NE Groundfish Survey S2021: The American Lobster Homarus americanus

Epizootic Shell disease in the Gulf of Maine and Georges Bank in Spring 2021

  • S2021 ESD survey was done without saving any lobsters but recording the data according to protocol.
    1. Therefor there was no possibility of determining False(+)
    2. Four files transmitted to JGK as a zip file quarantined to the kunkel home page directory on marlin.
    3. Here are three derivative files that show the structure of the data.
    4. The used data sets will be eventually shared when the analysis is completed and fully validated.
  • Five strings of 30 sequential trawls

    • developed R-script source("HBB_S2021_LatLonNandESDgrouped_black.R")
      1. Do Chi-Sq on the 5 ESD proportions.

    5 strings of 30 trawls with ESD tabled

    Chi Square Test of ESD even distribution in 5 sequential strings of 30 trawls

    Strings R O Y G B all 50
    k ESD 13 8 8 7 6 42
    non-ESD 911 140 265 480 534 2330
    n lobsters 924 148 273 487 540 2372
    • Chi Square with 4 df
    • Sum ((O-E)^2)/E
    • E = {42/2372; 2330/2372} = {0.01770658; 0.9822934}
      1. ER = {16.360880; 907.639120}
      2. EO = {2.620573; 145.379427}
      3. EY = {4.833895; 268.166105}
      4. EG = {8.623103; 478.376897}
      5. EB = {9.561551; 530.438449}
    • O = c(13,8,8,7,6,911,140,265,480,534)
    • E = c(16.36088, 2.620573, 4.833895, 8.623103, 9.561551, 907.63912, 145.379427, 268.166105, 478.376897, 530.438449)
    > O = t(matrix(OB,5,2))
    > O
         [,1] [,2] [,3] [,4] [,5]
    [1,]   13    8    8    7    6
    [2,]  911  140  265  480  534
    > E
              [,1]       [,2]       [,3]       [,4]       [,5]
    [1,]  16.36088   2.620573   4.833895   8.623103   9.561551
    [2,] 907.63912 145.379427 268.166105 478.376897 530.438449
    Expected cells > 5 is greater than 80%. Therefore Chis Square is applicable!
    > ChiSq=sum(((O-E)^2)/E)
     15.71728 with 4 df.
    
                   Probability less than the critical value
     ν           0.90      0.95     0.975      0.99     0.999
    --          -----     -----    ------     -----    ------
     1          2.706     3.841     5.024     6.635    10.828
     2          4.605     5.991     7.378     9.210    13.816
     3          6.251     7.815     9.348    11.345    16.266
     4          7.779     9.488    11.143    13.277  * 18.467
     5          9.236    11.070    12.833    15.086    20.515
    

    Therefore, P[ESD is distributed evenly in the NE Groundfish Survay GoM territory] < 1%

    Try Ten strings of 15 sequential trawls

    • developed R-script source("HBB_S2021_LatLonNandESD10groups15_black.R")
      1. Do Chi-Sq on the 10 ESD proportions.

    10 strings of 15 trawls with ESD tabled

    Chi Square Test of ESD even distribution in 10 sequential strings of 15 trawls

    N =  164 760 33 115 105 168 284 203 511 29  ... Sum = 2372 
    K =  6 7 1 7 0 8 3 4 6 0  ... Sum = 42 
    
    Observed:
          [,1]  [,2]  [,3]   [,4]   [,5] [,6] [,7] [,8]   [,9] [,10]
    [1,]     6     7    1      7      0     8    3    4     6     0
    [2,]   158   753   32    108    105   160  281  199    505   29
    
    Expected:
          [,1]  [,2]  [,3]   [,4]   [,5]  [,6] [,7]  [,8] [,9] [,10]
    [1,]   2.9  13.4  0.58   2.04    1.9   2.9  5.0   3.6    9   0.5
    [2,] 161.0 746.5  32.4  112.9  103.1 165.0  279 199.4  502  28.5
    Expected cells > 5 is less than 80% and 1 cell has observed n = 0. Therefore Chi-Square is not applicable!
    
    Expected:
               [,1]    [,2]      [,3]       [,4]       [,5]       [,6]       [,7]       [,8]       [,9]      [,10]
    [1,]   2.903879  13.457  0.584317   2.036256   1.859191   2.974705   5.028668   3.594435   9.048061  0.5134907
    [2,] 161.096121 746.543 32.415683 112.963744 103.140809 165.025295 278.971332 199.405565 501.951939 28.4865093
        
    > ChiSq=sum(((O-E)^2)/E)
    ChiSq = 32.11684 with 9 df
    Total number of ESD lobsters = 47 includes ESD lobsters south of Cape Cod!
    
    BUT expected cells > 5 is less than 80% and 1 cell has n = 0. Therefore Chi Square is not applicable!

    Bibliography

    1. Mary L. McHugh (2013). Lessons in biostatistics: The Chi-square test of independence. Biochemia Medica 23(2): 143-9.