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I have the following model model (R language). My goal model is Y~N*V but since this also happens for Y~N+V I thought I might change to the simpler model.

model <- lm(Y~N+V,data=data)

and running drop1(model,REML=FALSE) I get the output:

Model:
Y ~ N + V
       Df Sum of Sq       RSS    AIC
<none>               88592297 1027.7
N       5  30480453 119072749 1038.9
V       3  89885035 178477332 1072.1

The p-values are missing. I don't think it's that the model is oversaturated because the dataset has 72 observations and factor N has 6 levels, and factor V has 4 levels. So 1+4+6 < 72 observations in the dataset.

EDIT I manually selected the test to be used via drop1(model,test="F") and now it gives the p-values. But is this the correct test?

Pitouille
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KSHMR
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  • In case you have not seen this... it might help: https://stats.stackexchange.com/questions/4639/interpreting-the-drop1-output-in-r – Pitouille Dec 06 '21 at 14:37
  • Just saw your edit... it is difficult to answer your question as is. You will have to tell us more about what you want to achieve. – Pitouille Dec 06 '21 at 14:45

1 Answers1

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For linear models, as returned by the R function lm, the test based on the F-distribution is allright, because it uses a test statistic that is F-distributed. Note that for continuous variables, the p-values obtained by drop1 with the F-test (aka "Type II(I) ANOVA") are identical to the p-values based on the t-statistic returned by summary(lm).

For generalized linear models, as returned by the R function glm, you must use test="Chisq". The resulting p-values are based on log-likelihood differences, which are asymptotically $\chi^2$-distributed, according to an old result by Wilk from 1938. In principle, you could also use this for linear models, but as it is only an asymptotic approximation, there is no benefit in using it instead of the F-statistic based on $R^2$ differences.

cdalitz
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