tbl = devianceTest(mdl)
returns
an analysis of deviance table for the tbl
= devianceTest(mdl
)mdl
generalized
linear model. tbl
gives the result of a test of
whether the fitted model fits significantly better than a constant
model.

Generalized linear model, as constructed by 

Table containing two rows and four columns.

Deviance of a model M_{1} is twice the difference between the loglikelihood of that model and the saturated model, M_{S}. The saturated model is the model with the maximum number of parameters that can be estimated. For example, if there are n observations y_{i}, i = 1, 2, ..., n, with potentially different values for X_{i}^{T}β, then you can define a saturated model with n parameters. Let L(b,y) denote the maximum value of the likelihood function for a model. Then the deviance of model M_{1} is
$$2\left(\mathrm{log}L\left({b}_{1},y\right)\mathrm{log}L\left({b}_{S},y\right)\right),$$
where b_{1} are the estimated parameters for model M_{1} and b_{S} are the estimated parameters for the saturated model. The deviance has a chisquare distribution with n – p degrees of freedom, where n is the number of parameters in the saturated model and p is the number of parameters in model M_{1}.
If M_{1} and M_{2} are two different generalized linear models, then the fit of the models can be assessed by comparing the deviances D_{1} and D_{2} of these models. The difference of the deviances is
$$\begin{array}{l}D={D}_{2}{D}_{1}=2\left(\mathrm{log}L\left({b}_{2},y\right)\mathrm{log}L\left({b}_{S},y\right)\right)+2\left(\mathrm{log}L\left({b}_{1},y\right)\mathrm{log}L\left({b}_{S},y\right)\right)\\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\mathrm{log}L\left({b}_{2},y\right)\mathrm{log}L\left({b}_{1},y\right)\right).\end{array}$$
Asymptotically, this difference has a chisquare distribution
with degrees of freedom v equal to the number of
parameters that are estimated in one model but fixed (typically at
0) in the other. That is, it is equal to the difference in the number
of parameters estimated in M_{1} and M_{2}.
You can get the pvalue for this test using 1  chi2cdf(D,V)
, where D = D_{2} – D_{1}.