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Estimating Abundance Weight Sub-samplePowerPoint Presentation

Estimating Abundance Weight Sub-sample

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Estimating Abundance Weight Sub-sample

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- Used to estimate total number in a sample
- Method:
- Weigh a known number of individuals to get a mean weight
- Weigh the entire sample, then divide the total weight by mean weight to get total number of individuals.

- Example:
- 10 individuals weigh 68g, so mean weight = 68 / 10 = 6.8
- Total weight = 528, so total number = 528 / 6.8 = 77.6 individuals

- Important tool for estimating density, birth rate, and death rate for mobile animals.
- Method:
- Collect a sample of individuals, mark them, and then release them
- After a period, collect more individuals from the wild and count the number that have marks
- We assume that a sample, if random, will contain the same proportion of marked individuals as the population does
- Estimate population density

Marked animals in second sample (R)

Marked animals in first sample (M)

Total caught in second sample (C)

Total population size (N)

5

16

20

N

=

N = 64

=

= Proportions

- Marked and unmarked animals are captured randomly.
- Marked animals are subject to the same mortality rates as unmarked animals. The Peterson method assumes no mortality during the sampling period.
- Marked animals are neither lost or overlooked.

A fish biologist goes out and samples (sample 1) a population of trout. A total of 109 (M) trout were marked and released. At this time, a proportion of fish in the total population has a mark and we assume that this proportion remains constant. On a second sampling trip (sample 2), the biologist collected 177 (C) trout and 57 (R) of those were marked from the initial sample. How large is the population (N)?

R/C = M/N N=MC/R N = (109)(177)/57 N=338

Si Zi

Mi=

+ mi

Ri

Mi = Marked population size at time i

mi = Marked animals actually caught at time i

Ci =Total number of animals caught at time I Si / Ri = proportioncaught

Si = Total animals released at time i

Zi = Number of individuals marked before time i, not caught in the ith sample but caught in a sample after time i

Ri = Number of the Si individuals released at time i that are caught in a later sample

Estimate Pop. Size at Time 3

Si Zi

Mi=

+ mi

Ri

M3 = {(164)(39)/(54)} + 37 = 155.4

S3 = 164

C3 = 169

Z3 = 5+2+2+18+8+4 = 39

R3 = 33+13+8 = 54

m3 = 37

Population Estimation:

N3=M3C3/m3 = (155.4)(169)/37 = 710

Estimate Pop. Size at Time 4

Si Zi

Mi=

+ mi

Ri

M4 = {(202)(37)/(50)} + 56 = 205.5

S4 = 202

C4 = 209

Z4 = 2+2+8+4+13+8 = 37

R4 = 30+20 = 50

m4 = 56

Population Estimation:

N4=M4C4/m4 = (205.5)(209)/56 = 767

Estimating mortality:

We can compare the estimated number of marks in the wild versus a known amount to get mortality rates. For example, in year 3 we estimated that there were 155 marked individuals. We released a total of 132 newly marked individuals, for a total of 287 marked individuals. We estimated the number of marked individuals to be 206 for year 4. 206 is less than 287, so the survival rate is 206/287=0.718. Mortality is then 1-0.718=0.282.

Survival3 = Number Marks Estimated3 / Total Marks Released

Survival3 = 206 / 287 = 0.718

Mortality3 = 1 – Survival3 = 1 – 0.718 = 0.282

Estimating Natality:

We estimated that our mortality from year 3 4 was 28%, but our population estimation increased by 57 individuals from 710 to 767. Given an initial population of 710 (N3) and mortality of 28%, we should only have 511 (710-0.28*710) individuals in the population for year 4. However, our estimated population size in year 4 is 767, 256 more individuals than 511. So, we had 256 individuals added to the population!

N3 = 710, 28% Mortality3 = (710)(0.28) = 199

Expected N4 = 710 – 199 = 511

Estimated N4 = 767

Therefore, 767 – 511 = 256 new individuals

- Individuals evenly spread over a known area
- Use a known area quadrant to sample
- Determine the mean number per square area
- Multiply times total area to get total number of individuals