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Fish Population Assessment How many fish do we have?

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Presentation on theme: "Fish Population Assessment How many fish do we have?"— Presentation transcript:

1 Fish Population Assessment How many fish do we have?

2

3 Fish Population Assessment  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)

4 Plot Method Total population area Size of the plot Average number of fish per plot Population estimate

5 Plot Method - Estimated Variance Number of fish counted in i th plot Number of plots used

6 Plot Method - 95% confidence interval for s-1 df, p=0.05

7 Plot Method - Example  Pond area = 100 m 2  Size of plot = 1 m 2  Average number of fish per plot = 1.5  Pond area = 100 m 2  Size of plot = 1 m 2  Average number of fish per plot = 1.5

8 Mark and Recapture - Peterson Method (single) Number of fish initially marked & released Number of fish collected/examined in 2nd period Number of recaptures found in C Bailey modification

9 Mark and Recapture - Variance

10 Mark and Recapture - 95% confidence interval

11 Mark and Recapture - Example M = 550 C = 500 R = 157

12 Mark and Recapture - Example M = 550 C = 500 R = 157

13 Mark and Recapture - Schnabel Method Multiple episodes of mark and recapture CM = total captures X marked fish available for recapture R = recaptures of marked fish

14 Schnabel Method - Variance & 95% C.I. Then invert for 95% C.I. for N

15 Schnabel Method - example p. 137 (2nd ed.) PeriodRUnmarkedTotal C MCM 10150 00 22220322515033,750 3268611235339,536 …. 439…. Total254457,208

16 Schnabel Method - example p. 137 (2nd ed.) 95% C.I. = 1,602 - 2,049

17 Change in Ratio or Dichotomy Method Requirements:  1) two recognizable classes  Species  Sexes  Adults vs. juveniles  Age classes  2) different rates of exploitation Requirements:  1) two recognizable classes  Species  Sexes  Adults vs. juveniles  Age classes  2) different rates of exploitation

18 Change in Ratio or Dichotomy Method Two assumptions must be met:  1) All population change is due to harvest  No 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 harvest  No mortality, recruitment, migration  2) Figures for harvest must be reliable (need for GOOD data)

19 Change in Ratio or Dichotomy Method Two classes, X & Y X/Y 0 Total harvest Zero X harvested per Y Conducted by sport or commercial fisheries or artificial manipulation (selective removal)

20 Change in Ratio or Dichotomy Method  1. Total harvest C (C X, C Y )  2. Sample size before harvest n 1 (X 1,Y 1 )  3. Sample size after harvest n 2 (X 2,Y 2 )  1. Total harvest C (C X, C Y )  2. Sample size before harvest n 1 (X 1,Y 1 )  3. Sample size after harvest n 2 (X 2,Y 2 )

21 Change in Ratio or Dichotomy Method Proportion of X in first sample Proportion of X in second sample Population estimate for X

22 Change in Ratio or Dichotomy Method Population estimate for X + Y Population estimate for Y

23 Change in Ratio or Dichotomy Method - Example  Trout (T) and suckers (S)  Sample before harvest:  n 1 =90, T 1 =30, S 1 =60  Sample after harvest:  n 2 =58, T 2 =14, S 2 =44  Harvest between samples:  160 trout, 160 suckers  Trout (T) and suckers (S)  Sample before harvest:  n 1 =90, T 1 =30, S 1 =60  Sample after harvest:  n 2 =58, T 2 =14, S 2 =44  Harvest between samples:  160 trout, 160 suckers

24 Change in Ratio or Dichotomy Method - Example Proportions of trout in two samples

25 Change in Ratio or Dichotomy Method - Example Sucker estimate Trout estimate Trout and suckers combined

26 Removal Sampling - Zippin Method  3-pass removal  U 1 =number of fish removed on 1st pass  U 2 =number of fish removed on 2nd pass  U 3 =number of fish removed on 3rd pass  M=sum of all removals (U 1 +U 2 +U 3 )  t=number of removal passes (3)  C=weighted sum = (1 X U 1 )+(2 X U 2 )+(3 X U 3 )  3-pass removal  U 1 =number of fish removed on 1st pass  U 2 =number of fish removed on 2nd pass  U 3 =number of fish removed on 3rd pass  M=sum of all removals (U 1 +U 2 +U 3 )  t=number of removal passes (3)  C=weighted sum = (1 X U 1 )+(2 X U 2 )+(3 X U 3 )

27 Removal Sampling - Zippin Method Capture probability

28 Removal Sampling - Zippin Method Population estimate

29 Removal Sampling - Zippin Method - example Slimy sculpin in Garvin Brook t = 3 U 1 = 250 U 2 = 125 U 3 = 65 M = 440 C = (1) 250 + (2) 125 + (3) 65 = 695

30 Removal Sampling - Zippin Method - example Slimy sculpin in Garvin Brook

31 Removal Sampling - Zippin Method - example


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