# Capture - Recapture

## Capture - Recapture Revision

**Capture – Recapture**

**Capture-Recapture** is a **statistical** method used to estimate population sizes when it is too difficult to count the entire population.

For example, the number of ants that live in someone’s garden could be estimated. The sampler would **capture** a specific amount of ants, for example 20, and **mark** them with an ethical tag. The ants would then be released, and another time, the sampler would return and **capture** some more ants. In this second sample, the number of **marked** ants would be counted and this information would be used to estimate the overall population.

The more **marked** animals that are **recaptured**, the closer the overall population size is to the **captured** amount.

You will need to know how to estimate a population size using this method.

**Using Capture – Recapture**

The key equation for **capture-recapture** is as follows,

\dfrac{m}{n}=\dfrac{M}{N}

Where

M = size of first sample

N = size of total population

m = the number of previously** tagged** individuals that are** recaptured**

n = size of second sample

Hence, the equation may be written as,

\dfrac{\text{Number recaptured}}{\text{Size of second sample}}=\dfrac{\text{Size of first sample}}{\text{Size of population}}

This can be rearranged to find the size of the total population from the capture-recapture method:

N = \dfrac{n}{m}\times M\\

\text{Size of population}=\dfrac{\text{Size of second sample}}{\text{Number of recaptured}}\times\text{Size of first sample}

**Assumptions**

As with most **statistical** methods, there are certain properties we need to assume in order for the **capture – recapture** method to be as accurate as possible.

These include:

**Marked**and**unmarked**individuals have the same mortality rate (the**marking**does not cause individuals to be more likely to die)- The population is closed (there is no death or migration)
- The population must be given time between the first and second sample for the
**marked**and**unmarked**individuals to fully mix

**Example 1**

Emerson **captures** 20 deer from his local forest and **ethically marks them** with a tag. He then releases the deer back into the forest. A week later, Emerson returns and **captures** 50 deer and counts that 10 of these deer have the **ethical tag** he previously put on them.

Use the equation for **capture-recapture** to work out an estimate for the total deer population in Emerson’s local forest.

**[3 marks]**

We will use this equation,

N = \dfrac{n}{m}\times M\\

So, we need to work out what each of the variables equal,

N (total population) = unknown

n (size of second sample) =50 deer

m (number of **recaptured**) =10 deer

M (size of first sample) =20 deer

Therefore, the total population size can be estimated as:

\dfrac{50}{10}\times 20 = 100 deer

**Example 2**

A summer camp in America has a species of turtle living in their lake. They want to estimate how many turtles live in the lake. The campers **capture** some turtles and **mark** their shells. A day later, they **capture** some more turtles and count how many have **marked** shells.

Their results are summarised in the table below:

Estimate the total population size.

**[3 marks]**

We will use this equation,

N = \dfrac{n}{m}\times M\\

So, we need to work out what each of the variables equal,

N (total population) = unknown

n (size of second sample) =20 turtles

m (number of **recaptured**) =4 turtles

M (size of first sample) =12 turtles

Therefore, the total population size can be estimated as:

\dfrac{20}{4}\times 12 = 60 turtles

## Capture - Recapture Example Questions

**Question 1**: Ellie has a beach at the end of her road. She wants to estimate the number of jellyfish at this beach. One day, she captures 50 jellyfish and marks them ethically. The next day she captures 100 and counts how many have been previously marked.

What is wrong with Ellie’s experiment?

**[2 marks]**

A beach is not a closed population as the jellyfish can easily migrate in and out of this particular beach into the sea.

This means that the capture – recapture method cannot be used as the assumptions aren’t met.

**Question 2**: A nature reserve in Zambia specialises in conserving a lion population. The conservationists are trying to estimate the number of lions on the reserve by using the capture – recapture method.

They capture and mark 64 lions.

A week later, they capture 120 lions, and 48 of these have been marked.

The nature reserve needs to be home to at least 150 lions in order to be called a conversation zone. Work out whether the reserve can be called a conservation zone.

**[3 marks]**

Using the equation,

N = \dfrac{n}{m}\times M\\N = \dfrac{120}{48}\times 64\\

N = 160 lions

Therefore, this nature reserve has enough lions to be called a conservation zone.

**Question 3**: There are 150 otters living in a lake.

A geography class is learning the capture – recapture method. The class go to the lake and capture and mark 30 otters. The next day, the class return and capture 60 otters.

If the class have done the experiment well, work out how many otters they would expect to have recaptured.

**[4 marks]**

As we know N, it will be easier to use this equation,

\dfrac{m}{n}=\dfrac{M}{N}

Where

N = 150\\ M = 30\\ n = 60\\And m is to be found.

The equation can be rearranged to,

m=\dfrac{M}{N}\times n\\m=\dfrac{30}{150}\times 60\\

m=12

The class would expect to recapture 12 marked otters.