# Latent growth curve model with more time points than participants

I can't quite figure out if having more time points in a latent growth curve model reduces power or increases it. I don't have the data yet so I can't experiment. The study has around 300 measurement points and about 150 participants. I could collapse the measurement points into chunks to reduce the number of time points, but why throw away data if i don't have to..

First, yes you gain power if you have more time points. How much power depends on the correlation between measures - the lower the correlation, the greater the increase in power. (But also, the lower the correlation, the less power you have in the first place - so be careful what you wish for).

Second, you tagged this [sem] . You won't be able to fit this as a structural equation model. You need to fit as a multilevel model. These approaches are equivalent (see http://curran.web.unc.edu/files/2015/03/Curran2003.pdf ).

Third (and finally) you don't need the data to experiment, just make up some data. Here's an example in R. With this set of parameters, with 300 time points I get a standard error of 0.004, with 100 I get a standard error of 0.007. So with more time points, I gain more power.

library(lme4)

set.seed(1234)
n.ind <- 180 # number of people
n.t <- 300 # number of time points per person

slope <- 1  # average slope
sd.slope <- 1 # sd of the slopes (between people)
sd.resid <- 1 # sd of the resids

# generate slopes
slopes <- rnorm(n.ind, 1, 1)

d <- lapply(slopes, function(x) {
t <- 1:n.t  # time variable
# generate data for each person - their slope + residual
y <-   t * x + rnorm(n.t, mean = 0, sd = sd.resid)
return(data.frame(t = t, y = y))
})

# put i for individual into each data frame
for(i in 1:length(d)) {
d[[i]]\$i <- i
}

# combine data frames in d
dLong <- do.call(rbind, d)

# fit growth curve to all data
fit300 <- lmer(y ~ t + (1|i), data = dLong)
summary(fit300)

# fit to 100 times only
fit100 <-  lmer(y ~ t + (1|i), data = subset(dLong, t <= 100))
summary(fit100)