I am analyzing data (size/survival) of two groups of fish (Group1, Group2) mixed together in tanks with two different environmental variables applied at two levels to the tanks, e.g. +Predator/-Predator, Low-food/High-food, for a combined total of 4 treatments. There are 2 tanks per treatment and 50 fish per group in each tank to start. Survival and size of fish were measured after 30d in the tank. The way I'd planned on analyzing this was with generalized linear mixed models using the lmer4 package, so that the variables (e.g. size) were modelled as:

lmer (variable ~ FishGroup * PredatorPresence * FoodLevel + (1|Tank/FishGroup)) followed by ANOVA analysis.

However, due to the n=2 tanks the calculated standard deviation on the model predicted values is 0. Is there any way to run this sort of test on an n=2, or a way to do an appropriate analysis on the individual fish rather than the tanks (at least for size). Thanks.

  • $\begingroup$ I would try variable ~ FishGroup * PredatorPresence * FoodLevel + Tank+ (1|Tank:FishGroup) as a start; see here for some of the answers about what to do with zero-variance estimates ... $\endgroup$
    – Ben Bolker
    Apr 22 '16 at 3:19

I presume you have a total of 8 tanks for your Food and predator level combinations. In this case you will need to number the tanks from 1 to 8, as each tank is an individual unit of replication. Your design is somewhat similar to a split-plot design and the tanks represent the whole plots here. I think this is an issue with the tanks being labelled 1 and 2 for each treatment level combination and should be resolved once you relabel the values for the Tank variable

  • $\begingroup$ Thank you, I did have tanks labelled 1-8 and still had standard deviation on predicted values = 0. I will try running this as a split-plot design instead. $\endgroup$
    – RLeg
    Apr 21 '16 at 14:28
  • 1
    $\begingroup$ Can you post a output of your model here. What are the estimated variances ?? $\endgroup$ Apr 21 '16 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.