QUESTION 122 (5 marks)
A capture–recapture estimate uses M = 50, n = 80 and m = 16. An independent complete census later shows that the population was 500 during both capture sessions. All marks remained visible. Marked animals had been released beside baited traps and returned to those traps more often than unmarked animals.
a) Calculate the Lincoln estimate and compare it numerically with the census. [2 marks]
b) Explain how the stated behaviour accounts for the direction of the error. [2 marks]
c) Describe one change to the procedure that would help address this bias. [1 mark]
Practice marking scheme
Answer
The Lincoln estimate is 250, half the census. Elevated recapture probability increases m and biases the estimate downward.
Working
N = 50×80/16 = 250. The difference is 250 animals below the census. Attraction to release-site traps makes marked animals overrepresented in the recapture sample; a larger denominator lowers the estimate. Adequate mixing and suitable release/sampling locations could reduce this bias.
Marking criteria
- Calculates 250 animals. [1 mark]
- States 250 below, or 50% of, the census. [1 mark]
- Explains overrepresentation of marked animals in m. [1 mark]
- Links increased m to a smaller estimate. [1 mark]
- Suggests adequate mixing or a justified release/trap-location change. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.