# percentage of difference with estimated marginal means

I am calculating the difference between concentration of different metabolites according to the factor sex (male/female) and age group (<55 / >=55) from this linear model:

model <- lm(metabolite ~ agegroup + sex, data =df)
emmeans(model, "agegroup")

sex = 0:
agegroup emmean     SE   df lower.CL upper.CL
0  0.329 0.0291 2499    0.272    0.386
1  0.373 0.0327 2499    0.308    0.437

sex = 1:
agegroup emmean     SE   df lower.CL upper.CL
0 -0.490 0.0303 2499   -0.549   -0.430
1 -0.446 0.0377 2499   -0.520   -0.372

Confidence level used: 0.95


Now I want to calculate the percentage of difference of male vs. female by age group and plot it. I am trying this

mean_diff_sex <- as.data.frame(emmeans::contrast(emmeans(model, "sex")))

contrast   estimate         SE   df   t.ratio       p.value
1 0 effect -0.5206612 0.01737366 2499 -29.96842 7.576451e-169
2 1 effect  0.5206612 0.01737366 2499  29.96842 7.576451e-169


However, I am not sure if I can transform this estimate into the percentage of difference doing: ((2^estimate)-1)x100 or it would be better to using: Absolute difference / Average x 100

Thank you in advance for any help

However, you need to have your estimates on the log scale first. For example, fit the model with log(metabolite) as the response.