# How do I calculate probability of reference category in logistic regression?

I came across this quote from a paper by Pool and Doris (2021): "The 18–24 age group reported the highest levels of food insecurity, with 27.6% reporting being food insecure, as opposed to 8.2% of over 65-year-olds (odds ratio (OR) 2.45, 95% confidence interval (CI): 1.05; 5.75)." I'd like to understand how they calculated the probability of over-65s which was their reference group in the logistic regression they ran.

Example logistic regression output:

  Coefficients:
Estimate Std.  Error z value Pr(>|z|)
(Intercept) -2.457415 0.221487 -11.095  < 2e-16 ***
wave5       -0.510565 0.089884  -5.680 1.34e-08 ***
wave7       -0.360667 0.089068  -4.049 5.14e-05 ***
age_r2      -0.167394 0.163767  -1.022 0.306713
age_r3      -0.664402 0.171821  -3.867 0.000110 ***
age_r4      -0.898270 0.161779  -5.552 2.82e-08 ***
age_r5      -1.704232 0.162821 -10.467  < 2e-16 ***
age_r6      -2.909736 0.183640 -15.845  < 2e-16 ***


I want to find out what the probability of the outcome variable is for the reference categories e.g. wave4 and age_r1. Apologies if this is straightforward but I've really struggled to find an answer. How can I do this in R?

Reference: Pool, U. and Dooris, M., 2021. Prevalence of food security in the UK measured by the Food Insecurity Experience Scale. Journal of Public Health,.

• You could use the emmeans package to get the logit of each group. And then transform the logit to probability. May 13 at 1:06

From what you say, I assume that the combination (wave4, age_r1) means that all other $$X$$ variable values are 0. (Although it seems odd that there are only categories wave4, wave5 and wave7).
So, the estimated logit of the probability is $$-2.457415 +(0\times -0.510565) + (0 \times -0.360667) + (0 \times -0.167394) + (0 \times -0.664402) + (0 \times -0.898270) + (0 \times -1.704232) + (0 \times -2.909736)$$ $$= -2.457415$$
Thus the estimated probability itself is $$\frac{\exp(-2.457415)}{1+\exp(-2.457415)}$$ $$= 0.07889799$$
That is the easiest one. You can use the same trick to get the probabilities for all other combinations too, although there will be some "$$1\times$$" multipliers.