fish population assessment how many fish do we have?

Fish Population Assessment

Fish Population Assessment

How many fish do we have?How many fish do we have?

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)

Plot MethodPlot Method

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ˆ N =A

a× n

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A =

a =

n =

ˆ N =

Total population area

Size of the plot

Average number of fish per plot

Population estimate

Plot Method - Estimated Variance

Plot Method - Estimated Variance

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V ( ˆ N ) =A2

a×

V (n)

s×

A − (sa)

A

V (n) =(ni − n )2

s −1∑

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ni =

s =

Number of fish counted in ith plot

Number of plots used

Plot Method - 95% confidence interval

Plot Method - 95% confidence interval

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ˆ N ± t V ( ˆ N )

t for s-1 df, p=0.05

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

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100m2

1m2 ×1.5 =150

Mark and Recapture - Peterson Method (single)

Mark and Recapture - Peterson Method (single)

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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

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R

C=

M

N

N =MC

RBailey modification

Mark and Recapture - Variance

Mark and Recapture - Variance

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V ( ˆ N ) =M 2(C +1)(C − R)

(R +1)2(R + 2)

Mark and Recapture - 95% confidence intervalMark and Recapture -

95% confidence interval

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ˆ N ±1.96 V ( ˆ N )

Mark and Recapture - Example

Mark and Recapture - Example

M = 550 C = 500 R = 157

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ˆ N =550(500 +1)

(157 +1)=1,744

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V ( ˆ N ) =5502(501)(500 −157)

(157 +1)=13,096

Mark and Recapture - Example

Mark and Recapture - Example

M = 550 C = 500 R = 157

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ˆ N ±1.96 13,096 =1,744 ± 224

P(1,520 ≤ ˆ N ≤1,968) = 0.95

Mark and Recapture - Schnabel Method

Mark and Recapture - Schnabel Method

Multiple episodes of mark and recapture

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ˆ N =CM∑R∑

CM = total captures X marked fish available for recapture

R = recaptures of marked fish

Schnabel Method - Variance & 95% C.I.Schnabel Method - Variance & 95% C.I.

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V (1ˆ N

) =R∑

( CM)2∑

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1ˆ N

±1.96 V (1ˆ N

) Then invert for 95% C.I. for N

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

Schnabel Method - example p. 137 (2nd ed.)

Schnabel Method - example p. 137 (2nd ed.)

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ˆ N =CM∑R∑

=457,208

254=1,800

95% C.I. = 1,602 - 2,049

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

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)

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)

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)

Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method

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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

Change in Ratio orDichotomy MethodChange in Ratio orDichotomy Method

Population estimate for X + Y

Population estimate for Y

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ˆ N X +Y =ˆ N XP1

ˆ N Y = ˆ N X +Y − ˆ N X

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

Change in Ratio orDichotomy Method -

Example

Change in Ratio orDichotomy Method -

Example

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P1 =30

90= 0.333

P2 =14

58= 0.2414

Proportions of trout in two samples

Change in Ratio orDichotomy Method -

Example

Change in Ratio orDichotomy Method -

Example

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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

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)

Removal Sampling -Zippin Method

Removal Sampling -Zippin Method

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ˆ p =

ˆ p = 0.996784 + (−0.924031)R + (0.319563)R2 + (−0.390202)R3

R =C − M

M

Capture probability

Removal Sampling -Zippin Method

Removal Sampling -Zippin Method

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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

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

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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

Removal Sampling -Zippin Method - example

Removal Sampling -Zippin Method - example

Slimy sculpin in Garvin Brook

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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

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