Why is the p-value from the analysis of deviance table different from the estimated with pchisq()?

Question

I'm studying GLM models from the Lane (2002) paper and I am a bit confused with the analysis of deviance for the Gamma-GLM model.

In the paper, the p-value is lower than P < 0.001 but if we used the deviance reported as well as the degrees of freedom to calculate the p-value with the pchisq() function in R we get the following results:

> 1-pchisq(11.1057, 7)
[1] 0.1340744` 

and not the P <0.001 reported in the paper.

I've copied the data to replicate the GLM model here's the link ! and this is the code I used to generate the results:

test <- read_csv("data/test_glm_gamma.csv", col_types = cols())

model.test <- glm(soil ~ trt, family = Gamma(link = "log"), data = test)

anova(model.test, test = "Chisq")

which returns:

Analysis of Deviance Table

Model: Gamma, link: log

Response: cont

Terms added sequentially (first to last)


     Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    
NULL                    23    11.5897              
trt   7   11.106        16     0.4839 < 2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

with similar deviances as in the paper and I suspect similar P-value but this is not the 0.13 obtained before.

Is there any transformation done before the P-value is calculated? Or I am calculating the p-value in the wrong way? How do they get the <2.2e-16 in the deviance table?

Lane, PW (2002). Generalized linear models in soil science. European Journal of Soil Science, 53, 241–251. https://doi.org/10.1046/j.1365-2389.2002.00440.x

Answer 1

You need to pass the deviance scaled by the dispersion to pchisq :

p <- pchisq(anova(model.test)$Deviance[2]/
         summary(model.test)$dispersion, 
       anova(model.test)$Df[2], 
       lower.tail = FALSE)

p == anova(model.test, test = "Chisq")$`Pr(>Chi)`[2]
#[1] TRUE

You can study the code of stats:::stat.anova to see how the p-values are calculated for the different tests.