Test GLM model using null and model deviances

Question

I've built a glm model in R and have tested it using a testing and training group so am confident it works well. The results from R are:

Coefficients:
                            Estimate Std. Error  t value Pr(>|t|)    
(Intercept)               -2.781e+00  1.677e-02 -165.789  < 2e-16 ***
Coeff_A                    1.663e-05  5.438e-06    3.059  0.00222 ** 
log(Coeff_B)               8.925e-01  1.023e-02   87.245  < 2e-16 ***
log(Coeff_C)              -3.978e-01  7.695e-03  -51.689  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasibinomial family taken to be 0.9995149)

    Null deviance: 256600  on 671266  degrees of freedom
Residual deviance: 237230  on 671263  degrees of freedom
AIC: NA

All the p values for the coefficients are small as expected.

Looking at this question (Interpreting Residual and Null Deviance in GLM R), I should be able to calculate if the null hypothesis holds using the following equation:

p-value = 1 - pchisq(deviance, degrees of freedom)

Sticking this in gives:

1 - pchisq(256600, 671266)
[1] 1

So am I correct in thinking the null hypothesis cannot be rejected here, even though the p values for all coefficients are so small or have I misinterpreted how to calculate this?

Answer 1

There is a misunderstanding here. The difference between the null deviance and the model's deviance is distributed as a chi-squared with degrees of freedom equal to the null df minus the model's df. For your model, that would be:

1-pchisq(256600 - 237230, df=(671266 - 671263))
# [1] 0

By default, pchisq() gives the proportion of the distribution to the left of the value. To get the proportion more extreme than your difference, you can specify lower.tail = FALSE or subtract the result from $1$ (as you and I have done).