Estimating the number of fish that return to spawn using capture-recapture methods.
Point Estimate
As a first step, a method must be developed to transforms the sample quantities
into an estimate of the escapement.
Recall what is known from this experiment:
Known Statistics | Unknown Parameters |
n1 = number of fish initially tagged | p1 = proportion of all fish that are tagged |
n2 = number of carcasses examined | p2 = proportion of all carcasses that are searched |
m2 = number of tags recovered from carcasses examined | N = total number of fish returning to spawn (the
escapement) |
We wish to obtain an estimate of N, the total number of spawning fish, which is denoted .
Start by equating the observed ratio in our sample, to an unknown ratio in the population:
m2 number of carcasses with marks number of tags applied n1
-- = ------------------------------- = ----------------------- = --
n2 number of carcasses examined number of spawning fish N
This simple equation depends upon many assumptions. The most
important of these are:
- Tagged fish do not lose their tags.
- In each sample, every animal has the same chance of being captured.
- The population is closed, i.e., no other fish from other rivers
can enter the system, and none of the fish that pass the tagging site
spawn in other rivers.
- The tagged fish mix randomly with the untagged fish between the first
and second samples.
If these assumptions hold, the proportion of carcasses in the entire population that are tagged should be equal to the proportion of all fish tagged.
Finally, the proportion of carcasses that are tagged in the sample should be
roughly equal to the proportion in the population which then gives the left
side of the equation.
The point estimate
Rearranging the terms, we find that
number of carcasses examined x number of tags applied
number of spawning fish = -----------------------------------------------------
number of carcasses with marks
or, equivalently,
= |
n1n2
----
m2
|
To demonstrate that is a sensible estimator of N, we develop the following identities:
n1 = N x p1 | Total population x Proportion of fish tagged. |
n2 = N x p2 | Total population x Proportion of carcasses searched. |
m2 ~ N x p1 x p2 | Total population x Proportion of fish tagged x Proportion of carcasses searched. Note, that unlike the previous identities, this is only approximate because only a sample of carcasses can be searched |
If these identities are substituted into the above equation, we find that
= |
n1n2
----
m2
|
(Np1)(Np2)
= ---------- = N
(Np1p2)
|
Here is an example using actual 1994 data from the Chilko River.
n1 = | 2,213 | number of fish tagged |
n2 = | 59,665 | number of carcasses examined |
m2 = | 505 | number of tags recovered from carcasses examined |
Then
= |
n1n2
----
m2
|
(2,213)(59,665)
= --------------- = 261,462
505
|
i.e., it is estimated that approximately 261,000 fish returned to spawn in the Chilko river.
Questions
- Estimate the number of fish that return to spawn if the following
statistics were observed:
- n1=3,000; n2=60,000; m2=500
- n1=3,000; n2=30,000; m2=250
- n1=6,000; n2=60,000; m2=1,000
- Examine the assumptions made for this estimator.
- Suppose that a proportion of both tagged and untagged fish
"leave" the system. What effect does this have on the point estimate,
i.e. express n2 and m2 in terms of N, p1, p2, and theta
(where theta is the proportion that leave the system) and see
if the estimator still estimates N.
- Suppose that tagging is very stressful, and a certain fraction
die before reaching the spawning ground and the carcasses can never
be found. What effect does this have on the point estimate, i.e.,
express n2 and m2 in terms of N, p1, p2, and the fraction dying of
stress and see if the estimator still estimates N.
- Suppose that new fish enter the spawning ground without having
a chance of being tagged. What effect does this have on the estimator?
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