# Setting the intercept of a Binomial GLMM (glmer) to a set value

I have a run an experiment where subjects are given a choice between four objects. Their response is either correct or incorrect, however because there are four choices and only one is correct, it is expected that the correct choice will occur randomly at a frequency of 25% and the incorrect will be 75%.

A subset of the data looks like this:

Treatment Probe Subject FREQ Correct Incorrect
1 R         R     2B     100       120         0
2 T         T1    2B     100        24         0
3 T         T2    2B     83.3      20         4
4 T         T3    2B     66.7      16         8
5 T         T4    2B     33.3       8        16
6 T         T5    2B     20.8       5        19
7 R         R     8M     97.5     117         3
8 T         T1    8M     83.3      20         4
9 T         T2    8M     62.5      15         9
10 T         T3    8M     62.5      15         9


I am analysing the choice frequency for different Treatments using the following GLMM (glmer).

Model <- glmer(cbind(Correct, Incorrect)~Treatment+(1|Subject),
> family=binomial, data=DF1). What I really want to test is whether each treatment mean is different from 25% (the expected correct frequency, represented by the gray line on the figure). I can do this when the expectation is that correct and incorrect are expected at an equal frequency (50%) by updating the original model to change the intercept from a 'reference level coding' to 'level means coding'.

mod1A <- update(mod1, . ~ . - 1)


Does anyone know how I might change my model so the intercept is essentially set to a pre-selected value, in this case 25%?

Your null hypothesis is probability ($$\pi$$) of correct = 0.25 and alternative hypothesis is $$\pi > 0.25$$. The logistic regression was used.
$$\pi = 0.25$$ is equivalent to log odds = log(0.25/0.75) = -log(3) = -1.0986
So the original $$H_o: \pi = 0.25$$ can be converted as $$H_o:$$ log odds = -1.0986, and $$H_A:$$ is log odds > -1.0986.
So for each treatment, you can get $$X\hat\beta$$, which is log odds, and test if $$X\beta = -1.0986$$ as one-sided test.