Can unbalanced randomization improve power?

Brief public service announcement: simstudy update

When it became clear that I needed to explore the implications of unbalanced randomization for this project, I realized that the simstudy package, which supports much of the simulations on this blog, could not readily handle anything other than 1:1 randomization. I had to quickly rectify that shortcoming. There is a new argument ratio in the trtAssign function where you can now specify any scheme for any number of treatment arms. This is available in version 1.16, which for the moment can be found only on github (kgoldfeld/simstudy).

Here is an example based on a 1:2:3 allocation. I’m not sure if that would ever be appropriate, but it shows the flexibility of the new argument. One counter-intuitive aspect of this implementation is that the balance argument is set to TRUE, indicating that the allocation to the groups will be perfect, or as close as possible to the specified ratios. If balance is FALSE, the ratios are used as relative probabilities instead.

library(simstudy)
library(parallel)

RNGkind("L'Ecuyer-CMRG")
set.seed(16)

dx <- genData(600)
dx <- trtAssign(dx, nTrt = 3, balanced = TRUE, 
                ratio = c(1,2,3), grpName = "rx")

dx[, table(rx)]

## rx
##   1   2   3 
## 100 200 300

Unbalanced designs with a binary outcome

The outcome in the COVID-19 study is a composite binary outcome (at least one of a series of bad events has to occur within 30 days to be considered a failure). Here, I am considering the effect of different randomization schemes on the power of the study. We assumed in the usual care arm 40% of the patients would have a bad outcome and that the drug treatment would reduce the bad outcomes by 30% (so that 28% of the drug treatment arm would have a bad outcome).

If we generate a single data set under these assumptions, we can fit a logistic regression model to recover these parameters.

estCoef <- function(n, formula, ratio) {
  
  def <- defDataAdd(varname = "y", formula = formula, dist = "binary")
  
  dx <- genData(n)
  dx <- trtAssign(dx, grpName = "rx", ratio = ratio)
  dx <- addColumns(def, dx)
  
  coef(summary(glm(y~rx, family = binomial, data = dx)))
}

estCoef(n = 244*2, formula = "0.4 - 0.3 * 0.4 * rx", ratio = c(1, 1))

##               Estimate Std. Error   z value     Pr(>|z|)
## (Intercept) -0.4328641  0.1310474 -3.303111 0.0009561867
## rx          -0.4577476  0.1924533 -2.378487 0.0173838304

The probabilities of a bad outcome for the usual care group and drug treatment group are

c(usual = 1/(1 + exp(0.433)), drug = 1/(1+exp(0.433 + 0.458))) 

##     usual      drug 
## 0.3934102 0.2909035

Assessing power

In order to assess power, we need to generate many data sets and keep track of the p-values. The power is calculated by estimating the proportion of p-values that fall below 0.05.

Here is the analytic solution for a 1:1 ratio.

power.prop.test(p1 = .4, p2 = .7*.4, power = .80)

## 
##      Two-sample comparison of proportions power calculation 
## 
##               n = 243.4411
##              p1 = 0.4
##              p2 = 0.28
##       sig.level = 0.05
##           power = 0.8
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

The sample size estimate based on 80% suggests we would need 244 patients per arm, or 488 total patients. If we use this estimated \(n\) in a simulation for power (using 1000 datasets), we should be close to 80%:

est.power <- function(n, ratio, p1, reduction) {
  
  formula = paste(p1, "* (1 -", reduction, "* rx)")
  p.val <- estCoef(n, formula, ratio)["rx", 4]
  return(p.val)
  
}

pvals <- unlist(mclapply(1:1000, 
    function(x) est.power(488, c(1, 1), 0.4, 0.3)))

mean(pvals < 0.05)

## [1] 0.814

The power experiment

Now we are ready to evaluate the question that motivated all of this. If we start to change the ratio from 1:1 to 1:2, to 1:3, etc., what happens to the power? And does this pattern change based on the assumptions about failure rates in the usual care arm and the expected reductions in the treatment arm? Here is the code that will allow us to explore these questions:

res <- list()

for (p1 in c(.2, .3, .4, .5)) {
  for (r in c(.2, .25, .3)) {
    
    p2 <- (1- r) * p1
    n <- ceiling(power.prop.test(p1 = p1, p2 = p2, power = .80)$n)*2
    
    for (i in c(1:5)) {
      
      pvals <- mclapply(1:1000, function(x) est.power(n, c(1, i), p1, r))
      pvals <- unlist(pvals)
      
      dres <- data.table(n, control_p = p1, pct_reduction = r, 
                         control = 1, rx = i, power = mean( pvals < .05))
      
      res <- append(res, list(dres))
      
    }
  }
}

res <- rbindlist(res)

Repeating the power simulation for a variety of assumptions indicates that, at least in the case of a binary outcome, using an unbalanced design does not improve the quality of the research even though it might get more patients the drug treatment:

ggplot(data = res, aes(x = rx, y = power)) +
  geom_line(color = "blue") +
  facet_grid(control_p ~ pct_reduction, labeller = label_both) +
  theme(panel.grid = element_blank()) +
  scale_x_continuous(name = "ratio of treatment to control", 
                     breaks = c(1:5), labels = paste0(c(1:5),":1")) +
  scale_y_continuous(limits = c(.5,.9), breaks = c(.6, .7, .8))

Continuous outcomes

In the case of binary outcomes, reducing sample size in the control group reduces our ability to efficiently estimate the proportion of events, even though we may improve estimation for the treatment group by adding patients. In the case of a continuous outcome, we may be able to benefit from a shift of patients from one group to another if the variability of responses differs across groups. In particular, arms with more variability could benefit from a larger sample. Next time, I’ll show some simulations that indicate this might be the case.

Stay well.