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Age Length Key Age Length Key

Age Length Key - PowerPoint Presentation

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Age Length Key - PPT Presentation

Age Analysis 2 Importance Many stock parameters are annual rates eg growth amp mortality components of yield time must be recorded Usually as the age of the fish Age ID: 569214

length age fish key age length key fish lcat sample number reminder assigned concept category samplelen table distribution 50freq

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Slide1

Age Length KeySlide2

Age Analysis

2

Importance

Many stock parameters are

annual

rates

.

e.g., growth & mortality (components of yield)

time” must be recorded.

Usually as

the age of the fish.

Age

assessment is vitally

important.

It is also

EXPENSIVE!Slide3

Age Length Key

3

Concept – Overall

Have a

large

sample of

fish.

Measure length on all fish.

Choose a portion of sample to assign age.

Called the

age sample

.

Fish chosen either

in proportion to the number of fish in each length category

as a fixed number per each length category (more common)

Fish that are not aged are called the

length sample.

Develop relationship between age and length from fish in age sample.

two-way contingency table called an

age-length key

“Assign” age to

fish in length sample with age-length key .Slide4

Age Length Key

4

Concept – Age-Length Key

Age Sample

len age

22 1

24 1

26 2

27 2

30 1

32 2

34 2

39 3 40 1 44 2 46 3 49 3 52 2 53 2 55 3 56 3

Use 10-cm intervals for length categoriesMake raw contingency table AgeLCat 1 2 3 20 30 40 50 Convert to row-proportions table AgeLCat 1 2 3 20 30 40 50

LCat 20 20 20 20 30 30 30 30 40 40 40 40 50 50 50 50

2

2

2

2

2

2

0

0

1

1

1

1

0.00

0.00

0.50

0.50

0.50

0.50

0.50

0.50

0.25

0.25

0.25

0.25Slide5

Age Length Key

5

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) Age LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Use 10-cm intervals for length categoriesCreate length distributionLCat 20 30 40 50Freq 4 3 5 2

LCat 20 20 20 20 30 30 30 40 40 40 40 40 50 50Slide6

Age Length Key

6

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Length distribution (as a reminder)LCat 20 30 40 50Freq 4 3 5 2Identify number in each length category to be assigned each age20-cm  4*0.5 = 2 age-1  4*0.5 = 2 age-2  4*0 = 0 age-3

Randomly determine which fish are assigned these ages.LCat 20 20 20 20 30 30 30 40 40 40 40 40 50 50

2

2

1

1Slide7

Age Length Key

7

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Length distribution (as a reminder)LCat 20 30 40 50Freq 4 3 5 2Identify number in each length category to be assigned each age30-cm  3*0.25 = 0.75 age-1  3*0.5 =

1.5 age-2  3*0.25 = 0.75 age-3What to do now?LCat 20 20 20 20 30 30 30 40 40 40 40

40 50

50

2

2

1

1Slide8

Age Length Key

8

Age-Length Key – Fractionation

Fractionation

When

a

fraction of a fish

should

be assigned a given age.

Handling fractionation

Round all values

down

to integers.For remaining number of fish, choose random ages in proportion to the proportion in each age.Slide9

Age Length Key

9

Age-Length Key – Fractionation

Handling fractionation

Round all values down to integers.

30-cm

 3*0.25 =

0.75

=

0

age-1

 3*0.5 =

1.5 = 1 age-2  3*0.25 = 0.75 = 0 age-3For remaining number of fish ...two ages must be chosen such that age-1 has 25% chance, age-2 has 50% chance, and age-3 has 25% chance of being selected.e.g., 2, 1 was chosenthus, randomly assign 1 age-1, 2 age-2, & 0 age-3 for fish in the 30-cm length intervalSlide10

Age Length Key

10

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Length distribution (as a reminder)LCat 20 30 40 50Freq 4 3 5 2Identify number in each length category to be assigned each age30-cm  3*0.25 = 0.75 age-1  3*0.5 =

1.5 age-2  3*0.25 = 0.75 age-3Randomly determine which fish are assigned these ages.LCat 20 20 20 20 30 30 30 40 40 40 40

40 50

50

2

2

1

1

 1

 2

 0

1

2

2Slide11

Age Length Key

11

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Length distribution (as a reminder)LCat 20 30 40 50Freq 4 3 5 2Identify number in each length category to be assigned each age40-cm  5*0.25 = 1.25 = 1 age-1  5*0.25 = 1.25 = 1 age-2  5*0.5 = 2.5 = 2 age-3

Extra fish was chosen to be age-3.Randomize age assignments.LCat 20 20 20 20 30 30 30 40 40 40 40 40 50

50

2

2

1

1

 1

 1

 3

1

2

2

1

3

2

3

3Slide12

Age Length Key

12

Concept – Age-Length Key

Length Sample

len age

23

25

26

28

32

33

36

40 43 44 48 49 51 54

Age-Length Key (as a reminder) LCat 1 2 3 20 0.50 0.50 0.00 30 0.25 0.50 0.25 40 0.25 0.25 0.50 50 0.00 0.50 0.50Length distribution (as a reminder)LCat 20 30 40 50Freq 4 3 5 2Identify number in each length category to be assigned each age50-cm  2*0.5 = 1 age-2  2*0.5 =

1 age-3Randomize age assignments.LCat 20 20 20 20 30 30 30 40 40 40 40 40 50

50

2

2

1

1

1

2

2

1

3

2

3

3

3

2Slide13

Age Length Key

13

Utility

Lengths in age sample should cover range of lengths in length sample.

Age sample and length sample must be from the same population.

typically age sample is a subsample

“Extra” effort should be put in age, not length, sample.

Combine age sample and age-assigned length sample for further analysis.Slide14

Age Length Key

14

How – Apply the A-L Key

Use the

age.key()

function to assign ages to fish in a length sample given an A-L Key.

required arguments:

age-length key row proportions table as first argument.

data frame with length sample as second argument.

cl=

(name or number of column containing the measured lengths)

optional arguments: ca= (name or number of column that should receive the assigned ages)if this column does not exist it will be created & called “Age”. type= (A string indicating the type of randomization) type=“SR”  semi-random (default, method described here) type=“CR”  completely randomSlide15

Age Length Key

15

How – Apply the A-L Key

Demonstration with the age and lengths of spot (

Leiostomus

xanthurus

) from Virginia.

403 fish were collected

as many as 10 per 1-inch length category were aged from

otolith

thin sections.

72 fish were agedinterested in mean length-at-age and age distribution.Examine HandoutSummarize()lencat()table()prop.table()ageKey()