The first thing I can't figure out is what Cohen's $f$ is. I'll assume it's the square root of Cohen's $f^2$? It makes sense to make this into a regression, and then it's easier to work with variance explained like $\eta^2$ and $R^2$. According to Wikipedia, Cohen's $f^2$ is: $$f^2=\frac{R^2}{1-R^2}$$ so $$R^2=\frac{f^2}{f^2+1}=\frac{0.125^2}{1+0.125^2}=0.015$$ So that's one component.
Next, we can also figure out the correlations between the repeated measures. If $\rho=0.62$, we can first generate three columns of data from a multivariate normal distribution:
set.seed(12345)
n <- 200000
cor.mat <- matrix(.62, 3, 3)
diag(cor.mat) <- 1
dat <- as.data.frame(MASS::mvrnorm(n, rep(0, 3), cor.mat))
cov(dat) # Variances should be about 1, and covariances about .62
V1 V2 V3
V1 0.9999592 0.6187661 0.6203490
V2 0.6187661 1.0040113 0.6228203
V3 0.6203490 0.6228203 1.0043763
To simplify things, I'll assume the $f^2$ comes any one of the three time points. If I do that, then I have the following equation:
$$y=\beta\times treat+\epsilon$$
where $y$ is the data from any one time point, and for now $\epsilon$ is any column from the data frame above. Nicely, it has a variance of 1 like the columns of the data frame we just generated. If $f=0.125$, then $$0.125^2=\mathrm{Var}(\beta\times treat)=\beta^2\times\mathrm{Var}(treat)$$
Since you specified that one group has proportions 10% greater than the other, I'll assume a 40-60 split, with 40% in the treatment group (it could readily be the other way round). This is equivalent to $treat\sim Bern(0.4)$. And since $treat$ is Bernoulli, its variance is $0.4 \times 0.6=0.24$. So for $0.125^2=\beta^2\times\mathrm{Var}(treat)$, $$\beta=\frac{0.125}{sd(treat)}=\frac{0.125}{\sqrt{(0.24)}}=\frac{0.0625}{\sqrt{0.06}}$$
dat$treat <- rbinom(n, 1, .4)
mean(dat$treat); var(dat$treat)
[1] 0.39928
[1] 0.2398567
beta <- 0.0625 / sqrt(0.06)
dat$y1 <- beta * dat$treat + dat$V1
dat$y2 <- beta * dat$treat + dat$V2
dat$y3 <- beta * dat$treat + dat$V3
A quick check on things:
c(summary(lm(y1 ~ treat, dat))$r.squared,
summary(lm(y2 ~ treat, dat))$r.squared,
summary(lm(y3 ~ treat, dat))$r.squared)
[1] 0.01576065 0.01593496 0.01562384
Seems about right. Because the groups are unbalanced, you should analyze using lmer()/lme() from lme4/nlme as suggested in the anova() documentation. I'll use lmerTest so I can obtain p-values. First, I need to reshape the data to long format:
dat$ID <- 1:nrow(dat)
dat <- reshape(
dat, idvar = "ID", varying = list(c("y1", "y2", "y3")), direction = "long")
Then:
library(lmerTest)
res <- lmer(y1 ~ treat + (1 | ID), dat)
coef(summary(res))["treat", 5] # this is what we want
[1] 0
Now we can replicate this process at different sample sizes to get a sense of the power:
# Essential components
cor.mat <- matrix(.62, 3, 3)
diag(cor.mat) <- 1
beta <- 0.0625 / sqrt(0.06)
alpha <- .05
res <- t(replicate(1000, {
n <- sample(seq(50, 500, 50), 1)
dat <- as.data.frame(MASS::mvrnorm(n, rep(0, 3), cor.mat))
dat$treat <- rbinom(n, 1, .4)
dat$ID <- 1:nrow(dat)
dat <- reshape(
dat, idvar = "ID", varying = list(c("V1", "V2", "V3")), direction = "long")
dat$y <- dat$V1 + beta * dat$treat
res <- lmer(y ~ treat + (1 | ID), dat)
c(n, coef(summary(res))["treat", 5] < alpha)
}))
colnames(res) <- c("n", "power")
res.agg <- aggregate(power ~ n, res, mean)
plot(power ~ n, res.agg)
These are the barebones of the kind of power analysis you can conduct. With this, you have a general sense of the sample sizes you need to be at for 90% power; it's a number above 300. You can further narrow down the sample size in the power analysis.
Is the Cohen's $f$ computed using an average of the three scores, how was this between effect generated specifically? That can affect data generation and the procedure above. Also, if you use lmerTest, you should explore the options for obtaining p-values.
