Drawing values for a predictor once

We will assume that values for our predictor come from a normal distribution. Since I am making this point without any context, let’s just assume a standard normal distribution.

set.seed(1238)

n <- 20
x <- rnorm(n, mean = 0, sd = 1)

We have now drawn our predictor and will use this in a power simulation. Next, we need to specify additional parameters for the simple regression model

with .

beta <- c(beta0 = 1, beta1 = .5)
se <- 1

We will now run a power simulation to find out how many participants we need to find an effect of .

pval <- replicate(2000, {
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m0 <- lm(y ~ 1)
  m1 <- lm(y ~ x)
  anova(m0, m1)[2, "Pr(>F)"]
  }
)

mean(pval < 0.05)

[1] 0.458

This results in a power that is not sufficient, yet, so let us adjust our N and rerun it.

n <- 25
x <- rnorm(n, mean = 0, sd = 1)

pval <- replicate(2000, {
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m0 <- lm(y ~ 1)
  m1 <- lm(y ~ x)
  anova(m0, m1)[2, "Pr(>F)"]
  }
)

mean(pval < 0.05)

[1] 0.7965

OK, this looks good. So we need to collect 25 subjects, right? Looks like the power increased quite a bit for just 5 more subjects. Let’s make sure and rerun it.

set.seed(12)

n <- 25
x <- rnorm(n, mean = 0, sd = 1)

pval <- replicate(2000, {
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m0 <- lm(y ~ 1)
  m1 <- lm(y ~ x)
  anova(m0, m1)[2, "Pr(>F)"]
  }
)

mean(pval < 0.05)

[1] 0.6125

Obviously, the power is strongly influenced by our predictor and what values it takes. What happens when we draw the predictor within each iteration of the power simulation? And are there differences in parameter recovery for these two approaches?

Drawing values for the predictor in each iteration

set.seed(1450)

n <- 25

pval <- replicate(2000, {
  # Draw predictor
  x <- rnorm(n, mean = 0, sd = 1)
  # Simulate dependent variable
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m0 <- lm(y ~ 1)
  m1 <- lm(y ~ x)
  anova(m0, m1)[2, "Pr(>F)"]
  }
)

mean(pval < 0.05)

[1] 0.6365

With this approach we see that N = 25 is not enough to find our effect with a power of 80%.

n <- 37

pval <- replicate(2000, {
  # Draw predictor
  x <- rnorm(n, mean = 0, sd = 1)
  # Simulate dependent variable
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m0 <- lm(y ~ 1)
  m1 <- lm(y ~ x)
  anova(m0, m1)[2, "Pr(>F)"]
  }
)

mean(pval < 0.05)

[1] 0.81

Power estimation with this approach is more stable than when we draw the predictor only once. But what does this mean for parameter recovery?

Parameter recovery

n <- 200

# Draw predictor
x <- rnorm(n, mean = 0, sd = 1)

out <- replicate(2000, {
  # Simulate dependent variable
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m <- lm(y ~ x)
  c(coef(m), sigma = sigma(m))
  }
)

boxplot(t(out))

rowMeans(out)

(Intercept)           x       sigma 
  0.9986695   0.5022547   0.9995984 

n <- 200

out <- replicate(2000, {
  # Draw predictor
  x <- rnorm(n, mean = 0, sd = 1)
  # Simulate dependent variable
  y <- beta[1] + beta[2] * x + rnorm(n, mean = 0, sd = se)
  m <- lm(y ~ x)
  c(coef(m), sigma = sigma(m))
  }
)

boxplot(t(out))

rowMeans(out)

(Intercept)           x       sigma 
  0.9983379   0.5017602   0.9996439 

Conlusions

But that makes the conclusion easy. Then I would recommend drawing the predictor in each iteration of the power simulation since this makes the power estimation much more stable. This is, of course, the easiest version of this simulation. What happens if one draws more than one predictor or if the predictors are correlated?

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] htmlwidgets_1.6.4 compiler_4.6.1    fastmap_1.2.0     cli_3.6.6        
 [5] tools_4.6.1       htmltools_0.5.9   otel_0.2.0        yaml_2.3.12      
 [9] rmarkdown_2.32    knitr_1.52        jsonlite_2.0.0    xfun_0.62        
[13] digest_0.6.39     rlang_1.3.0       evaluate_1.0.5   

The example

Suppose you roll a volleyball from the back boundary line to the middle line under the net. The ball will travel a distance of 9m. It will start fast, but gradually slow down to come to a stop under the net. The top of Figure 1 shows the distance-time function. For example, after the first second, the ball has travelled 5m, after another second, it has hit the 8m mark, and within the third second, it only travels 1m to its final resting point, so it has constantly slowed down. The bottom of Figure 1 shows the instantaneous velocity. The ball starts out with 6m/s, but within the first second, it slows down to 4m/s. After three seconds, the velocity is 0m/s.

Code

s <- function(t) 6*t - t^2
v <- function(t) 6 - 2*t

par(mfrow = c(2, 1), mai = c(.6, .6, .1, .1), mgp = c(2, .7, 0))
curve(s, 0, 3, xname = "t", col = "darkblue",
      xlab = "Time, t (in seconds)", ylab = "Travelled distance (m), s(t)",
      panel.first = {
        grid(lty = 1)
        abline(h = c(0, 9), lty = 2)
})
text(1.4, 0.5, "Back boundary line")
text(1.4, 8.5, "Middle line, net")
text(0.2, 4.5, "Ball movement", srt = 90)
arrows(0, 0.1, 0, 8.9, 0.05)
curve(v, 0, 3, xname = "t", col = "darkblue",
      xlab = "Time, t (in seconds)",
      ylab = "Instantaneous velocity (m/s), v(t)",
      panel.first = {
        polygon(c(0, 3, 0), c(0, 0, 6), col = "lightblue")
        grid(lty = 1)
})

Formally, the distance-time function displayed in Figure 1 is (it shows the travelled distance over time), and the instantaneous velocity is its derivate (it shows how much the travelled distance changes over time). In general, for well-behaved functions, there is the relation which can be interpreted in the context of the example as follows: If we “sum up” (integrate) small time increments multiplied by the velocity the ball had at this time, we should get the total distance that the ball has travelled. And indeed, the ball has travelled 9m. And there are two equivalent ways to get the solution, via the antiderivative and via basic geometry: The area of the triangle under the curve of is By whatever way we arrive there, we see that the integral of the instantaneous velocity amounts to the total travelled distance.

Power simulation for ANCOVA (two time points)

This power simulation for the original pre/post design assumed that the data will be analyzed with an ANCOVA model (based on the paper by Vickers and Altman 2001). This is the original power simulation based on values found for previous data collected by this PhD student.

n <- 87

# Draw plausible pre values and setup data frame
dat <- data.frame(
  pre = rnorm(n, mean = 12.5, sd = 3.5),
  grp = factor(rep(1:3, each = n/3), labels = c("g1", "g2", "g3"))
)

# Specify the minimum relevant average treatment effects and set the remaining
# parameters to plausible values

# mu = mean learning outcome (expected pre-post effect in reference group)
# b  = slope between pre and post values
# a2 = expected difference between group 2 and group 1
# a3 = expected difference between group 3 and group 1
# se = sigma of ANCOVA model

beta <- c(mu = 3, b = .7, a2 = 1.5, a3 = 3)
means <- model.matrix(~ pre + grp, dat) %*% beta
se <- 2

# Estimate power
pval <- replicate(2000, {
  post <- means + rnorm(n, sd = se)
  m <- lm(post ~ pre + grp, dat)
  summary(m)$coef["grpg2", "Pr(>|t|)"]
})

mean(pval < 0.05)

[1] 0.8015

I will not go into detail, why this is the appropriate analysis for this design. Maybe this will become another post at some point. Read the paper by Vickers and Altman (2001) in the meantime.

Power simulation for mixed-effects model (three time points)

Unfortunately, when we add another time point into the mix, we now have to switch the analysis to a mixed-effects model with a group x time interaction. It was also much harder for the PhD student to term her hypothesis in terms of model parameters. So, I made her draw the expected pattern and add mean values to the plot. This is what she came up with.

n <- 3

design <- data.frame(id = factor(rep(1:n, each = 3)),
                  time = factor(rep(c("pre", "post", "delayed"), times = n),
                                levels = c("pre", "post", "delayed")),
                  grp = factor(rep(1:3, each = 3, times = n / 3),
                               labels = c("g1", "g2", "g3"))
)


means <- c(5.5, 6.8, 6.5, 5.5, 6.8, 6.5, 5.5, 8.5, 8)

lattice::xyplot(means ~ time, data = design, groups = grp,
                type = "a",
                ylim = 0:10,
                auto.key = TRUE)

We can now use these means and the “design” data frame to calculate the fixed model parameters with

mm <- model.matrix(~ grp * time, data = design)
(beta <- as.vector(solve(mm) %*% means))

[1] 5.5 0.0 0.0 1.3 1.0 0.0 1.7 0.0 1.5

Now, we still need the variance terms for the random effects. For simplicity reasons, we still stick to a random intercept model here. So we need to specify a residual variance and a person variance.

sp <- 1.5
se <- 2

The power simulation then looks something like this.

library("lme4")

n <- 120

dat <- data.frame(id = factor(rep(1:n, each = 3)),
                  time = factor(rep(c("pre", "post", "delayed"), times = n),
                                levels = c("pre", "post", "delayed")),
                  grp = factor(rep(1:3, each = 3, times = n / 3),
                               labels = c("g1", "g2", "g3"))
)


mm <- model.matrix(~ grp * time, data = dat)
means <- as.vector(mm %*% beta)


pval <- replicate(1000, {
  y <- means +                                        # structural part
       rep(rnorm(n, mean = 0, sd = sp), each = 3) +   # person effect
       rnorm(n * 3, mean = 0, sd = se)                # random error
  m0 <- lmer(y ~ grp + time + (1 | id), data = dat, REML = FALSE)
  m1 <- lmer(y ~ grp * time + (1 | id), data = dat, REML = FALSE)
  # Likelihood-Ratio-Test
  anova(m0, m1)["m1", "Pr(>Chisq)"]
  }
)

mean(pval < 0.05)

[1] 0.774

For more examples how to run power simulations in R see Wickelmaier (2026).

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] lme4_2.1-0   Matrix_1.7-6

loaded via a namespace (and not attached):
 [1] digest_0.6.39     fastmap_1.2.0     xfun_0.62         lattice_0.23-1   
 [5] splines_4.6.1     knitr_1.52        htmltools_0.5.9   rmarkdown_2.32   
 [9] Rdpack_2.6.6      cli_3.6.6         nloptr_2.2.1      grid_4.6.1       
[13] reformulas_0.4.4  compiler_4.6.1    rbibutils_2.4.1   boot_1.3-32      
[17] tools_4.6.1       nlme_3.1-171      minqa_1.2.8       evaluate_1.0.5   
[21] Rcpp_1.1.2        yaml_2.3.12       otel_0.2.0        rlang_1.3.0      
[25] jsonlite_2.0.0    htmlwidgets_1.6.4 MASS_7.3-66      

Data simulation

set.seed(1109)

# Set effect coding to get the treatment effect for an "average" study
options(contrasts = c("contr.sum", "contr.poly"))

# Simulate data for 10 studies
n <- c(20, 50, 30, 26, 10, 20, 30, 100, 20, 46)
mu <- 1                                          # mean in control group
d_true <- 1                                      # "true" effect

What excatly did we do here? We assume that we conducted 10 different studies with a different number of participants ranging from 10 to 100. We have two groups: Control (ctr) vs. Treatment (trt) and want to compare the treatment effect (the difference in the DV between the two groups). We set the mean in the control group to and assume a true treatment effect .

sd <- 1

The next thing we need to decide on is the variation in the data. We set the varianz in the data to .

d_obs <- rnorm(length(n), mean = d_true, sd = 2 / sqrt(n))

Next we draw the effect sizes for each of the studies. We assume that they come from a normal distribution with . Doing it like this is in line with a fixed-effect model for meta-analysis. Since we are in the “theoretical” mindframe that all studies originated in the same lab under very similar conditions this might be a reasonable choice? Go ahead and adjust the simulation so the underlying model corresponds to a random-effects model for meta-analysis. Do the results change? (I did not try it.)

I must admit that I have no good idea how to draw the effect sizes for the study. I decided to weigh the variance by so that for bigger the variance is smaller and, therefore, the probability to draw a value closer to the “true” value is higher. I added this later on. When you leave this out and draw all effect sizes with same variance, then the meta-analysis has a much wider confidence intervall for the estimate of the treatment effect. I guess that makes sense. Well, I guess doing it with a weighted variance is in line with a frequentist view and also with the underlying assumptions for meta-analysis.

dat_list <- list()

for (i in 1:10) {
  dat0 <- data.frame(id = factor(1:n[i]),
                     group = factor(rep(c("ctr", "trt"), each = n[i] / 2),
                                    levels = c("trt", "ctr")),
                     resp = c(rnorm(n[i] / 2, mean = mu, sd = sd),
                              rnorm(n[i] / 2, mean = mu + d_obs[i], sd = sd)),
                     n = n[i],
                     study = factor(paste0("study", sprintf("%02d", i))))
  dat_list[[i]] <- dat0
}

rm(dat0)

# Integrate into one dataframe
dat_all <- do.call("rbind", dat_list)

bwplot(resp ~ factor(group, levels = c("ctr", "trt")) | study, data = dat_all,
       layout = c(5, 2))

Analyze the data

# Analyzing all data in one model
m1 <- lm(resp ~ group, data = dat_all)
summary(m1)

Call:
lm(formula = resp ~ group, data = dat_all)

Residuals:
     Min       1Q   Median       3Q      Max 
-3.04447 -0.70303  0.01982  0.79411  2.80768 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   1.4982     0.0539  27.798   <2e-16 ***
group1        0.4772     0.0539   8.854   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.011 on 350 degrees of freedom
Multiple R-squared:  0.183, Adjusted R-squared:  0.1806 
F-statistic: 78.39 on 1 and 350 DF,  p-value: < 2.2e-16

# Estimated treatment effect
coef(m1)["group1"] * 2

   group1 
0.9543521 

This ignores completely that the data were collected in different studies. Let us add a covariate study to the model and see what the estimated treatment effect for an “average” study looks like.

# Adding study as a covariate to the model
m2 <- lm(resp ~ group + study, data = dat_all)
summary(m2)

Call:
lm(formula = resp ~ group + study, data = dat_all)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.78958 -0.70356  0.00348  0.72203  2.72573 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.44624    0.06319  22.888  < 2e-16 ***
group1       0.47718    0.05280   9.038  < 2e-16 ***
study1      -0.44367    0.20795  -2.133  0.03360 *  
study2       0.10850    0.14033   0.773  0.43997    
study3       0.49978    0.17367   2.878  0.00426 ** 
study4       0.24583    0.18490   1.330  0.18455    
study5      -0.01158    0.28722  -0.040  0.96786    
study6       0.22640    0.20795   1.089  0.27706    
study7      -0.31949    0.17367  -1.840  0.06669 .  
study8       0.17633    0.10883   1.620  0.10609    
study9      -0.38748    0.20795  -1.863  0.06328 .  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.9906 on 341 degrees of freedom
Multiple R-squared:  0.2361,    Adjusted R-squared:  0.2137 
F-statistic: 10.54 on 10 and 341 DF,  p-value: 1.442e-15

# Estimated treatment effect
coef(m2)["group1"] * 2

   group1 
0.9543521 

The effect is exactly the same… Is this always true? Why would I even add the covariate then?

# Doing a metaanalysis on these data
datm    <- aggregate(resp ~ study + group, data = dat_all, FUN = mean)
datm$sd <- aggregate(resp ~ study + group, data = dat_all, FUN = "sd")[, 3]
datm$n  <- aggregate(resp ~ study + group, data = dat_all, FUN = "length")[, 3]

es <- escalc("MD",
             m1i  = datm$resp[datm$group == "trt"],
             m2i  = datm$resp[datm$group == "ctr"],
             sd1i = datm$sd[datm$group == "trt"],
             sd2i = datm$sd[datm$group == "ctr"],
             n1i  = datm$n[datm$group == "trt"],
             n2i  = datm$n[datm$group == "ctr"])

res <- rma(yi, vi, data=es)

forest(res)

Conclusions

# Compare true and estimated treatment effects
est <- c(d_true, res$beta, coef(m2)[2] * 2)
cilb <- c(NA, res$ci.lb, confint(m2)[2, 1] * 2)
ciub <- c(NA, res$ci.ub, confint(m2)[2, 2] * 2)
cbind(est, cilb, ciub)

             est      cilb     ciub
       1.0000000        NA       NA
       0.9767077 0.7728563 1.180559
group1 0.9543521 0.7466444 1.162060

par(mai = c(.6, .7, .1, .1), mgp = c(2.4, 1, 0))
plot(est, axes = F, ylim = c(-1,3), pch = 16, cex = 2, xlab = "", 
     ylab = "Mean difference")

arrows(1:3, cilb, 1:3, ciub, code = 3, length = .05, angle = 90)

axis(1, at = 1:3, labels = c("true", "ma", "lm"))
axis(2)

abline(h = est[1], lty = 3)
abline(h = 0, lty = 3, col = "red")
box()

The results (for this one draw of 10 studies!) show that the meta-analysis and the regression model with study as covariate show pretty similar estimates for the treatment effect. The confidence intervalls look also similar. I guess, the main advantage of the meta-analysis is the beautiful plot that you can create with metafor out of the box… I am still not sure why we even need the covariate in this. Guess, I have to think about this some more.

Do it ten times for 10 studies

I repeated the code from above ten times to get an idea how stable the results are. Here are the plots. Looks like the estimate obtained via meta-analysis is a bit more “uncertain”. Guess, that is a good thing.

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] metafor_5.2-1       numDeriv_2016.8-1.1 metadat_1.6-0      
[4] Matrix_1.7-6        lattice_0.23-1     

loaded via a namespace (and not attached):
 [1] digest_0.6.39     fastmap_1.2.0     xfun_0.62         knitr_1.52       
 [5] htmltools_0.5.9   rmarkdown_2.32    mathjaxr_2.0-0    cli_3.6.6        
 [9] grid_4.6.1        compiler_4.6.1    tools_4.6.1       nlme_3.1-171     
[13] evaluate_1.0.5    yaml_2.3.12       otel_0.2.0        rlang_1.3.0      
[17] jsonlite_2.0.0    htmlwidgets_1.6.4

Simulate skewed data

I found an R package that lets you simulate data from a “skew normal distribution” (I guess, that’s an oxymoron, but what the heck): fGarch.

With the function fGarch::rsnorm() we can simulate skewed data. The argument xi controls the extend of the skewness. Let us have a look.

library("fGarch")

xi <- 2

x <- rsnorm(1000, mean = 0, sd = 1, xi = xi)
hist(x, freq = FALSE, ylim = c(0, .5), border = "white", main = "")
curve(dsnorm(x, xi = xi), -2.5, 4.5, add = TRUE, lwd = 2)

Testing for normality

The first step in an analysis procedure like this is to decide which test to use when testing for normality. There are many different options. Honestly, I have never given too much thought to this. I learned at some point that most of these tests have bad statistical properties and luckily I was never in s situation where I had to apply these tests. Two common choices are the Kolmogorov-Smirnov test and the Shapiro-Wilk test of normality. Both are available in the stats package in R. I ran a small simulation and was a bit suprised by the results. (I am sure that shows my ignorance on the topic and that there are much better choices for testing normality. However, these are common choices that I regularly encounter in statistics consultation.)

compare_normtest <- function(xi, n_group, test = "Shapiro", all_skewed = TRUE) {
  pval <- replicate(2000, {
    if (all_skewed) {
      resp1 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)
    } else {
      resp1 <- rsnorm(n_group, mean = 0, sd = 1, xi = 1)
    }
    resp2 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)
    resp3 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)
    dat <- data.frame(cond = rep(c("c1", "c2", "c3"), each = n_group),
                      resp = c(resp1, resp2, resp3))
    ifelse(test == "Shapiro",
           shapiro.test(dat$resp)$p.value,
           ks.test(dat$resp, "pnorm", mean = 0, sd = 1)$p.value)
  })
  mean(pval < .05)
}

comb <- expand.grid(xi = c(1, 3, 5),
                    n_group = c(10, 50, 100),
                    test = factor(c("Shapiro", "Kolmogorov-Smirnov")),
                    all_skewed = c(TRUE, FALSE))

comb$pwr <- mapply(compare_normtest, n_group = comb$n_group,
                   xi = comb$xi, test = comb$test, all_skewed = comb$all_skewed)

lattice::xyplot(pwr ~ n_group | factor(all_skewed,
                                       labels = c("one normal group",
                                                  "all groups skewed")) + test,
                data = comb, groups = xi,
                type = c("g", "b"),
                xlab = "N in each group",
                ylab = "Power",
                auto.key = list(space = "top", columns = 3, title = expression(xi)))

Based on this plot I am inclined to go with the Shapiro test here. But go on and try it out with other tests testing the normality assumption yourself!

One-way ANOVA vs Kruskal-Wallis rank sum test

Next, I will simulate data for three independent groups. I will simulate data for each of these groups separately, but always assuming the same size of skewness and the same sample size N. I will let vary from 1 (normal distribution) to 10 (very skewed) and N take the values 10, 100, and 500. To each data set, we will fit an ANOVA with aov(), a Kruskal-Wallis rank sum test with kruskal.test() which is a nonparametric test meaning its only assumption is that the data are continuous but not which distribution it follows. Additionally, we will first test for normality of the data and then decide if we do an ANOVA or a Kruskal-Wallis rank sum test based on the results. Hence, we condition our analysis on the data.

compare_parametric <- function(xi, n_group = 20, test, all_skewed = TRUE) {
  pval <- replicate(1000, {
    if (all_skewed) {
      resp1 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)
    } else {
      resp1 <- rsnorm(n_group, mean = 0, sd = 1, xi = 1)
    }
    resp2 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)
    resp3 <- rsnorm(n_group, mean = 0, sd = 1, xi = xi)

    dat <- data.frame(cond = factor(rep(c("c1", "c2", "c2"), each = n_group)),
                      resp = c(resp1, resp2, resp3))
    if (test == "ANOVA") {
      anova(lm(resp ~ cond, dat))["cond", "Pr(>F)"]
    } else if (test == "Kruskal") {
      kruskal.test(resp ~ cond, dat)$p.value
    } else {
      ifelse(shapiro.test(dat$resp)$p.value >= .05,
             anova(lm(resp ~ cond, dat))["cond", "Pr(>F)"],
             kruskal.test(resp ~ cond, dat)$p.value)
    }
  })
  mean(pval < .05)
}

comb <- expand.grid(xi = 1:10, n_group = c(10, 100, 500),
                    test = c("ANOVA", "Kruskal", "Conditioned"),
                    all_skewed = c(TRUE, FALSE))

comb$pwr <- mapply(compare_parametric,
                   xi = comb$xi,
                   n_group = comb$n_group,
                   test = comb$test,
                   all_skewed = comb$all_skewed)

lattice::xyplot(pwr ~ xi | factor(all_skewed,
                                  labels = c("one normal group",
                                             "all groups skewed")) +
                           factor(n_group,
                                  labels = paste("N =", unique(n_group))),
                data = comb,
                groups = test,
                type = c("g", "b"),
                xlab = expression(xi),
                ylab = "Type I Error Rate",
                ylim = c(0, 0.5),
                auto.key = list(space = "top", columns = 3,
                                title = "N in each group"))

So what did we do here again? We tested for a group effect for a one-way design. We did not have an effect here, so the Type I Error Rate should not exceed , no matter what sample size we have.

I would have expected that the error rate would actually increase for the ANOVA with increasing violation of the normality assumption (that means increasing ). We can see that this is not the case. For the ANOVA the error rate stays at 5%, as it should. Interestingly, this is also TRUE for the Kruskal-Wallis rank sum test and for the conditioned test, when data in all groups is skewed. If data in one group is normal, then the Type I Error Rate increases for the Kruskal-Wallis rank sum test and this also implies an increase for the conditioned test. This is very pronounced for large N. I think this is very suprising and do not have an explanation for it.

But we do have an overall result: Stick to the ANOVA that you preregistered and don’t bother with testing for normality.

I only looked at one way to violate normality here. Another way could be to look at bimodal distributions or draw data from other distributions than the normal distribution. I do expect the results to look similar, but have not tried it, yet.

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] fGarch_4052.93

loaded via a namespace (and not attached):
 [1] fBasics_4052.98     digest_0.6.39       fastmap_1.2.0      
 [4] Matrix_1.7-6        xfun_0.62           lattice_0.23-1     
 [7] spatial_7.3-19      knitr_1.52          gbutils_0.5.1      
[10] htmltools_0.5.9     rmarkdown_2.32      Rdpack_2.6.6       
[13] cli_3.6.6           cvar_0.6.1          grid_4.6.1         
[16] timeDate_4052.112   compiler_4.6.1      rbibutils_2.4.1    
[19] tools_4.6.1         evaluate_1.0.5      yaml_2.3.12        
[22] otel_0.2.0          timeSeries_4052.112 rlang_1.3.0        
[25] jsonlite_2.0.0      htmlwidgets_1.6.4  

Power simulation

pwrFun <- function(d, model, raneff_class = TRUE, nrep = 1000) {
  pval <- replicate(nrep, {
    # switch between data generating moel
    if (raneff_class) {
       y <- 5 + d * lme4::dummy(dat$group) +
         rnorm(18, mean = 0, sd = .5)[dat$class] +
         rnorm(90, mean = 0, sd = .5)
    } else {
       y <- 5 + d * lme4::dummy(dat$group) + rnorm(90, mean = 0, sd = .5)
    }
    # switch between analysis model
    if (model == "LMM") {
       m0 <- lme4::lmer(y ~ 1     + (1 | class), data = dat, REML = FALSE)
       m1 <- lme4::lmer(y ~ group + (1 | class), data = dat, REML = FALSE)
       anova(m0, m1)["m1", "Pr(>Chisq)"]
    } else if (model == "ttest") {
       t.test(y ~ group, data = dat, var.equal = TRUE)$p.value
    }
    }
  )
  # calculate power
  mean(pval < 0.05)
}

pwrFun(0.2, model = "LMM",   raneff_class = TRUE)

[1] 0.151

pwrFun(0.2, model = "ttest", raneff_class = TRUE)

[1] 0.404

We can see that the power to detect a group effect of 0.2 for for the LMM and the t-test differs quite a bit when the data generating model includes a random intercept for class.

pwrFun(0.2, model = "LMM",   raneff_class = FALSE)

[1] 0.415

pwrFun(0.2, model = "ttest", raneff_class = FALSE)

[1] 0.434

If we generate the data without a random intercept for class – meaning the model does not know anything of the classes – then the power for the t-test and the LMM look very similar.

comb <- expand.grid(d = c(0, 0.2, 0.4, 0.6),
                    raneff_class = c(TRUE, FALSE),
                    model = factor(c("LMM", "ttest"), levels = c("ttest", "LMM")))

comb$pwr <- mapply(pwrFun,
                   d = comb$d,
                   model = comb$model,
                   raneff_class = comb$raneff_class)

lattice::xyplot(pwr ~ d | factor(raneff_class,
                                 labels = c("w/o class effect", "w class effect")),
                data = comb,
                groups = model,
                type = c("b", "g"),
                auto.key = list(corner = c(1, 0)))

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] lme4_2.1-0        digest_0.6.39     fastmap_1.2.0     Matrix_1.7-6     
 [5] xfun_0.62         lattice_0.23-1    splines_4.6.1     knitr_1.52       
 [9] htmltools_0.5.9   rmarkdown_2.32    Rdpack_2.6.6      cli_3.6.6        
[13] nloptr_2.2.1      grid_4.6.1        reformulas_0.4.4  compiler_4.6.1   
[17] rbibutils_2.4.1   boot_1.3-32       tools_4.6.1       nlme_3.1-171     
[21] minqa_1.2.8       evaluate_1.0.5    Rcpp_1.1.2        yaml_2.3.12      
[25] otel_0.2.0        rlang_1.3.0       jsonlite_2.0.0    htmlwidgets_1.6.4
[29] MASS_7.3-66      

A single well formatted R script with a header

If you just want to provide the analysis code for a single research paper then a single R script is often enough. This R script will include only the code needed to exactly reproduce the results presented in your paper. You will probably have done more analyses, e.g., some cross checks on your data or you looked at some contingency tables or similar. However, if those did not end up in your paper, do not include them in this script.

You should add a header to this script which contains some meta data. This is a simple example:

# author_year_journal.R
# Analysis accompanying Author (Year). Title of my paper. *Journal*, Vol., pages
#
# Input:  mydata.csv
# Output: figure1.png
#         figure2.png
#
# Contents: (1) Data preparation
#           (2) Descriptive statistics
#           (3) Hypothesis tests
#
# Last mod: 2026-08-31, Your Name

Like I said, this script should only include the code to reproduce your results. Do not add any fancy data manipulations or similar. Rather, create a data file mydata.csv that is already in the correct format. If you do several analyses that need different data formats, it might make sense to do the reformatting within the script, but always consider the readability of the script. You can always write two analysis scripts for two data sets. (This is advice that reproducibilty experts will probably challenge and rightfully though. But we are not experts, yet.)

Load packages at the beginning of your script

Load packages at the very beginning of your script and only once. For a reproducible script, try to use as few packages as possible. If another person wants to tun your script, it will irk them when they have to install a bunch of packages to do so.

We will add the following to our script right after the header.

library("lattice")
library("lme4")

Use package::function()

If you use a single function from a package you might consider not loading the package, but using package::function() instead. The lattice package, e.g., is always pre-installed with R. If you just want to use a single function for one plot from it, there is no need to load the whole package. You can maybe just use lattice::xyplot(...).

It is sometimes also useful to explicitly state a function call like this, even if you loaded the whole package, since it makes it explicit from which package the function comes from.

Follow a style guide / consistent style

R has no mandatory or commonly accepted style guide. (A style guide consists of certain rules or suggestions how code should be organized.) However, Hadley Wickham and Google developed style guides which are now widely accepted:

• https://google.github.io/styleguide/Rguide.html
• https://style.tidyverse.org/

It is always a good idea to follow a style guide and not “create” your own rules (if you deviate, be consistent!). A style guide helps with

• Keeping code clean which is easier to read and interpret
• Making it easier to catch and fix mistakes
• Making it easier for others to follow and adapt your code
• Preventing possible problems, e.g., avoiding dots in function names

The following rules are based on the style guide by Hadley Wickham. I highly suggest that you follow these rules in (all) your scripts.

# Good
x[, 1]

# Bad
x[,1]
x[ ,1]
x[ , 1]

## Parantheses
# Good
mean(x, na.rm = TRUE)

# Bad
mean (x, na.rm = TRUE)
mean( x, na.rm = TRUE )

## Infix operators
# Good
height <- (feet * 12) + inches
mean(x, na.rm = TRUE)

# Bad
height<-feet*12+inches
mean(x, na.rm=TRUE)

# Good
sqrt(x^2 + y^2)
df$z
x <- 1:10

# Bad
sqrt(x ^ 2 + y ^ 2)
df $ z
x <- 1 : 10

# Good
package?stats
?mean

# Bad
package ? stats
? mean

# Good
list(
  total = a + b + c,
  mean  = (a + b + c) / n
)

# Also fine
list(
  total = a + b + c,
  mean = (a + b + c) / n
)

# Good
mean(1:10, na.rm = TRUE)

# Bad
mean(x = 1:10, , FALSE)
mean(, TRUE, x = c(1:10, NA))

# Good
rep(1:2, times = 3)
cut(1:10, breaks = c(0, 4, 11))

# Bad
rep(1:2, t = 3)
cut(1:10, br = c(0, 4, 11))

Use sensible variable and function names

Variable and function names should use only lowercase letters, numbers, and _. Use underscores (so called snake case) to separate words within a name.

# Good
day_one
day_1

# Bad
DayOne
dayone

Generally, variable names should be nouns and function names should be verbs. Strive for names that are concise and meaningful.

# Good
day_one

# Bad
first_day_of_the_month
djm1

Avoid re-using names of common functions and variables.

# Bad
T <- FALSE
c <- 10
mean <- function(x) sum(x)

Use |> not %>%

I do not use a lot of funtions from the tidyverse since I learned R before it was around. I also do not use a lot pipes in my code – same reason. However, when you use pipes, make sure that you use the native R pipe |> and not the tidyverse pipe %>%. The native pipe works with base R, for the tidyverse one you need to load extra packages. This itself is a good enough reason to not use it anymore.

Test your script with source()

After you wrote your short and well structured script, close your R session (in RStudio you can use the shortcut Ctrl + Shift + F10). In a new and clean R session navigate to your script (e.g., with setwd()) and then run

source("myscript.R")

There should be no error and all output, like figures, should be created and saved.

This is the first minimal step to get a script reproducible. Do not ever skip it.

Let a colleague test the script

After you sourced your script and everything looked fine, zip the folder that contains the script and the data set(s) used in it and send it together with a copy of the accompanying paper to a colleague. Ask them to review and run the code. They should be able to

• reproduce all results in the paper
• create all the figures in the paper
• understand what the code does

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] htmlwidgets_1.6.4 compiler_4.6.1    fastmap_1.2.0     cli_3.6.6        
 [5] tools_4.6.1       htmltools_0.5.9   otel_0.2.0        yaml_2.3.12      
 [9] rmarkdown_2.32    knitr_1.52        jsonlite_2.0.0    xfun_0.62        
[13] digest_0.6.39     rlang_1.3.0       evaluate_1.0.5   

Model with one categorical factor

Let us simulate some data first, for subjects and a group predictor with levels ctr for the control group and trt for the treatment group.

dat <- data.frame(id    = factor(1:10),
                  group = factor(c("ctr", "trt")),
                  resp  = c(rnorm(5, mean = 1), rnorm(5)))

aggregate(resp ~ group, data = dat, FUN = mean)

  group       resp
1   ctr  0.2609670
2   trt -0.1312885

Per default R uses so-called dummy coding to represent factors. This means that the reference category (here ctr) is set to 0 and the second category is set to 1. R takes the first factor level as reference category in dummy coding.

contrasts(dat$group)

    trt
ctr   0
trt   1

You can check the default settings of your R session with

options()$contrasts

        unordered           ordered 
"contr.treatment"      "contr.poly" 

The first entry "contr.treatment" indicates dummy coding.

lm1 <- lm(resp ~ group, data = dat)
summary(lm1)

Call:
lm(formula = resp ~ group, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.5820 -0.4338 -0.1752  0.6928  1.6374 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)   0.2610     0.4451   0.586    0.574
grouptrt     -0.3923     0.6295  -0.623    0.551

Residual standard error: 0.9953 on 8 degrees of freedom
Multiple R-squared:  0.0463,    Adjusted R-squared:  -0.07292 
F-statistic: 0.3883 on 1 and 8 DF,  p-value: 0.5505

coef(lm1)[1]    # mean ctr

(Intercept) 
   0.260967 

sum(coef(lm1))  # mean trt

[1] -0.1312885

contrasts(dat$group) <- "contr.sum"
contrasts(dat$group)

    [,1]
ctr    1
trt   -1

lm2 <- lm(resp ~ group, dat)
summary(lm2)

Call:
lm(formula = resp ~ group, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.5820 -0.4338 -0.1752  0.6928  1.6374 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.06484    0.31473   0.206    0.842
group1       0.19613    0.31473   0.623    0.551

Residual standard error: 0.9953 on 8 degrees of freedom
Multiple R-squared:  0.0463,    Adjusted R-squared:  -0.07292 
F-statistic: 0.3883 on 1 and 8 DF,  p-value: 0.5505

sum(coef(lm2))    # mean ctr

[1] 0.260967

-diff(coef(lm2))  # mean trt

    group1 
-0.1312885 

mean(dat$resp)

[1] 0.06483924

Model with two categorical factors

Let us again simulate some data first. This time for subjects, a group predictor like before and an additional factor study with two levels s1 and s2.

dat <- data.frame(id    = factor(1:20),
                  group = factor(rep(c("ctr", "trt"), each = 10)),
                  study = factor(rep(c("s1", "s2"), each = 5)),
                  resp  = c(rnorm(10, mean = 1),
                            rnorm(5, mean = 0.5),
                            rnorm(5)))

aggregate(resp ~ group + study, data = dat, FUN = mean)

  group study       resp
1   ctr    s1  1.0575195
2   trt    s1  0.9622200
3   ctr    s2  1.2211857
4   trt    s2 -0.2307511

mean(dat$resp)

[1] 0.7525435

lattice::xyplot(resp ~ group, data = dat, groups = study, type = "a",
                auto.key = list(corner = c(0, 1)))

Both factors are now dummy coded, since we did not tell R to do otherwise.

contrasts(dat$group)

    trt
ctr   0
trt   1

contrasts(dat$study)

   s2
s1  0
s2  1

lm3 <- lm(resp ~ group * study, data = dat)
summary(lm3)

Call:
lm(formula = resp ~ group * study, data = dat)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.59696 -0.61529  0.00399  0.56974  1.86454 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)  
(Intercept)        1.0575     0.4582   2.308   0.0347 *
grouptrt          -0.0953     0.6480  -0.147   0.8849  
studys2            0.1637     0.6480   0.253   0.8038  
grouptrt:studys2  -1.3566     0.9165  -1.480   0.1582  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.025 on 16 degrees of freedom
Multiple R-squared:  0.2826,    Adjusted R-squared:  0.1481 
F-statistic: 2.101 on 3 and 16 DF,  p-value: 0.1404

coef(lm3)[1]

(Intercept) 
    1.05752 

coef(lm3)[1] + coef(lm3)[2]

(Intercept) 
    0.96222 

coef(lm3)[1] + coef(lm3)[3]

(Intercept) 
   1.221186 

coef(lm3)[1] + coef(lm3)[2] + coef(lm3)[3] + coef(lm3)[4]   # sum(coef(lm3))

(Intercept) 
 -0.2307511 

contrasts(dat$group) <- "contr.sum"
contrasts(dat$study) <- "contr.sum"

lm4 <- lm(resp ~ group * study, dat)
summary(lm4)

Call:
lm(formula = resp ~ group * study, data = dat)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.59696 -0.61529  0.00399  0.56974  1.86454 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)   
(Intercept)     0.7525     0.2291   3.285  0.00467 **
group1          0.3868     0.2291   1.688  0.11075   
study1          0.2573     0.2291   1.123  0.27796   
group1:study1  -0.3392     0.2291  -1.480  0.15822   
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.025 on 16 degrees of freedom
Multiple R-squared:  0.2826,    Adjusted R-squared:  0.1481 
F-statistic: 2.101 on 3 and 16 DF,  p-value: 0.1404

coef(lm4)[1] + coef(lm4)[2] + coef(lm4)[3] + coef(lm4)[4]

(Intercept) 
    1.05752 

coef(lm4)[1] - coef(lm4)[2] + coef(lm4)[3] - coef(lm4)[4]

(Intercept) 
    0.96222 

coef(lm4)[1] + coef(lm4)[2] - coef(lm4)[3] - coef(lm4)[4]

(Intercept) 
   1.221186 

coef(lm4)[1] - coef(lm4)[2] - coef(lm4)[3] + coef(lm4)[4]   # sum(coef(lm3))

(Intercept) 
 -0.2307511 

Setting contrasts globally

You can also change the settings for the contrast coding globally in R.

options(contrasts = c("contr.treatment", "contr.poly"))

options()$contrasts

[1] "contr.treatment" "contr.poly"     

contrasts(factor(1:3))

  2 3
1 0 0
2 1 0
3 0 1

options(contrasts = c("contr.sum", "contr.poly"))

options()$contrasts

[1] "contr.sum"  "contr.poly"

contrasts(factor(1:3))

  [,1] [,2]
1    1    0
2    0    1
3   -1   -1

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] htmlwidgets_1.6.4 compiler_4.6.1    fastmap_1.2.0     cli_3.6.6        
 [5] tools_4.6.1       htmltools_0.5.9   otel_0.2.0        yaml_2.3.12      
 [9] rmarkdown_2.32    grid_4.6.1        knitr_1.52        jsonlite_2.0.0   
[13] xfun_0.62         digest_0.6.39     rlang_1.3.0       lattice_0.23-1   
[17] evaluate_1.0.5   

Example

This is an example where three different media are compared to each other: A Multi-Touch Table (MTT) and two different Virtual Reality conditions (VR1 and VR2). These are all presented to every subject (within-factor mediatype). Additionally, we have a between-factor interaction with levels passive and active. Subjects can either actively interact for all media types or they can just passively look at things presented to them with the three different media types. The dependent varible is immersion. Subjects should feel more immersed when actively interacting and Virtual Reality should also lead to more immersion than a Multi-Touch Table.

library(lme4)

# Data simulation for a 2 (between) x 3 (within) design
set.seed(1046)

n   <- 100
dat <- data.frame(id          = factor(rep(1:n, each = 3)),
                  mediatype   = factor(c("MTT", "VR1", "VR2")),
                  interaction = factor(rep(c("active", "passive"), each = 3),
                                       levels = c("passive", "active")))

## Set parameters
beta  <- c(.5, .2, .3, .4, .7, .4)
sigma <- .8
spi   <- .5
theta <- spi / sigma

## Simulate data
dat$y <- simulate(~ 1 + mediatype + interaction + mediatype:interaction +
                  (1 | id),
                  newparam = list(beta = beta, theta = theta, sigma = sigma),
                  newdata = dat)$sim_1

## Visualize simulated data
datm <- aggregate(y ~ mediatype + interaction, data = dat, FUN = mean)

colors <- c("#3CB4DC", "#91C86E")

par(mai = c(.8, .8, .1, .1), mgp = c(2.4, 1, 0))
plot(y ~ I(1:3), data = datm[datm$interaction == "passive", ],
     xlab = "Media type",
     ylab = "Immersion",
     xlim = c(0.5, 3.5),
     ylim = c(-2, 4),
     axes = FALSE,
     type = "n")
axis(1, at = 1:3, labels = c("MTT", "VR1", "VR2"))
axis(2)
box()
points(y ~ jitter(as.numeric(mediatype)), data = dat,
       col = colors[dat$interaction])
points(y ~ I(1:3), datm[datm$interaction == "passive", ], col = colors[1],
       type = "b", pch = 16, cex = 2)
points(y ~ I(1:3), datm[datm$interaction == "active", ], col = colors[2],
       type = "b", pch = 16, cex = 2)
text(1.45, 0.4, "passive", col = colors[1])
text(1.4, 1.6, "active", col = colors[2])

Linear mixed-effects model

The following model will be fitted to the data.

with and

## Fit linear mixed-effects model
m1 <- lmer(y ~ 1 + mediatype + interaction + mediatype:interaction + (1| id),
           data = dat)

tab <- summary(m1)$coef
ci  <- confint(m1)

Implied model formulae

The two categorical predictors are dummy coded. For interaction the level passive is set as reference category. The dummy variable is therefore 0 for passive and 1 for active.

contrasts(dat$interaction)

        active
passive      0
active       1

The three level factor mediatype is represented by two dummy variables, namely and . Since MTT is the reference category, the coding scheme is as folows.

contrasts(dat$mediatype)

    VR1 VR2
MTT   0   0
VR1   1   0
VR2   0   1

When we plug these dummy variables into the model formula from above, we get the following equations for each data point of the aggregated data set.

Compare to plot

Session info

sessionInfo()

R version 4.6.1 (2026-06-24 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 26100)

Matrix products: default
  LAPACK version 3.12.1

locale:
[1] LC_COLLATE=German_Germany.utf8  LC_CTYPE=German_Germany.utf8   
[3] LC_MONETARY=German_Germany.utf8 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.utf8    

time zone: Europe/Berlin
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] lme4_2.1-0   Matrix_1.7-6

loaded via a namespace (and not attached):
 [1] digest_0.6.39     fastmap_1.2.0     xfun_0.62         lattice_0.23-1   
 [5] splines_4.6.1     knitr_1.52        htmltools_0.5.9   rmarkdown_2.32   
 [9] Rdpack_2.6.6      cli_3.6.6         nloptr_2.2.1      grid_4.6.1       
[13] reformulas_0.4.4  compiler_4.6.1    rbibutils_2.4.1   boot_1.3-32      
[17] tools_4.6.1       nlme_3.1-171      minqa_1.2.8       evaluate_1.0.5   
[21] Rcpp_1.1.2        yaml_2.3.12       otel_0.2.0        rlang_1.3.0      
[25] jsonlite_2.0.0    htmlwidgets_1.6.4 MASS_7.3-66