Detection probabilities for single-season multi method model

[ALAdams]

Hi,

I am trying to analyse my data using a single-season multi-method approach with 2 detection methods, 6 surveys across 71 sites. I am only interested in determining the detection probability and how these differ between the two detection methods. As such I have a constant model, psi(.) theta(.) p(.), and one which p differs with method, psi(.)theta(.)p(Method).

The model output for a species for the method model produces different p values for the two detection methods across the different surveys. E.G.:
Site estimate Std.err 95% conf. interval
p[1-1] 1 GRASM10 : 0.6667 0.2722 0.1535 - 0.9566
p[1-2] 1 GRASM10 : 1.0000 0.0000 1.0000 - 1.0000
p[2-1] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[2-2] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[3-1] 1 GRASM10 : 0.5000 0.2500 0.1235 - 0.8765
p[3-2] 1 GRASM10 : 1.0000 0.0000 1.0000 - 1.0000
p[4-1] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[4-2] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[5-1] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[5-2] 1 GRASM10 : 0.2853 0.2908 0.0238 - 0.8672
p[6-1] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000
p[6-2] 1 GRASM10 : 0.0000 0.0000 0.0000 - 1.0000

Therefore, what is the best way to report the p for each of the two methods (method 1 and method 2). Should I average the 3 values for each method or use the median?

Thank-you for any input anyone can offer!

[jhines]

Hi,

The model output suggests that your model was psi(.)theta(.)p(method*survey), or a different detection rate for each method and survey. To build the survey-constant, method-specific model, the design matrix for p should b:

-,b1,b2
p[1-1] 1 0
p[1-2] 0 1
p[2-1] 1 0
p[2-2] 0 1
p[3-1] 1 0
p[3-2] 0 1
p[4-1] 1 0
p[4-2] 0 1
p[5-1] 1 0
p[5-2] 0 1
p[6-1] 1 0
p[6-2] 0 1

This will give you an average detection probability over all 6 surveys for each method to report. Even if you had lots of data, the estimates from this model would be the ones you want to report if you wanted to report overall estimates for each method, as these are weighted according to the amount of data in each survey.

Cheers,

Jim