Capture-Recapture

Match the marked proportion in the sample to the marked proportion in the whole population.

1. The marked proportion in the later sample is \frac4{20}=\frac15.
2. Assume the same proportion of the population is marked. The 12 marked beetles represent one fifth of the population.
3. The estimated population is 12\times5=60 beetles.

Consider whether the two marked counts refer to separate groups of counters.

1. The 6 recaptured marked counters belong to the original group of 18 marked counters. They are not a new group to add.
2. Capture-recapture uses the marked proportion in the later sample to estimate the whole population, including counters never sampled.

Divide the marked count by the total sample size.

1. The proportion is \frac8{40}=\frac15, which is also 0.2.

The initially marked count is compared with the marked count in the sample.

1. Let the estimated population be N. The marked proportions give \frac{24}{N}=\frac6{45}.
2. Rearrange to N=\frac{24\times45}{6}=180.
3. The population estimate is 180 moths.

Think about how many different newts the researcher has definitely seen.

1. The researcher marked 70 different newts, so the pond holds at least 70 newts. An estimate of 60 is impossible.
2. (The capture-recapture estimate is \frac{70\times40}{16}=175 newts.) Adding the two catches counts the marked newts twice, so 110 is not the method.

Calculate the capture-recapture estimate before rounding.

1. Assume the fraction of the second sample that is marked equals the fraction of the whole population that is marked: \frac{12}{50}=\frac{35}{N}.
2. Rearrange: N=\frac{35\times50}{12}=\frac{1750}{12}=145.83\ldots
3. To the nearest whole animal, the estimate is 146.

Rearrange the proportion equation to make the initially marked number the unknown.

1. Let M be the initially marked number. Then \frac{M}{200}=\frac8{50}.
2. Therefore M=200\times\frac8{50}=32.
3. There were 32 animals marked initially.

Use the population estimate to find the estimated marked fraction.

1. The estimated marked fraction is \frac{45}{300}=\frac3{20}.
2. If the later sample size is S, then \frac6S=\frac3{20}.
3. Cross-multiplying gives 3S=120, so S=40.

Multiply the sample size by the estimated marked proportion.

1. The estimated marked proportion is \frac{60}{480}=\frac18.
2. The expected marked count in a sample of 80 is 80\times\frac18=10.

The sample should represent marked and unmarked animals fairly.

1. The marked proportion in a representative sample can estimate the marked proportion in the population.
2. This requires marked and unmarked animals to have equal capture chances. Catching all animals or changing the population is not required.

Decide whether the marked proportion in the later sample is too high or too low.

1. Marked animals are over-represented, so the later sample has an unusually large marked proportion.
2. The usual method treats this proportion as applying to the whole population.
3. This makes the estimated population too small.

Consider what happens to the recaptured-marked count in the denominator.

1. Lost marks make the observed marked count in the later sample too small.
2. The formula still uses the larger original marked total, so dividing by this smaller recapture count inflates the estimate.
3. The estimate is likely to be too large.

Consider whether greater reliability guarantees an exact count.

1. Even a random sample may have a marked fraction that differs from the population fraction.
2. A larger representative sample generally reduces this random variation, but does not guarantee an exact estimate.

Work out how many lizards in the sample are marked first.

1. 15\% of 80 is 0.15\times80=12 marked lizards in the sample.
2. Assume the marked fraction in the sample equals the marked fraction in the population: \frac{12}{80}=\frac{30}{N}.
3. Rearrange: N=\frac{30\times80}{12}=200 lizards.
4. (Or: 15\% of the population is marked, so 30 is 15\% of N and N=30\div0.15=200.)

Add both sample totals and both marked counts before applying the method.

1. The combined sample contains 20+30=50 animals and 3+7=10 marked animals.
2. The marked sample fraction is \frac{10}{50}=\frac15.
3. The population estimate is 30\times5=150.

Estimate the whole population first, then distinguish it from the unmarked part.

Part (a)

1. Use \frac{48}{N}=\frac{16}{72}.
2. The estimate is N=\frac{48\times72}{16}=216 animals.

Part (b)

1. The whole population contains the 48 marked animals, not just the 16 recaptured ones.
2. Subtract the whole marked group: 216-48=168 unmarked animals.

Find the required marked fraction in the later sample, then apply it to the estimated whole population.

1. The required marked fraction is \frac3{24}=\frac18.
2. The same fraction should be marked in the estimated population of 96.
3. The researcher should mark 96\times\frac18=12 animals.

The sample size includes both the marked and the unmarked animals.

1. The later sample contains 25+75=100 animals.
2. The marked fraction is \frac{25}{100}=\frac14.
3. The initial 60 marked animals represent an estimated quarter of the population, giving 60\times4=240.

How many marked crabs are still on the beach when the second sample is taken?

1. The method compares the marked crabs on the beach with the marked fraction in the sample. Only 50-10=40 marked crabs are still on the beach.
2. Marked fraction in the sample: \frac{8}{60}. So \frac{40}{N}=\frac{8}{60}.
3. Rearrange: N=\frac{40\times60}{8}=300 crabs.
4. Using 50 instead of 40 would give 375, which is too high, because 10 of those marked crabs could not be caught.

For Wednesday, work out how many birds in the population carry a mark by then.

Part (a)

1. On Tuesday 30 birds in the population are marked, and \frac{6}{40} of the catch is marked. So \frac{30}{N}=\frac{6}{40}.
2. N=\frac{30\times40}{6}=200 birds.

Part (b)

1. On Tuesday 40-6=34 unmarked birds were caught and marked, so by Wednesday 30+34=64 birds are marked.
2. On Wednesday \frac{12}{45} of the catch is marked, so \frac{64}{N}=\frac{12}{45}.
3. N=\frac{64\times45}{12}=240 birds.
4. (Using only the 30 birds marked on Monday would wrongly give \frac{30\times45}{12}=112.5.)