Miller and Ralph 



Chapter 20 



Inland Habitat Relationships in California 



percent. That is, murrelets are present, but we accept a 5 

 percent probability that they are not detected. Data from 

 previous surveys have been used in the discussion below 

 (table 1). From the data provided by Rob Hewlett, Steve 

 Kerns, Kim Nelson, and our studies, we determined the 

 number of survey mornings needed to meet this level of 

 confidence at sites having various levels of detection rates. 



In the following example, we assumed murrelets are 

 present in the relatively homogeneous stand of old-growth 

 timber to be surveyed. Each survey consists of one person 

 observing from a station for one morning. 



The method for examining our data was: 



P=l-(l-p) 

 where 



P is the probability of at least one detection, 



p is the proportion of surveys with at least one detection, 

 that is, the number of surveys with at least one detection, 

 divided by the number of surveys, and 



n is the number of surveys required to detect at least 

 one bird. 



To determine the number of surveys needed if we want 

 to be 95 percent certain (P = 0.95) we are not missing birds 

 which are present, we solve for n: 



Table 1 Detection rate at stations with low rates, and the percent of surveys 

 with detections 



In (1-P) 



n> 



In (!-/>) 

 where 



In is the natural log. 



We tested our survey sample size from 19 sites (table 1) 

 with relatively low average detection rates and a minimum 

 of seven survey mornings. The mean detection rate per 

 morning was divided into four categories, 0.4 to 2.5, 2.6 to 

 5.0, 5.1 to 7.5, and 9.4 to 16.6 detections. We used the 

 average percent of surveys with detections within each 

 category to estimate p. 



In the 0.4 to 2.5 category, the percent of survey mornings 

 with detections varied from 13 percent to 75 percent, with an 

 average of 48 percent of the mornings with detections. The 

 calculation is as follows: 



n > 



In (1-0.95) 

 In (1-0.48) 



= 4.58, or 5 surveys. 



In the 2.6 to 5.0 detection range, the percent of surveys 

 with detections varied over a smaller range, from 63 percent 

 to 91 percent, an average of 81 percent. Using the average 

 number, the calculation is: 



n > 



In (1-0.95) 

 In (1-0.81) 



= 1.80, or 2 surveys. 



In the 5.1 to 7.5 detection range, the percent of surveys 

 with detections varied from 65 percent to 88 percent, an 



average of 78 percent. The calculation as above was 1.98 or 

 a minimum of 2 surveys. 



The highest detection range used for this calculation 

 was 9.4 to 16.6 birds per morning, an average of 85 percent 

 of survey mornings with at least 1 detection. The calculation 

 resulted in 1.75, or 2 surveys. 



From these data we can conclude that in areas with 

 mean detection rates as low as 0.4 to 2.5 per survey (and 

 presumably low occupancy rates as well), a minimum of 

 five survey mornings will detect birds if they are present, 

 with a 95 percent probability. In areas of detection rates 

 from 9.4 to 16.6, the number of surveys necessary to 

 prevent a false negative is about two. Using this formula, 4 

 surveys would be required to detect birds in areas with a 

 mean of 1.0 to 2.5 detections per survey. We can then 

 conclude that a suggested survey rate of four surveys per 

 stand, will detect birds in excess of 95 percent of the time, 

 and will likely detect all but the smallest populations 99 

 percent of the time. 



Assumptions 



There are several assumptions we have made in using 

 these methods. We list them below and discuss each. 



We assume that the amount of canopy cover at a station 

 will have no effect on detection probability (P). 



214 



USDA Forest Service Gen. Tech. Rep. PSW-152. 1995. 



