Capture-Recapture

## Capture-Recapture

If the size of the population is not know, it can be estimated using the capture-recapture method.

A number of items are taken from the population and marked or tagged in some way. The number marked or tagged is set as N, and the items returned to the population.

A second sample is taken of sample size M and the number of marked or tagged items identified as m.

The proportion of the originally marked items to the proportion of marked items in the second capture are assumed to be the same, to give us te estimate. Writing these as fractions:

frac(text(originally marked))(text(total population)) = frac(text(retrieved marked))(text(second sample size))

or

frac(text(n))(text(N)) = frac(text(m))(text(M))

where

n is marked number on first sample

N is total population

m is number marked on second sample

M is total number retrieved on second sample.

## Example 1

A biologist is undertaking a study of ants nests. The biologist retrieves 100 ants and marks them up, and releases them back into the nest. A second sample is taken. There are 80 ants in the second sample, of which 6 are marked.

The biologists says that there are exactly 1,333 ants in the nest. Explain why the biologist is wrong, and why.

Make sure that you explain why the biologist is wrong.

Answer: The capture-recapture method provides an estimate, and is based on the fact that the proportion of marked ants in the second sample represents the proportion in the entire population.

## Example 2

On a remote Scottish Island, a colony of gulls is being monitored. Last year, 68 gulls were ringed (small rings around their legs). This year, 72 gulls were captured and released, of which 18 had rings from the previous year. What is the estimated population of the colony of gulls, assuming that the population is the same this year as it was last year?

Estimated population = frac(text(n))(text(N)) = frac(text(m))(text(M))

frac(68)(text(N)) = frac(18)(72)

Rearranging and solving:

N = frac((68 xx 72))(18) = 272