3. Computation of standard deviation (s) for prevalence index (PI): 



N-1 



For example: 



Transect P^ ^ (P'i - P'm) (P'i " P'm^ ^ 



1 2.85 2.98 -0.13 0.0169 



2 3.16 2.98 0.18 0.0324 



3 2.93 2.98 -0.05 0.0025 



0.0518 



s = 



0.0518 / 0.0518 



3-1 



Vo 



0259 = 0.161 



4. Computation of standard error (sx) of the prevalence index: 



s o.^6^ 0.16I 



sx = —- = —== = = 0.093 



N V3 1.73 



Since 0.093 does not exceed 0.20, no additional transects are needed. 



5. Record mean prevalence index value. 



P'm= 2.98 



Since 2.98 is less than 3.0, the area has hydrophytic vegetation, if the wetland 

 hydrology criterion is met, then the area is a wetland. 



D-4 



OU.S. Governraenc PnnCing Office: i <>8<>- >^i- 108 /OOi-'Jq 



