I fit this mixed model with beta for the response variable: photochemical efficiency or Fv/Fm and the predictor variables are categorical:

Family: beta  ( logit )
Formula:          FvFm ~ hora * temperatura + (1 | Experimento)
Data: d

     AIC      BIC   logLik deviance df.resid 
  -730.4   -711.2    371.2   -742.4      174 

Random effects:

Conditional model:
 Groups      Name        Variance Std.Dev.
 Experimento (Intercept) 0.0099   0.0995  
Number of obs: 180, groups:  Experimento, 5

Dispersion parameter for beta family ():  262 

Conditional model:
                    Estimate Std. Error z value Pr(>|z|)    
(Intercept)          0.26960    0.04821    5.59 2.24e-08 ***
hora4                0.03181    0.02627    1.21   0.2260    
temperatura20        0.05626    0.02630    2.14   0.0324 *  
hora4:temperatura20 -1.44971    0.03859  -37.56  < 2e-16 ***

I have been trying to interpret a posteriori test, emmeans results with type="response", so I get the odds ratios (exp) of the estimated marginal means for all possible comparison groups.

> emmGrid<-emmeans(m2, specs=pairwise~temperatura*hora, type="response")
> emmGrid
 temperatura hora response      SE  df asymp.LCL asymp.UCL
 2           0       0.567 0.01184 Inf     0.544     0.590
 20          0       0.581 0.01175 Inf     0.558     0.604
 2           4       0.575 0.01179 Inf     0.552     0.598
 20          4       0.251 0.00928 Inf     0.233     0.270

Confidence level used: 0.95 
Intervals are back-transformed from the logit scale 

 contrast                                  odds.ratio     SE  df null z.ratio p.value
 temperatura2 hora0 / temperatura20 hora0       0.945 0.0249 Inf    1  -2.139  0.1408
 temperatura2 hora0 / temperatura2 hora4        0.969 0.0254 Inf    1  -1.211  0.6200
 temperatura2 hora0 / temperatura20 hora4       3.903 0.1101 Inf    1  48.268  <.0001
 temperatura20 hora0 / temperatura2 hora4       1.025 0.0270 Inf    1   0.929  0.7894
 temperatura20 hora0 / temperatura20 hora4      4.128 0.1167 Inf    1  50.149  <.0001
 temperatura2 hora4 / temperatura20 hora4       4.029 0.1138 Inf    1  49.346  <.0001

P value adjustment: tukey method for comparing a family of 4 estimates 
Tests are performed on the log odds ratio scale

I have doubts about this: 1-Is it ok to do this test for glmmtmb beta? 2-if it is possible: how to interpret these odd. ratios?, from what I have read I understand that an odd ratio of 1 indicates no change therefore for odds.ratio close to 4 will indicate that it is four times more likely to occur, only those are significant. 3-odd ratio as shown below the output is in logit scale? Thank you very much!!, any ideas, please post... fran


1 Answer 1


Going over each of your questions:

Is it ok to do this test for glmmtmb beta? I've never used this combination myself but most likely yes. glmmTMB fits GLMMs via standard maximum likelihood and the fact that emmeans knows what to do with those objects makes me relatively confident that the estimates are accurate.

How to interpret these odd. ratios? First, recall that odds are a transformation of a risk or probability $p$ via $odds=p/(1-p)$. As probability increases so do odds, but their proportion is not linear. You are correct that an odds ratio > 1 means higher odds (and thereby higher probability) and < 1 is the opposite, but the nuance I want to make is that an odds ratio of 4 is not 4 times the risk, it is higher odds. You need a baseline risk to convert that into a risk ratio, and usually odds ratios are larger than risk ratios (but that depends on the absolute value of the risk involved).

Odd ratio as shown below the output is in logit scale? No, an odds ratio implies it was back-transformed from the log-odds scale. Only inference was generated in that scale, and back-transformed where indicated. In addition you requested marginal means in the response scale, so those are in fact probabilities (or risks).

As a sanity check you can manually calculate some of the parameters yourself: risk at temperatura2 hora4 is $0.575$ (odds $\approx 1.353$), at temperatura20 hora4 it is $0.251$ (odds $\approx 0.335$). Their ratio is $\approx 4.037$ which matches the contrast reasonably well. To restate my above point note that the risk ratio is only $\approx 2.291$.

  • $\begingroup$ Thanks @PBulls, consult how did you calculate the risk ratio?? $\endgroup$ Nov 9, 2023 at 19:33
  • $\begingroup$ By simply dividing the observed responses, i.e. $0.575/0.251$. I'm not sure how you'd go about getting a standard error or confidence interval for this quantity from such model though. $\endgroup$
    – PBulls
    Nov 9, 2023 at 19:38

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