Deviance test for binomial distribution

# Let A et B being observations from binomial distribution:
A <- c(10, 10, 9, 7, 3, 2, 3)
B <- c(2, 5, 6, 6, 10, 15, 20)

# And the prediction of proportion being pred:

pred <- c(0.9, 0.85, 0.75, 0.4, 0.3, 0.2, 0.1)

# For example prod could be the result of a glm, or other model.

# Number of parameters obtained from observations and used to estimate pred

parameter <- 2

# Then the degrees of freedom are:

df <- length(pred)-parameter

LnL <- sum(dbinom(x=A, size = A+B, 

                  prob = pred, log = TRUE))

LnLSat <- sum(dbinom(x=A, size = A+B, 

                     prob = A/(A+B), log = TRUE))

deviance <- -2*(LnL - LnLSat)

p <- 1-pchisq(deviance, df=df)

# The larger is p, the more the observations could have been obtained from the model used to calculate pred

# Do not use p<0.05 as a criteria; p is like a proxy

plot(x=pred, y=A/(A+B), bty="n", las=1, xlim=c(0, 1), ylim=c(0,1))

text(x=0.8, y=0.2, labels = paste("p=", round(p, digits = 2)))