Capture–recapture and sampling assumptions

(a) 250

1. The proportion tagged in the sample estimates the proportion in the lake.
2. $$\frac{8}{50} = \frac{40}{N} \;\Rightarrow\; N = \frac{40 \times 50}{8} = 250$$

• M1 $\frac{40 \times 50}{8}$ or an equivalent proportion.
• A1 Correct answer: 250.

(a) The population does not change between the samples, and every rabbit is equally likely to be caught (tagged rabbits mix back in).

1. No rabbits are born, die, arrive or leave between the two catches.
2. Tagged rabbits mix evenly and are as likely to be caught as untagged ones; tags are not lost.

• C1 The population is closed (no births, deaths or migration) between samples.
• C1 Each rabbit is equally likely to be caught (tags not lost, tagged rabbits mix fully).

(b) Fewer tagged rabbits are recognised, so the estimate is too high.

1. $N = \frac{\text{tagged} \times \text{sample}}{\text{tagged in sample}}$: a smaller denominator makes $N$ larger.

• C1 The estimate would be too large, because fewer tagged animals are counted in the second sample.

(a) $171$

1. Equate the marked proportion in the second catch to that in the population.
   $$7/40=30/N$$
2. Rearrange the proportional equation.
   $$30\times 40/7$$
3. Therefore $171$.

• P1 Equate the marked proportion in the second catch to that in the population.
• P1 Rearrange the proportional equation.
• A1 Correct answer: $171$

(b) The population stays approximately closed between catches, and marked fish mix fully and are as likely to be caught as unmarked fish.

1. The population stays approximately closed between catches, and marked fish mix fully and are as likely to be caught as unmarked fish.

• C1 Correct conclusion with supporting reasoning: The population stays approximately closed between catches, and marked fish mix fully and are as likely to be caught as unmarked fish.

(a) $197$

1. Scale the second sample by the reciprocal of its marked proportion.
   $$35/(8/45)$$
2. Therefore $197$.

• M1 Scale the second sample by the reciprocal of its marked proportion.
• A1 Correct answer: $197$

(b) The recorded marked recapture count becomes too low. Dividing by this smaller count makes the population estimate too high.

1. The recorded marked recapture count becomes too low. Dividing by this smaller count makes the population estimate too high.

• C1 Correct conclusion with supporting reasoning: The recorded marked recapture count becomes too low. Dividing by this smaller count makes the population estimate too high.

(a) $222$

1. Use marked first sample times second sample divided by marked recaptures.
   $$40\times 50/9$$
2. Therefore $222$.

• M1 Use marked first sample times second sample divided by marked recaptures.
• A1 Correct answer: $222$

(b) $\frac{9}{2}$

1. The estimate varies inversely with the marked recapture count.
   $$9/2$$
2. Therefore $\frac{9}{2}$.

• M1 The estimate varies inversely with the marked recapture count.
• A1 Correct answer: $\frac{9}{2}$

(c) With fixed sample sizes, no actual integer recapture count could give exactly that proposed doubled estimate.

1. With fixed sample sizes, no actual integer recapture count could give exactly that proposed doubled estimate.

• C1 Correct conclusion with supporting reasoning: With fixed sample sizes, no actual integer recapture count could give exactly that proposed doubled estimate.

(a) $60$

1. Rearrange population = (marked × second sample) ÷ marked in the second sample.
   $$300\times 12/60$$
2. Therefore $60$.

• M1 Rearrange population = (marked × second sample) ÷ marked in the second sample.
• A1 Correct answer: $60$

(b) Some marks may have been lost, or marked squirrels may avoid the traps, so too few marked squirrels are recaptured.

1. Some marks may have been lost, or marked squirrels may avoid the traps, so too few marked squirrels are recaptured.

• C1 Correct conclusion with supporting reasoning: Some marks may have been lost, or marked squirrels may avoid the traps, so too few marked squirrels are recaptured.

(a) $250$

1. Multiply the number marked by the second sample size and divide by the marked recaptures.
   $$50\times 40/8$$
2. Therefore $250$.

• M1 Multiply the number marked by the second sample size and divide by the marked recaptures.
• A1 Correct answer: $250$

(b) $250$

1. Use the same method with Group 2’s sample.
   $$50\times 100/20$$
2. Therefore $250$.

• M1 Use the same method with Group 2’s sample.
• A1 Correct answer: $250$

(c) Group 2, because its second sample is larger, so chance variation has less effect.

1. Group 2, because its second sample is larger, so chance variation has less effect.

• C1 Correct conclusion with supporting reasoning: Group 2, because its second sample is larger, so chance variation has less effect.

(a) 440

1. Only $60 - 5 = 55$ ringed birds were alive to be caught.
2. $$N = \frac{55 \times 72}{9} = 440$$

• P1 Using 55 ringed birds.
• P1 $\frac{55 \times 72}{9}$.
• A1 Correct answer: 440.

(b) It assumes more ringed birds were available to catch than there really were.

1. With 60, the ringed proportion in the population looks smaller than it is, so $N$ comes out larger: 480 instead of 440.

• C1 Using 60 overstates the ringed birds available, making $N$ too large.