A unit of Lasting Forests |
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Such estimates are:
Procedures:
| A powerful estimator is available for population abundance using estimates of animal home range. One or two animals per home-range area should approximate the total animals within an area and contribute to convergence on a useful population estimate.(These equations are based on data from many studies in many areas. The high R-square values are impressive.) |
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Where H = home range in hectares and W is weight in gram: H = 0.002 W1.02 herbivores H = 0.059W0.92 omnivores H = 0.11W1.36 carnivores And taken together, H = 0.008W1.08 with r2 = 0.6 H = f (1/precipitation, latitude) H = f(sex, age) related to H = f(weight) Food consumption/day in grams Carnivores Where FCc is food consumption by carnivores then: FCc = 1.7W0.68 ± 0.02 approximately FCc = 1.7 W0.66 grams Birds FCb = 0.51W0.63 ± 0.11 approximately FCb = 0.5 W0.66 |
| From: Harestad, A. S. and F. L. Bunnell. 1979. Home range and body weight - a reevaluation. Ecology 60(2):389-402. |
| Concept: If you see more on some days, how long will it take to see them all? |
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Observations: 100, 98, 118, 112, 113, 62, 105 Select largest number A = 118 Select next largest B = 113 N = 2A-B = (2 x 118) - 113 = 236 - 113 = 123 NL = 118 (at least these were all seen) NU = A + ((1.0 - a / a) (A - B)) NU = 118 + )(1 - 1.05)/0.05)(118 - 113) NU = 118 + (19)(5) = 213 |
| Premise: total marked animals (M) are to the unknown number (N) of animals, as the marked animals (m) are to the total animals in a sample. Thus: N = Mn/m |
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The standard error is ([M2n(n-m)] / m3)0.5 Where M = 27, n = 23, m = 17; then SE = 4.5 Since N = 36.5, the upper limit of the estimate is (36.5 + 2(4.5)) = 45.4 and the lower limit (2 standard errors) is (36.5 - 9.0) = 27.5 The population, 95 times out of 100, will lie between 27 and 46 animals. |
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N = (M(n+1) / (m +1)) - 1 SE = ((M2 (n+1)(n-m)) / ((M + 1)2 (m + 2)))0.5 |
| Example: Pheasants |
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100 cock "crows" per hour before the hunt or in an early survey 3 removed Therefore: 70 crows heard later (30 "lost") X = 10 There were 10 pheasant cocks originally. Similar analyses can be made with track counts after deer hunts. |
Resisting the desire or pressure to estimate populations is counterintuitive for the person who gets into the field of faunal resource management because of their love of animals or desire to "work with" animals. Estimating abundance needs to be resisted. Failing that, converge on an estimate.
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Last revision January 17, 2000.