and would like some advice on how or whether to perform contrasts to work out what levels of Factor1 (Age) and Factor 2
(Test) are significantly different from each other. I have not been able to clarify from my reading what is the appropriate approach to take when dealing with a significant interaction term. I am also not sure as to how one interprets a model when all the coefficients are non-significant but the chi-square ANOVA shows a highly significant interaction term.
I have graphed up the data as dot plots and there is definitely evidence of changes in proportions in later ages.
I want to provide evidence for when and for which tests there was a 'significant' change in behavior.
> snapper2
age test prefer avoid
1 1 LR 15 14
2 1 SD 15 13
3 1 SG 17 14
4 1 SW 14 14
5 2 LR 17 14
6 2 SD 16 19
7 2 SG 20 10
8 2 SW 15 21
9 3 LR 10 16
10 3 SD 14 10
11 3 SG 14 9
12 3 SW 13 15
13 4 LR 12 11
14 4 SD 14 11
15 4 SG 13 12
16 4 SW 11 14
17 5 LR 4 12
18 5 SD 8 8
19 5 SG 0 18
20 5 SW 10 6
21 6 LR 0 6
22 6 SD 3 4
23 6 SG 0 5
24 6 SW 5 3
>
dotplot(age~prefer/avoid,data=snapper2,group=snapper2$test,cex=1.5,pch=19,ylab="age",auto.key=list(space="right",title="Tests"))
> out2 <- glm(cbind(prefer,avoid) ~ age*test, data=snapper2,
family=binomial)
> summary(out2)
Call:
glm(formula = cbind(prefer, avoid) ~ age * test, family = binomial,
data = snapper2)
Deviance Residuals:
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 6.899e-02 3.716e-01 0.186 0.8527
age2 1.252e-01 5.180e-01 0.242 0.8091
age3 -5.390e-01 5.483e-01 -0.983 0.3256
age4 1.802e-02 5.589e-01 0.032 0.9743
age5 -1.168e+00 6.866e-01 -1.701 0.0890 .
age6 -2.575e+01 9.348e+04 0.000 0.9998
testSD 7.411e-02 5.307e-01 0.140 0.8890
testSG 1.252e-01 5.180e-01 0.242 0.8091
testSW -6.899e-02 5.301e-01 -0.130 0.8964
age2:testSD -4.401e-01 7.260e-01 -0.606 0.5444
age3:testSD 7.324e-01 7.846e-01 0.933 0.3506
age4:testSD 8.004e-02 7.863e-01 0.102 0.9189
age5:testSD 1.024e+00 9.301e-01 1.102 0.2707
age6:testSD 2.532e+01 9.348e+04 0.000 0.9998
age2:testSG 3.738e-01 7.407e-01 0.505 0.6138
age3:testSG 7.867e-01 7.832e-01 1.004 0.3152
age4:testSG -1.321e-01 7.764e-01 -0.170 0.8649
age5:testSG -2.568e+01 8.768e+04 0.000 0.9998
age6:testSG 2.121e-02 1.334e+05 0.000 1.0000
age2:testSW -4.616e-01 7.249e-01 -0.637 0.5242
age3:testSW 3.959e-01 7.662e-01 0.517 0.6054
age4:testSW -2.592e-01 7.858e-01 -0.330 0.7415
age5:testSW 1.678e+00 9.386e-01 1.788 0.0737 .
age6:testSW 2.626e+01 9.348e+04 0.000 0.9998
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 5.4908e+01 on 23 degrees of freedom Residual deviance: 2.6113e-10 on 0 degrees of freedom
AIC: 122.73
Number of Fisher Scoring iterations: 23
> anova(out2, test="Chisq")
Analysis of Deviance Table
Model: binomial, link: logit
Response: cbind(prefer, avoid)
Terms added sequentially (first to last)
Df Deviance Resid. Df Resid. Dev Pr(>Chi)
NULL 23 54.908
age 5 11.235 18 43.673 0.0469115 *
test 3 1.593 15 42.079 0.6608887
age:test 15 42.079 0 0.000 0.0002185 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
cheers
Andy
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