Dear R Community,
I am running GLM's within the "MASS" library. My data are overdispersed and
I am accounting for the overdispersion by using an ANOVA 'F' test instead of
ANOVA 'Chisq'. You will have to forgive me because I am new at this, but I
am not sure if R is conducting an ANOVA 'F' test appropriately. I was
hoping to explain this using my data results below:
anova(glm.model,test="F")
Analysis of Deviance Table
Model:binomial,link:logit
Response: ytrips
Terms added squentially (first to last)
DF Deviance
Resid. Df Resid. Dev. F Pr(>F)
NULL
125 1008.55
WaterLevel 1 0.14
124 1008.42 0.1365 0.7118
RiverFlowObserved 1 13.34
123 995.07 13.3427 0.0002594
WSBacklog 1 12.47
122 982.61 12.4675 0.0004141
factor(WDEW) 1 83.13
121 899.48 83.1258 <2.2e-16
ChickDays 1 157.22
120 742.26 157.2225 <2.2e-16
Salmon 1 6.91
119 735.34 6.9143 0.0085509
factor(Year) 3 370.31
116 365.03 123.4375 <2.2e-16
WSBacklog:factor(WDEW) 1 9.06
115 355.97 9.0644 0.0026063
I think the F-values in the results are over-inflated for an F-test.
For an F-test,
F-observed = (Deviance/Df numerator)/("MSE"=Residual Deviance/Residual Df)
When I calculate F-observed for one of my main effects, such as
'WSBacklogfactor(WDEW)' above I get:
F-observed = (9.06/1)/(355.97/115)=2.92 NOT 9.0644(as shown in the results
above)
It seems that the F-test in 'R' is not dividing by the "MSE".
Does anyone have any thoughts on this? Or can let me know where I am going
wrong?
Thank you!
~Christina
Christina J. Maranto
University of Washington
Department of Zoology
Box 351800
Seattle, WA 98195
(206) 618-2956
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