I'm analysing chick survival between 3 different years using a glm with quasibinomial error structure. Hence, my response variable is a cbind of fledged chicks and dead chicks, and one of my explanatory variables is Year (2013,2014,2015). After finding out that Year has a significant effect, I wanted to know how chick survival changed between the years according to my model.

So I ran a 'glht' with Tukey:


and got this:

Linear Hypotheses:
                  Estimate Std. Error z value Pr(>|z|)  
2014 - 2013 == 0 0.6131290    0.2421515   2.532   0.0304 *
2015 - 2013 == 0 0.6139173  0.2450897   2.505   0.0327 *
2015 - 2014 == 0 0.0007884  0.2324065   0.003   1.0000  
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- single-step method)

After that I transformed the logg odds to proportions:


and I got this:

2014 - 2013 2015 - 2013 2015 - 2014 
  0.6486542   0.6488339   0.5001971

Does this mean that in 2013 64% more chicks survived? According to my raw data this can't be true. You can see the mean proportions of fledged/hatched chicks for 2013, 2014 and 2015 here:

> mean((SurvivalData$Fledglings/SurvivalData$Hatchlings)[SurvivalData$Year=="2013"])
    [1] 0.6028452
    > mean((SurvivalData$Fledglings/SurvivalData$Hatchlings)[SurvivalData$Year=="2014"])
[1] 0.6393909
> mean((SurvivalData$Fledglings/SurvivalData$Hatchlings)[SurvivalData$Year=="2015"])
[1] 0.7186566

What did I do wrong or what did I miss?

Thanks a lot in advance!

  • 1
    $\begingroup$ The estimates from glht are log odds ratios because it gives you differences between logits. Your transformation is wrong, because it works for logits but not for log odds ratios. Write down the maths and you should see that. $\endgroup$
    – Roland
    Commented Jul 16, 2015 at 15:37
  • $\begingroup$ Asking for help with interpretations of results from a statistical model isn't really a programming questions and as such isn't appropriate for Stack Overflow. if you need help with statistical questions, you should be asking over on Cross Validated instead. $\endgroup$
    – MrFlick
    Commented Jul 16, 2015 at 16:03
  • $\begingroup$ Thanks Roland, I looked into it again and tried to widen my mathematical knowledge with the help of the internet. :) My conclusion is that the first transformation I made 1/(1+1/exp(coef(summary(SurvivalYear)))) should already give me a proportion. But this makes the result even weirder. Forgive me if I'm being too stupid here.... $\endgroup$ Commented Jul 16, 2015 at 16:19
  • $\begingroup$ Thanks, MrFlick for 'migrating' my question. I did not know better... $\endgroup$ Commented Jul 16, 2015 at 16:19
  • $\begingroup$ I edited the code in my question, but I'm still not getting to an correct answer.... can anybody help? $\endgroup$ Commented Jul 16, 2015 at 18:11

1 Answer 1


Putting my comment into an answer:

The estimates from glht are log odds ratios because they give you differences between logits. You can see this easily, if you write down the maths:

$\ln{\frac{p_{2014}}{1-p_{2014}}} - \ln{\frac{p_{2013}}{1-p_{2013}}} = \ln{\frac{p_{2014}(1-p_{2013})}{p_{2013}(1-p_{2014})}}$

The estimate of 0.6131 means that the odds (i.e., $\frac{p}{1-p}$) of a chick surviving in 2014 were almost twice as high as in 2013: $\exp(0.6131) = 1.846146$

(Assuming I understand correctly, how you specified the dependent. Possible those are the odds that they died instead.)

  • $\begingroup$ Thanks for this explanation, but this is exactly the math I did. Turning the odds 1.846146 into a probability (or proportion) gives me 0.6486542 (as seen in the output above). I assumed that this couldn't be right because looking at the mean survival rates there is only a difference of roughly 4-8% between the years. How can the model then say that the odds of surviving in 2014 was twice as high as in 2013? Sorry, I'm very very new to the world of R (and glms in general), so I might be misunderstanding some concepts. Danke für deine Geduld! ;) $\endgroup$ Commented Jul 16, 2015 at 20:05
  • $\begingroup$ Again: 1.846 is not an odds value. It's a ratio of odds. $\endgroup$
    – Roland
    Commented Jul 17, 2015 at 6:59
  • $\begingroup$ I see. So is there any way to get the information I need, i.e. "in 2013 x% more chicks survived compared to 2014"? $\endgroup$ Commented Jul 17, 2015 at 10:36
  • $\begingroup$ No, you can say only "the ratio of surviving to dying [or dying to surving, it's not clear to me how you defined the odds] chicks in 2014 was 84.6 % higher than in 2013". Btw., it would be better to not only give the estimate, but also a confidence interval. $\endgroup$
    – Roland
    Commented Jul 17, 2015 at 10:44

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