Mark–Recapture Estimate with Dye Loss and Age Bias
An ecology club wants to estimate the number of freshwater shrimp in a small pond. On Monday they catch 60 shrimp, mark each with a spot of nontoxic dye, and release them. On Tuesday they catch 80 shrimp and find that 12 carry the mark. The club uses the simple mark–recapture estimate N = (marked released × second sample) / marked recaptured. Before publishing, a teacher notes that a pilot trial showed the dye becomes hard to see on about 20% of marked shrimp within 24 hours, and that shrimp may learn to enter the traps for food. The pond is closed, with no immigration or emigration during the two days. The club also recorded that the Tuesday sample contained many more juveniles than the Monday sample. They want a defensible estimate and a statement of uncertainty, not just a single number.
1. Calculate the uncorrected estimate and the corrected estimate accounting for 20% mark loss, stating the assumptions.
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Uncorrected: N = 60 × 80/12 = 400.
The 20% mark loss means only 80% of marked shrimp in the population are recognisable. If 12 recognisable marked shrimp are caught, the number of actually marked shrimp in the sample is about 12/0.80 = 15. Equivalently, the recognisable marked population is 0.80 × 60 = 48, so N = 48 × 80/12 = 320. The corrected point estimate is 320.
This assumes: the population is closed; marked and unmarked shrimp are equally likely to be caught; marks are lost randomly and independently of capture; the dye loss rate is the same in the pond as in the pilot; and the 80% retention applies over the whole interval. If any of these fail, 320 is biased.
2. The Tuesday sample has more juveniles. Does this invalidate the estimate? Explain with reference to the assumptions, and suggest how to improve sampling.
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More juveniles on Tuesday is a warning, not automatically fatal. The estimate requires that the marked proportion in the Tuesday sample represents the whole population. If juveniles are more catchable than adults, or if the Monday marked sample was not representative by age, the sample is biased. For example, if marked adults are less likely to be recaught than unmarked juveniles, the recapture fraction is too low and N is overestimated; the reverse causes underestimation. Also, the change in age composition between days suggests the trap or the shrimp behaviour changed.
Improvements: use several recapture days, stratify by size/age, randomise trap placement, use identical trapping effort, and check whether marked and unmarked shrimp differ in catchability. Report separate estimates for juveniles and adults if possible, rather than one pooled number.
3. Another student says: because the corrected estimate is 320, the true number is 320. Evaluate this statement.
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320 is a point estimate under a model, not a census. The recapture count is only 12 (or 15 after correction), so random sampling error is substantial. A rough relative standard error for a mark–recapture estimate is about 1/√m, where m is the number of marked recaptures; with m ≈ 12, that is about 29%, so an approximate 95% interval is roughly 320 ± 180 (very rough). Systematic errors from mark loss, trap behaviour and age selectivity may be larger than the sampling error.
A defensible statement would be: our best estimate is about 320 shrimp, but the true number is uncertain; repeat sampling and tests of equal catchability are needed before using this for management. The word true implies an exact count that this method cannot provide.