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Fish Population Assessment
Fish Population Assessment
How many fish do we have?How many fish do we have?
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Fish Population Assessment
Fish Population Assessment
Estimating population size
Estimating population size
1) Plot method 2) Mark and recapture
(Peterson method) 3) Mark and recapture
(Schnabel method) 4) Change in ratio or
dichotomy method 5) Removal sampling
(Zippin method)
1) Plot method 2) Mark and recapture
(Peterson method) 3) Mark and recapture
(Schnabel method) 4) Change in ratio or
dichotomy method 5) Removal sampling
(Zippin method)
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Plot MethodPlot Method
€
ˆ N =A
a× n
€
A =
a =
n =
ˆ N =
Total population area
Size of the plot
Average number of fish per plot
Population estimate
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Plot Method - Estimated Variance
Plot Method - Estimated Variance
€
V ( ˆ N ) =A2
a×
V (n)
s×
A − (sa)
A
V (n) =(ni − n )2
s −1∑
€
ni =
s =
Number of fish counted in ith plot
Number of plots used
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Plot Method - 95% confidence interval
Plot Method - 95% confidence interval
€
ˆ N ± t V ( ˆ N )
t for s-1 df, p=0.05
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Plot Method -Example
Plot Method -Example
Pond area = 100 m2
Size of plot = 1 m2
Average number of fish per plot = 1.5
Pond area = 100 m2
Size of plot = 1 m2
Average number of fish per plot = 1.5
€
100m2
1m2 ×1.5 =150
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Mark and Recapture - Peterson Method (single)
Mark and Recapture - Peterson Method (single)
€
ˆ N =M(C +1)
(R +1)
M =
C =
R =
Number of fish initially marked & released
Number of fish collected/examined in 2nd period
Number of recaptures found in C
€
R
C=
M
N
N =MC
RBailey modification
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Mark and Recapture - Variance
Mark and Recapture - Variance
€
V ( ˆ N ) =M 2(C +1)(C − R)
(R +1)2(R + 2)
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Mark and Recapture - 95% confidence intervalMark and Recapture -
95% confidence interval
€
ˆ N ±1.96 V ( ˆ N )
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Mark and Recapture - Example
Mark and Recapture - Example
M = 550 C = 500 R = 157
€
ˆ N =550(500 +1)
(157 +1)=1,744
€
V ( ˆ N ) =5502(501)(500 −157)
(157 +1)=13,096
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Mark and Recapture - Example
Mark and Recapture - Example
M = 550 C = 500 R = 157
€
ˆ N ±1.96 13,096 =1,744 ± 224
P(1,520 ≤ ˆ N ≤1,968) = 0.95
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Mark and Recapture - Schnabel Method
Mark and Recapture - Schnabel Method
Multiple episodes of mark and recapture
€
ˆ N =CM∑R∑
CM = total captures X marked fish available for recapture
R = recaptures of marked fish
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Schnabel Method - Variance & 95% C.I.Schnabel Method - Variance & 95% C.I.
€
V (1ˆ N
) =R∑
( CM)2∑
€
1ˆ N
±1.96 V (1ˆ N
) Then invert for 95% C.I. for N
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Schnabel Method - example p. 137 (2nd ed.)
Schnabel Method - example p. 137 (2nd ed.)
Period R Unmarked
TotalC
M CM
1 0 150 150 0 0
2 22 203 225 150 33,750
3 26 86 112 353 39,536
…. …. …. …. 439 ….
Total 254 457,208
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Schnabel Method - example p. 137 (2nd ed.)
Schnabel Method - example p. 137 (2nd ed.)
€
ˆ N =CM∑R∑
=457,208
254=1,800
95% C.I. = 1,602 - 2,049
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
Requirements:
1) two recognizable classes Species SexesAdults vs. juvenilesAge classes
2) different rates of exploitation
Requirements:
1) two recognizable classes Species SexesAdults vs. juvenilesAge classes
2) different rates of exploitation
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
Two assumptions must be met:
1) All population change is due to harvestNo mortality, recruitment, migration
2) Figures for harvest must be reliable (need for GOOD data)
Two assumptions must be met:
1) All population change is due to harvestNo mortality, recruitment, migration
2) Figures for harvest must be reliable (need for GOOD data)
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
Two classes, X & YTwo classes, X & Y
X/Y
0Total harvest
Zero X harvested per Y
Conducted by sport orcommercial fisheries orartificial manipulation(selective removal)
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
1. Total harvest C (CX , CY)
2. Sample size before harvest n1 (X1 ,Y1)
3. Sample size after harvest n2 (X2 ,Y2)
1. Total harvest C (CX , CY)
2. Sample size before harvest n1 (X1 ,Y1)
3. Sample size after harvest n2 (X2 ,Y2)
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
€
P1 =X1
n1
P2 =X2
n2
ˆ N X =P1(CX − P2C)
P1 − P2
Proportion of X in first sample
Proportion of X in second sample
Population estimate for X
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Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method
Population estimate for X + Y
Population estimate for Y
€
ˆ N X +Y =ˆ N XP1
ˆ N Y = ˆ N X +Y − ˆ N X
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Change in Ratio orDichotomy Method -
Example
Change in Ratio orDichotomy Method -
Example Trout (T) and suckers (S) Sample before harvest:
n1=90, T1=30, S1=60
Sample after harvest: n2=58, T2=14, S2=44
Harvest between samples: 160 trout, 160 suckers
Trout (T) and suckers (S) Sample before harvest:
n1=90, T1=30, S1=60
Sample after harvest: n2=58, T2=14, S2=44
Harvest between samples: 160 trout, 160 suckers
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Change in Ratio orDichotomy Method -
Example
Change in Ratio orDichotomy Method -
Example
€
P1 =30
90= 0.333
P2 =14
58= 0.2414
Proportions of trout in two samples
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Change in Ratio orDichotomy Method -
Example
Change in Ratio orDichotomy Method -
Example
€
ˆ N T =0.333(160 − (0.2414 × 320))
0.333 − 0.2414= 300
ˆ N T +S =300
0.333= 900
ˆ N S = 600 Sucker estimate
Trout estimate
Trout and suckers combined
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Removal Sampling -Zippin Method
Removal Sampling -Zippin Method
3-pass removal U1=number of fish removed on 1st pass
U2=number of fish removed on 2nd pass
U3=number of fish removed on 3rd pass
M=sum of all removals (U1+U2+U3) t=number of removal passes (3) C=weighted sum = (1 X U1)+(2 X U2)+(3
X U3)
3-pass removal U1=number of fish removed on 1st pass
U2=number of fish removed on 2nd pass
U3=number of fish removed on 3rd pass
M=sum of all removals (U1+U2+U3) t=number of removal passes (3) C=weighted sum = (1 X U1)+(2 X U2)+(3
X U3)
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Removal Sampling -Zippin Method
Removal Sampling -Zippin Method
€
ˆ p =
ˆ p = 0.996784 + (−0.924031)R + (0.319563)R2 + (−0.390202)R3
R =C − M
M
Capture probability
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Removal Sampling -Zippin Method
Removal Sampling -Zippin Method
€
ˆ N =M
1− (1− ˆ p )t
SE( ˆ N ) =ˆ N ( ˆ N − M)M
M 2 −[ ˆ N ( ˆ N − M)(tˆ p )2 /(1 − ˆ p )]
95%C.I. = ˆ N ± 2SE( ˆ N )
Population estimate
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Removal Sampling -Zippin Method - example
Removal Sampling -Zippin Method - example
Slimy sculpin in Garvin Brookt = 3U1 = 250U2 = 125U3 = 65M = 440
C = (1) 250 + (2) 125 + (3) 65 = 695
€
R =C − M
M=
695 − 440
440= 0.5795
ˆ p = 0.996784 + (−0.924031)0.5795 + (0.319563)0.57952 + (−0.390202)0.57953
= 0.4927
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Removal Sampling -Zippin Method - example
Removal Sampling -Zippin Method - example
Slimy sculpin in Garvin Brook
€
ˆ N =M
1 − (1 − ˆ p )t =440
1 − (1 − 0.4927)3 = 506
SE( ˆ N ) = 8.71≅ 9
95%CI = 506 ±18
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Removal Sampling -Zippin Method - example
Removal Sampling -Zippin Method - example
Slimy Sculpin Removal Sampling
y = -0.4943x + 249.64
R2 = 0.9999
0
50
100
150
200
250
300
0 100 200 300 400 500 600
Fish removed in all previous passes
U