1. Age Length Key

2. 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!

3. 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.

4. 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 categories Make raw contingency table Age LCat 1 2 3 20 30 40 50 Convert to row-proportions table Age LCat 1 2 3 20 30 40 50 LCat 20 30 40 50 22 2 2 220 0 11 11 0.00 0.50 0.25

5. 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 0.50 Use 10-cm intervals for length categories Create length distribution LCat 20 30 40 50 Freq 4 3 5 2 LCat 20 30 40 50

6. 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 0.50 Length distribution (as a reminder) LCat 20 30 40 50 Freq 4 3 5 2 Identify number in each length category to be assigned each age 20-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 30 40 50 2 2 1 1

7. 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 0.50 Length distribution (as a reminder) LCat 20 30 40 50 Freq 4 3 5 2 Identify number in each length category to be assigned each age 30-cm  3*0.25 = 0.75 age-1  3*0.5 = 1.5 age-2  3*0.25 = 0.75 age-3 What to do now? LCat 20 30 40 50 2 2 1 1

8. 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.

9. 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-3 –For 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 chosen thus, randomly assign 1 age-1, 2 age-2, & 0 age-3 for fish in the 30-cm length interval

10. 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 0.50 Length distribution (as a reminder) LCat 20 30 40 50 Freq 4 3 5 2 Identify number in each length category to be assigned each age 30-cm  3*0.25 = 0.75 age-1  3*0.5 = 1.5 age-2  3*0.25 = 0.75 age-3 Randomly determine which fish are assigned these ages. LCat 20 30 40 50 2 2 1 1  1  2  0 1 2 2

11. 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 0.50 Length distribution (as a reminder) LCat 20 30 40 50 Freq 4 3 5 2 Identify number in each length category to be assigned each age 40-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 30 40 50 2 2 1 1  1  3 1 2 2 1 3 2 3 3

12. 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 0.50 Length distribution (as a reminder) LCat 20 30 40 50 Freq 4 3 5 2 Identify number in each length category to be assigned each age 50-cm  2*0.5 = 1 age-2  2*0.5 = 1 age-3 Randomize age assignments. LCat 20 30 40 50 2 2 1 1 1 2 2 1 3 2 3 3 3 2

13. 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.

14. 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 random

15. 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 aged –interested in mean length-at-age and age distribution. Examine Handout –Summarize() –lencat() –table() –prop.table() –ageKey()