# Interpretation of BayesFactor BF01 for two sample t-test correct?

I want to test a nullinterval with bayes. So the hypothesis I have is on the absence of an effect. Group 1 and 2 do not differ in their mean (at least not beyond an effect size of $$\delta > 0.2$$). However I am very unsure if my rural understanding of Bayesian Analasis brought me to the correct interpretation of my results. I followed the instructions proided in the vignette for the R package BayesFactor available here and also in this nice guide. However I am not sure if I am interpreting the resulting Bayes Factor correctly?

Here is some Simlulation Data

# Loading Packages
set.seed(333)
n1 <- 70
n2 <- 50
example <- data.frame(
treatment = factor(c(
rep("Treatment A", n1),
rep("Treatment B", n2)
)),
outcome = c(
rnorm(n = n1, mean = 7, sd = 18),
rnorm(n = n2, mean = 13, sd = 15)
)
)


And here the nullinterval bayes t-test

bfInterval <- ttestBF(
formula = outcome ~ treatment,
data = example,
nullInterval = c(-0.2, 0.2)
)


Is my understanding now correct that the variable bfinterval no contains two bayes factors?

• One for the null-interval $$|\delta| < 0.2$$ compared to the null-point and
• One for the complement $$|\delta| > 0.2$$ of the null-interval to the null-point

And the complement (that there would be an effect > 0.2) is more likely, because BF $$BF_1 = 0.18$$ is farther away from 1, than the BF BF $$BF_0 = 0.96$$ for the null-interval? Would that be correct?

> bfInterval
Bayes factor analysis
--------------
[1] Alt., r=0.707 -0.2<d<0.2    : 0.9674333 ±0%
[2] Alt., r=0.707 !(-0.2<d<0.2) : 0.185251  ±0.01%

Against denominator:
Null, mu1-mu2 = 0
---
Bayes factor type: BFindepSample, JZS


Now if I want to know, if, given my data, the hypothesis that there is no effect $$|\delta| < 0.2$$ is more likely than the complement (there is an effect, $$|\delta| > 0.2$$ ) I can just divide the two BF since they have the same denominator correct?

> bfInterval[1] / bfInterval[2]
Bayes factor analysis
--------------
[1] Alt., r=0.707 -0.2<d<0.2 : 5.222284 ±0.01%

Against denominator:
Alternative, r = 0.707106781186548, mu =/= 0 !(-0.2<d<0.2)
---
Bayes factor type: BFindepSample, JZS


So this ($$BF_{01} = 5.2$$) would now mean, given my data, it is ~5 times more likely that there is no effect (bigger than 0.2) than that there is one, correct?

Also, beyond my insecurities in the interpretation I would like to know if the floor effect in the observed data could be problematic?