Prior checks

Independent-samples t test and ANOVA

The independent samples t test and ANOVA functions should provide the same answers with the default prior settings.

# Create data
x <- rnorm(20)
x[1:10] = x[1:10] + .2
grp = factor(rep(1:2,each=10))

dat = data.frame(x=x,grp=grp)

t.test(x ~ grp, data=dat)

## 
##  Welch Two Sample t-test
## 
## data:  x by grp
## t = 0.5, df = 17, p-value = 0.6
## alternative hypothesis: true difference in means between group 1 and group 2 is not equal to 0
## 95 percent confidence interval:
##  -0.793  1.255
## sample estimates:
## mean in group 1 mean in group 2 
##           0.411           0.180

If the prior settings are consistent, then all three of these numbers should be the same.

as.vector(ttestBF(formula = x ~ grp, data=dat))

## Alt., r=0.707 
##         0.431

as.vector(anovaBF(x~grp, data=dat))

##   grp 
## 0.431

as.vector(generalTestBF(x~grp, data=dat))

##   grp 
## 0.431

Regression and ANOVA

In a paired design with an additive random factor and and a fixed effect with two levels, the Bayes factors should be the same, regardless of whether we treat the fixed factor as a factor or as a dummy-coded covariate.

# create some data
id = rnorm(10)
eff = c(-1,1)*1
effCross = outer(id,eff,'+')+rnorm(length(id)*2)
dat = data.frame(x=as.vector(effCross),id=factor(1:10), grp=factor(rep(1:2,each=length(id))))
dat$forReg = as.numeric(dat$grp)-1.5
idOnly = lmBF(x~id, data=dat, whichRandom="id")

summary(aov(x~grp+Error(id/grp),data=dat))

## 
## Error: id
##           Df Sum Sq Mean Sq F value Pr(>F)
## Residuals  9   49.1    5.46               
## 
## Error: id:grp
##           Df Sum Sq Mean Sq F value Pr(>F)   
## grp        1   25.3   25.33    17.1 0.0025 **
## Residuals  9   13.3    1.48                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

If the prior settings are consistent, these two numbers should be almost the same (within MC estimation error).

as.vector(lmBF(x ~ grp+id, data=dat, whichRandom="id")/idOnly)

## grp + id 
##     23.1

as.vector(lmBF(x ~ forReg+id, data=dat, whichRandom="id")/idOnly)

## forReg + id 
##        23.2

Independent t test and paired t test

Given the effect size $\hat{\delta}=t\sqrt{N_{eff}}$, where the effective sample size N_eff is the sample size in the one-sample case, and $$ N_{eff} = \frac{N_1N_2}{N_1+N_2} $$ in the two-sample case, the Bayes factors should be the same for the one-sample and two sample case, given the same observed effect size, save for the difference from the degrees of freedom that affects the shape of the noncentral t likelihood. The difference from the degrees of freedom should get smaller for a given t as N_eff → ∞.

# create some data
tstat = 3
NTwoSample = 500
effSampleSize = (NTwoSample^2)/(2*NTwoSample)
effSize = tstat/sqrt(effSampleSize)

# One sample
x0 = rnorm(effSampleSize)
x0 = (x0 - mean(x0))/sd(x0) + effSize

t.test(x0)

## 
##  One Sample t-test
## 
## data:  x0
## t = 3, df = 249, p-value = 0.003
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  0.0652 0.3143
## sample estimates:
## mean of x 
##      0.19

# Two sample
x1 = rnorm(NTwoSample)
x1 = (x1 - mean(x1))/sd(x1)
x2 = x1 + effSize

t.test(x2,x1)

## 
##  Welch Two Sample t-test
## 
## data:  x2 and x1
## t = 3, df = 998, p-value = 0.003
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.0656 0.3138
## sample estimates:
## mean of x mean of y 
##  1.90e-01  4.98e-18

These (log) Bayes factors should be approximately the same.

log(as.vector(ttestBF(x0)))

## Alt., r=0.707 
##          1.72

log(as.vector(ttestBF(x=x1,y=x2)))

## Alt., r=0.707 
##          1.77

Paired samples and ANOVA

A paired sample t test and a linear mixed effects model should broadly agree. The two are based on different models — the paired t test has the participant effects substracted out, while the linear mixed effects model has a prior on the participant effects — but we’d expect them to lead to the same conclusions.

These two Bayes factors should be lead to similar conclusions.

# using the data previously defined
t.test(x=dat$x[dat$grp==1],y=dat$x[dat$grp==2],paired=TRUE)

## 
##  Paired t-test
## 
## data:  dat$x[dat$grp == 1] and dat$x[dat$grp == 2]
## t = -4, df = 9, p-value = 0.003
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -3.48 -1.02
## sample estimates:
## mean difference 
##           -2.25

as.vector(lmBF(x ~ grp+id, data=dat, whichRandom="id")/idOnly)

## grp + id 
##     23.1

as.vector(ttestBF(x=dat$x[dat$grp==1],y=dat$x[dat$grp==2],paired=TRUE))

## Alt., r=0.707 
##          18.9

This document was compiled with version 0.9.12-4.6 of BayesFactor (R version 4.4.1 (2024-06-14) on x86_64-pc-linux-gnu).