Open-Population Mark-Recapture Models - Introduction to Ecological Sampling - page 120

is only one group of animals. R code necessary to fit the 36 models described
in Section 8.6 is available on the web site.
One of the examples considered by Lebreton et al . (1992) concerned the
European dipper ( Cinclus cinclus ) in eastern France. In this case, there are
mark-recapture records for 294 birds, consisting of 141 males and 153 females.
There were seven samples taken at yearly intervals, from 1981 to 1987, and
there is the additional information that the survival probabilities for years
1982-1983 and 1983-1984 may have been affected by a major flood.
8.8.1 Constant Probability of Capture
Consider the fitting of the model which posits that the probability of a
marked bird being recaptured was the same for each of the years 2 to 7 when
this was possible, and the probability of surviving from one year to the next
varied with time but was the same for males and females. This is the model
( p constant, ϕ time). It must be stressed that this model is only used here to
illustrate the model-fitting process. It would not be an obvious choice for the
first model to fit to these data at the start of a serious analysis.
For this model, the estimated constant probability of capturing a marked
bird in any year is p = exp(2.220)/{1 + exp(2.220)} = 0.902, with an estimated
standard error of 0.029, while the estimated survival rates for years 1 to 6,
with standard errors in parentheses, are
ϕ 1 = exp(0.336 + 0.178)/{1 + exp(0.336 + 0.178)} = 0.626 (0.113)
ϕ 2 = exp(0.336 - 0.520)/{1 + exp(0.336 - 0.520)} = 0.454 (0.067)
ϕ 3 = exp(0.336 - 0.423)/{1 + exp(0.336 - 0.423)} = 0.478 (0.059)
ϕ 4 = exp(0.336 + 0.172)/{1 + exp(0.336 + 0.172)} = 0.624 (0.058)
ϕ 5 = exp(0.336 + 0.102)/{1 + exp(0.336 + 0.102)} = 0.608 (0.055)
and
ϕ 6 = exp(0.336)/{1 + exp(0.336)} = 0.583 (0.058).
These estimates relate to the survival from 1981 to 1982, 1982 to 1983, and so
on up to the survival from 1986 to 1987.
There are seven estimated parameters in this model. The maximized log-
likelihood is −329.87, and the AIC of Equation (8.25) is 673.73. TEST 2 and
TEST 3 give an overall statistic of 11.76 with 10 df, which is not at all signifi-
cantly large. There is therefore no evidence of lack of fit of the basic JS model
for these data.