Biology · Introductory biology · Worked example
Estimate population size with mark–recapture
Biologists mark 60 turtles in a pond and release them. Two weeks later they catch 75 turtles, 15 of them marked. Estimate the population with the Lincoln–Petersen and Chapman formulas, and say how precise the estimate is.
Set up the proportion
Marked turtles are 15 of the 75 caught, one fifth. If the same fraction of the whole population is marked, the 60 marked turtles are one fifth of N.
Apply the Lincoln–Petersen formula
The same answer comes straight from N̂ = MC/R.
Apply Chapman’s correction
Chapman’s formula gives a slightly smaller, less biased estimate.
Gauge the uncertainty
For Chapman’s estimate the standard error is about 53.6 turtles, so the approximate 95% interval is 288.75 ± 1.96 × 53.6, about 184 to 394 turtles. Fifteen recaptures pin the population down only roughly.
Check the assumptions
The estimate assumes that no turtles were born, died or moved in or out during the two weeks, that marks stayed visible, that marked turtles mixed back in, and that every turtle was equally likely to be caught.
Result
About 300 turtles by Lincoln–Petersen, or 289 by Chapman’s correction; the approximate 95% interval runs from about 184 to 394.
Your turn
A biologist marks 120 fish; a later catch of 100 fish includes 8 marked ones. Estimate the population with the Lincoln–Petersen formula.
Show the answer and explanation
About 1,500 fish; Chapman’s correction gives about 1,357.
N̂ = MC/R = 120 × 100/8 = 1,500. With only 8 recaptures the estimate is very uncertain.
Keep exploring
The mark–recapture tool opens with these counts and shows the Chapman estimate, its standard error and the approximate interval. It lists each sampling assumption until you confirm it.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
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