I have a question regarding the interpretation of an ordinal regression in R with the clm function.
I am new to this and used this formula:
RegADHD <- clm(ADHD$Score ~ ADHD$Group)
Score is a ordered factor from 0-3 (0= no symptoms, 3= maximal symptoms)
Group is a group variable with 3 categories (no exposure, less exposure, severe exposure).
With summary I get this:
Coefficients:
Estimate Std. Error z value Pr(>|z|)
ADHD$Group3 1.0422 0.4647 2.242 0.0249 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Threshold coefficients:
Estimate Std. Error z value
0|1 1.3612 0.3720 3.659
1|2 2.5735 0.4465 5.764
2|3 3.3582 0.5367 6.258
(2 observations deleted due to missingness)
But I don´t really understand what is the interpretation of this coefficient?
More specifically, I struggle because I get only one value (for group 3). These are the contrasts of group:
2 3
1 0 0
2 1 0
3 0 1
To get the odds I did:
round(exp(RegADHD$beta), 3)
This is 2.835
with(RegADHD, table(ADHD$Score, ADHD$Group))
leads to
1 2 3
0 0 34 30
1 0 7 11
2 0 1 5
3 0 1 5
Does this mean that the probability of having a higher score (in ADHD) increases 2.835 times with being in the severe exposure group, compared with no exposure? But then I don´t know anything about less exposure... I don´t understand why I get only one coefficient!
I hope anyone can help me, I am thankful for every advice!
Groupvariable. It looks like it has only two levels. I also suspect you want some output other than that given bysummary. To get an anova-like table, trylibrary(car); library(RVAideMemoire); Anova.clm(RegADHD). To get differences between groups, trylibrary(emmeans); emmeans(RegADHD, ~ Group); pairs(emmeans(RegADHD, ~ Group))– Sal Mangiafico Oct 07 '18 at 15:22with(RegADHD, table(ADHD$Score, ADHD$Group), you have no observations for the first Group. – Sal Mangiafico Oct 07 '18 at 16:44Anova.clmandemmeans, probably along with the group medians. That's the way I would look at this, like I would a typical anova with post-hoc. But the information fromsummarymay be more meaningful for you. ... When you say do another one and compare both, not sure what you would intend to compare. – Sal Mangiafico Oct 07 '18 at 19:05anova(model1, model2)will conduct a likelihood-ratio test on the two models. I think the type-I SS question comes into play here only by which models you compare. That is, by which models you put in to theanovafunction, you are deciding whether you are testing each term sequentially or not sequententially. This link addresses this in a slightly different context. – Sal Mangiafico Oct 08 '18 at 12:04