interpreting z score in occu output

[cbpozz...@alaska.edu]

I am currently running a single season model using occu() function. I have created a suite of models and used modSel() to pick the top ranking AIC model. I am now interested in exploring that specific model in more detail. Specifically I am curious about the significance of each covariate within the model. Here is the summary output form my top ranking model...

Call:

occu(formula = ~dist + method + dsls ~ habitat + snow, data = wt)

Occupancy:

                     Estimate    SE      z         P(>|z|)

(Intercept)            -0.316 0.387 -0.816  0.41428

habitatShrub         0.504 0.465  1.085  0.27804

habitatTundra     -1.622 0.609 -2.664 0.00773

snow                   -0.463 0.231 -2.005 0.04498

Detection:

               Estimate    SE      z  P(>|z|)

(Intercept)      -0.279 0.421 -0.661 5.08e-01

dist              0.803 0.176  4.570 4.87e-06

methodtransect   -1.189 0.486 -2.448 1.44e-02

dsls              0.243 0.136  1.789 7.36e-02

AIC: 564.0724 

How should I interpret the P values for individual covariates and what exactly is the z score that is given? In addition, are there other techniques that are commonly used to test model fit? I have tried parboot() but am not entirely sure how to interpret that data either. Clearly I have some stats to brush up on, but any help to get me pointed in the right direction would be much appreciated! 

Casey

[Nicholas Som]

Hi Casey,

 

In short, the z score is each estimate divided by its standard error (SE), and standardizes these to a standard normal distribution (a normal distribution with mean = 0 and sd = 1). Under the null hypothesis that each coefficient = 0, the p-value represents the probability, given the null is true and for your model, that you would obtain such a coefficient estimate with the data you have. Small p-values represent a low probability of obtaining such a coefficient if in fact the true value is 0, so folks tend take this as evidence that the true coefficient isn't really 0, and hence is associated with your response (for example: a site being occupied). Based on your question, I'd suggest reading pages 43 - 47 of The Statistical Sleuth (ISBN: 0534386709) which gives a nice intro to p-values and their interpretation.

 

Making sure your model fits your data well is also a good idea, and precludes spending time interpreting p-values. Maybe spending some time with more examples, and perhaps reading MacKenzie and Baily's 2004 paper "Assessing the fit of site-occupancy models" (Journal of Agricultural, Biological, and Environmental Statistics 9(3): 300-318) will get you moving in the right direction.

 

Good luck,

Nicholas Som

 

 

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Nick

[Kery Marc]

Hi Casey,

for non-normal generalized linear models such as occupancy models, you better use a likelihood ratio test (LRT) to test for the significance of a term: you fit a model with the term and another model without it and the difference in the deviance between the two models is the likelihood ratio test statistic, which under the Null hypothesis is distributed as chisquare with degrees of freedom equal to the difference in the d.f. between the two models (you may have to read this twice ...).

I think you can use function anova() or perhaps aov() or even lrt() to provide this test for two models fit using occu().

Kind regards  --  Mar

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From: unma...@googlegroups.com [unma...@googlegroups.com] on behalf of Nicholas Som [nick.or...@gmail.com]
Sent: 17 October 2014 01:06
To: unma...@googlegroups.com
Subject: Re: [unmarked] interpreting z score in occu output