Trap-Happy Lizards and a Biased Lincoln–Petersen Estimate
On a 4-ha island, conservation biologists study a lizard population. In week 1 they capture 150 adults, mark them with unique toe clips, and release them. In week 2 they capture 120 adults; 72 are marked. A pilot study found that marked lizards are 1.8 times more likely to re-enter baited traps than unmarked lizards. Trapping is closed-no births, deaths, or migration during the week-and toe clipping does not affect survival. The same 20 baited trap sites are used in both weeks. The team needs a population estimate for a permit application and must decide whether the naive Lincoln–Petersen estimate is acceptable. They also have limited funds for a third capture session and want to know whether a different design would reduce bias.
1. Calculate the naive Lincoln–Petersen population estimate and state its direction of bias.
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The naive Lincoln–Petersen estimate is N = (M × C) / R, where M is the number marked in week 1, C is the number captured in week 2, and R is the number of marked animals in week 2. Here N = (150 × 120) / 72 = 18,000 / 72 = 250 lizards. This estimate assumes that marked and unmarked animals have equal capture probabilities in week 2. The pilot study shows that marked lizards are 1.8 times more likely to be recaptured, so marked animals are overrepresented in the second sample. R is therefore too high, and because R is in the denominator, the naive estimate is biased downward. The true population is likely larger than 250; the naive figure should not be submitted as an unbiased estimate without correction or qualification.
2. Using the 1.8-fold trap-happy ratio, derive a bias-corrected estimate and compare it with the naive estimate.
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Let p be the capture probability of an unmarked lizard in week 2. A marked lizard has capture probability 1.8p. If the true population is N and M = 150 are marked, the expected number captured is C = p[(N − M) + 1.8M] = p(N + 0.8M). The expected number of marked captures is R = 1.8pM. The observed marked fraction is R/C = 72/120 = 0.6. Therefore 0.6 = 1.8M / (N + 0.8M). With M = 150, 0.6 = 270 / (N + 120). Solving: 0.6(N + 120) = 270; N + 120 = 450; N = 330. The bias-corrected estimate is about 330 lizards, compared with the naive 250. This correction depends on the 1.8 ratio being accurate and on the closed-population assumptions; it should be presented with a range or confidence interval rather than as an exact count.
3. Recommend a sampling design that would reduce this bias without relying on a single recapture session.
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A single baited-trap recapture session cannot separate trap response from true capture probability. A better design would use a robust or Pollock-type design with multiple secondary capture occasions within primary periods, so that capture probability and temporary emigration can be estimated. Traps should be mixed with unbaited traps or alternate trap types, and trap sites should be rotated to reduce learned trap attraction. Camera-based mark–resight, if individuals can be identified, provides an independent capture method and can be analyzed with mark–resight models. If only one more session is affordable, the team should at least randomize trap placement, use the same effort, and record whether marked animals are approaching traps; the estimate should be reported as a minimum or with an explicit trap-response sensitivity analysis rather than as a definitive population size.