1
$\begingroup$

I have a question regarding the use of random effects in order to account for a violation of the assumption independent samples. I have this discussion with my supervisor and we disagree about this point. I am wondering what other people think about it:

Imagine an experiment with 3 treatments with 10 replicates for each treatment. So I have 30 units from which I draw samples. From every unit I take a sample; the number of individuals of species X. I do this every day for the duration of one week. So at the end I have 30 times 7 samples. I want to test the effect of the treatment and build a linear mixed effect model. Now comes the question.

I would think that I need to add unit as a random effect because every 7 samples come from the same unit and are thus depending on each other. Each unit might have some random variation, adding unit as random effect would make most sense to me. My supervisor says it would be better to add time as a random effect, because each sample is only replicated at each of the times. But what if you are actually interested in the effect of time and want to add it as a fixed effect, it cannot be random and fixed simultaneously, right?

I guess she is right (that's why she is my supervisor), but I feel that not adding replicate would not take care of the violation of that specific assumption.

Please let me know your thoughts about this.

$\endgroup$
  • 1
    $\begingroup$ 1) What is a replicate? Is it like a "case"/"observation", such as an individual in a survey - a unit of random sample? 2) Where is the place of the treatment factor? Do you have 30 "replicates" after the treatment and another 30 "replicates" as control group. Or what? $\endgroup$ – ttnphns May 25 '14 at 9:33
  • $\begingroup$ Sounds to me like you have 210 replicates, with day of the week as a control variable. $\endgroup$ – Peter Flom May 25 '14 at 10:37
  • $\begingroup$ I have changed the example a bit in the hope to make my point a bit clearer. $\endgroup$ – Robbie May 25 '14 at 13:05

Your Answer

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

Browse other questions tagged or ask your own question.