cebert wrote:Hello once again,
and again thanks for the reply and the advice. This helped me a lot, but I now have one additional question concerning this topic: When using the (unconditional) 'by hand'-calculated CI's for my model averages, what about the corresponding SE? Shall I use the 'unconditional' SE calculated by program MARK in the regular model averaging function or do I have to calculate the SE also 'by hand' in some manner? Or am I again galloping off in some wrong direction? I did not find anything concerning this question in the corresponding section of chapter 14.
If you didn't find it, you might want to look harder next time. Its right in the book (chapter 14), in 2 different places (thus doubling the chances someone would find it - or, in the spirit of things, increasing the latent detection probability by 2 - all other things being equal).
1. p. 33 - first paragraph after the equation at the top of the page:
In fact, MARK handles this calculation of the unconditional variance for you - you would simply
need to take the reported unconditional SE and square it to get the unconditional variance. But - you
need to calculate the CI by hand.
2. p. 34 (about half-way down the page):
Note that if we were to fit these models in MARK, the unconditional SE for the model averaged
abundance would be reported as 26.9045 - if we square this value, we get (26.9045)^2 = 723.901. Again,
the unconditional SE - and thus the variance - reported by MARK is correct (i.e., you do not need to
calculate the SE - or variance - by hand). However, the CI as reported by MARK is not correct - this,
you need to do by hand.
So, to finish this off - the unconditional SE reported by MARK for the SE of the model-averaged abundance is correct. You can use it as is, or square it to get an estimate of the unconditional variance. But, if you want the correct 95% CI for the model-averaged estimate, you need to do it by hand, as discussed at length in Chapter 14.