Your interpretation of the mean exposure parameter is correct: it's the number of units per subject. In your case, the rate parameters $\exp(\beta_0)$ and $\exp(\beta_1)$ specify the pill consumption rate per day, so the unit is 1 day.
One way to think about this is in terms of patient days. Under some fairly strong assumptions (more on this below), you get the same number of patient days in a study of $n$ patients who are followed up for $m$ days each, and a study with $n \times m$ patients who are followed up for 1 day. Consequently, you get the same power with both study designs.
You can verify this with G*Power by checking that power is ~0.8 under the following two settings:
- sample size $n$ = 1175, mean exposure = 1
- sample size $n$ = 1175 / $m$ days, mean exposure = $m$ days
while the other input parameters are fixed at the values specified in the question: one-sided test, base rate $\exp(\beta_0)$ = 2, $\exp(\beta_1) = 0.9$, $\alpha$ = 0.05, $X$ distribution = Binomial with $\pi$ = 0.5.
Here are the results for $m$ days = {1, 31, 62, 91}, ie. follow-up is 1 day, 1 month, 2 months and 3 months.
#> days patients power patient_days
#> 1 1175 0.800 1175
#> 31 38 0.801 1178
#> 62 19 0.801 1178
#> 91 13 0.802 1183
So what are the conditions to get the same power by following up 13 patients for 3 months as you would get by following up 1,175 patients for 1 day?
One assumption is independence within patient: the number of pills patient $i$ consumes in a day are iid $\operatorname{Poisson}(\lambda_i)$.
The second assumption is that the daily pill consumption rate doesn't vary by patient: $\lambda_i = \exp(\beta_0)$ if patient $i$ is in the control group and $\lambda_i = \exp(\beta_0 + \beta_1)$ if patient $i$ is in the treatment group. If this assumption is violated, the Poisson counts will be over-dispersed and the power will be lower than planned. This will happen even if the observations are independent.
Let's demonstrate this with a simulation. I use the simstudy package to simulate Poisson data with $\lambda_i \sim \operatorname{normal}\big(\exp(\beta_0 + \beta_1 x), \sigma^2_{\text{patient}}\big)$ where $x$ = 0 for the control group and $x = 1$ for the treatment group. I estimate the power by replicating the study 1,000 times: each time I simulate count data under the alternative $\exp(\beta_1) = 0.9$, fit a Poisson GLM and check whether the confidence interval for $\beta_1$ excludes 0. The last three columns correspond to $\sigma_{\text{patient}}$ = {0, 0.05, 0.1}.
#> days patients expected_power `0` `0.05` `0.1`
#> 1 1175 0.800 0.797 0.807 0.799
#> 31 38 0.801 0.794 0.792 0.729
#> 62 19 0.801 0.793 0.733 0.716
#> 91 13 0.802 0.788 0.729 0.69
You can see that when the patient rate $\lambda_i$ varies about the mean rate $\exp(\beta_0 + \beta_1 x)$ — a rather reasonable supposition — we lose power by recruiting fewer patients even though we follow them up for a longer period of time.
R code to estimate power for a Poisson regression. NB: The simulation takes ~30min.
library("broom")
library("simstudy")
library("tidyverse")
simulate_poisson_data <- function(patients, days, exp0, exp1,
sd.patient = 0, sd.day = 0,
prob.x = 0.5, seed = NULL) {
beta0 <- log(exp0)
beta1 <- log(exp1)
def.patient <- defData(
varname = "x",
dist = "binary",
formula = prob.x,
id = "patient"
)
def.patient <- defData(def.patient,
varname = "lambda0",
dist = "normal",
formula = "..beta0 + ..beta1 * x",
# Set `variance = 0` to let each patient mean equal to the group mean
variance = sd.patient^2
)
def.patient <- defData(def.patient,
varname = "days",
# Comment out `dist = ...` to observe the same number of days per patient
# dist = "noZeroPoisson",
formula = "..days"
)
def.day <- defDataAdd(
varname = "lambda",
dist = "normal",
formula = "lambda0",
# Set `variance = 0` to let each day mean equal to the patient mean
variance = sd.day^2
)
def.day <- defDataAdd(def.day,
varname = "y",
dist = "poisson",
formula = "lambda",
link = "log"
)
set.seed(seed)
dt.patient <- genData(patients, def.patient)
dt.day <- genCluster(
dt.patient,
cLevelVar = "patient",
numIndsVar = "days",
level1ID = "patient_day"
)
dt.day <- addColumns(def.day, dt.day)
dt.day
}
estimate_poisson_rate <- function(dd, alternative = c("two.sided", "less", "greater"),
conf.level = 0.95) {
alternative <- match.arg(alternative)
fit <- glm(
y ~ x,
data = dd,
family = poisson
)
bx.hat <- tidy(fit, conf.int = TRUE)[2, ]
if (alternative == "less") {
lower <- -Inf
upper <- bx.hat$estimate + qnorm(conf.level) * bx.hat$std.error
} else if (alternative == "greater") {
lower <- bx.hat$estimate - qnorm(conf.level) * bx.hat$std.error
} else {
lower <- bx.hat$conf.low
upper <- bx.hat$conf.high
}
list(lower = lower, upper = upper)
}
study_poisson_counts <- function(patients, days, exp0, exp1,
sd.patient = 0, sd.day = 0,
prob.x = 0.5,
alternative = c("two.sided", "less", "greater"),
conf.level = 0.95, seed = NULL) {
dd <- simulate_poisson_data(patients, days, exp0, exp1,
sd.patient = sd.patient, sd.day = sd.day,
prob.x = prob.x, seed = seed
)
# Check for overdispersion: var(y|x) > mean(y|x)
dd[, list(mean.y = mean(y), var.y = var(y)), by = x]
conf.int <- estimate_poisson_rate(dd,
alternative = alternative, conf.level = conf.level
)
(conf.int$lower > 0) || (conf.int$upper < 0)
}
exp0 <- 2
exp1 <- 0.9
study <- tribble(
~days, ~patients, ~power,
1, 1175, 0.8000978,
31, 38, 0.8009854,
62, 19, 0.8009854,
91, 13, 0.8024570
)
study %>%
mutate(
patient_days = patients * days
)
set.seed(1234)
out <- study %>%
expand_grid(
sd = c(0, 0.05, 0.1)
) %>%
rename(
expected_power = power
) %>%
mutate(
actual_power = pmap_dbl(list(patients, days, sd), \(patients, days, sd) {
excludes0 <- replicate(
1000,
study_poisson_counts(
patients, days, exp0, exp1,
sd.patient = sd,
alternative = "less"
)
)
mean(excludes0, na.rm = TRUE)
})
)
out %>%
pivot_wider(
names_from = sd,
values_from = actual_power
)
Best Answer
As an aside type I error rate $\alpha$ is neither a rate nor an error. I prefer to call it type I assertion probability.
The answer to your question is no. But you could use slightly different wording for the parameter that needs to go into the power calculation (besides $\alpha, n, \sigma$). Call it the effect you would be embarrassed to miss. This may harmonize with the minimal biologic effect in many cases.
For many problems, possibly including yours, power is a less useful concept than estimating the likely information yield of the experiment. A simple way to think of information yield is the margin of error in estimating a key quantity of interest. This quantity may be a difference in means between two groups, and correspond to half the width of, say, a 0.95 confidence interval. This approach is usually called planning for precision, and I delve into this in BBR. Instead of a "difference not to miss" you have to specify an acceptable margin of error. This is not necessarily a trivial task but is a bit easier than agreeing on a biologic effect not to miss. And it is more useful in my view, because if you run the experiment and get p=0.12 you have no idea what to conclude, but the compatibility (confidence) interval has meaning whether you accept or reject $H_0$.