Semipalmated sandpipers were counted during rapid surveys on

Semipalmated sandpipers were counted during rapid surveys on 152 plots. The total count was 1, 242 individuals.

a. What is average # of Sanpipers per plot (y)

Answer: 1,242/152=8.17

I understand question a, but not b. also I am struggling to find a formula for question b. This is for Wildlife techniques class.

Solution

Detection probability (in intensive survey) = n1/ N = 1

Detection probability (in rapid survey) = n2/ N = 2

where,

n1= number of sandpipers seen in intensive survey = 101

n2 = number of sandpipers seen in rapid survey = 72

N = [(n1+1) (n2+1)/ m2+1]-1

m = number of sandpipers seen in both surveys = (101-72)=29

N = [102*73/30]-1 = 247.2

So, detection probability in intensive survey = 1 = 101/247.2 = 0.41

Detection probability in rapid survey = 2 = 72/247.2 = 0.29

Semipalmated sandpipers were counted during rapid surveys on 152 plots. The total count was 1, 242 individuals. a. What is average # of Sanpipers per plot (y) A

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