Catherine Lee, PhD

Catherine Lee, PhD

Associate Professor, Biostatistician. I find that nothing is ever straight-forward!

2x2 factorial design: two trials for the price of one

Published:

Just reminding myself why 2x2 factorial designs give “two trials for the price of one.”

Below, it is demonstrated that a 2x2 factorial design with $N=4n$ subjects has the same precision (and thus power) for estimating the treatment effect of $A$ (and also of estimating the treatment effect of $B$) as a single RCT with $N=4n$ subjects evaluating treatment $A$ only. (I verified via simulations, where I fit linear regression models to estimate the treatment effects, and the power is indeed the same.)

2x2 factorial design with sample size $N=4n$

Assumptions:

  • Two treatments under consideration, $A$ and $B$.
  • 2x2 factorial design with $n$ subjects assigned to each treatment combination (total sample size $N=4n$).
  • Normally distributed outcome modeled by $y = \beta_0 + \beta_A I(A=1) + \beta_B I(B=1) + \epsilon$, where $\epsilon$ are i.i.d. $N(0,\sigma^2)$. (Note: this assumes no interaction.)

Variance for estimate of $\beta_A$, which targets the efficacy of treatment $A$

Note: $\beta_A = 0.5\cdot[E(y|A=1,B=0)-E(y|A=0,B=0)+E(y|A=1,B=1)-E(y|A=0,B=1)]$. Each expectation can be estimated by the mean of the outcomes ($y$) within each treatment combination, i.e., a mean of $n$ values with resulting variance $\sigma^2/n$.

It follows that the variance of the estimated $\beta_A$ is $0.25\cdot[4 \cdot \sigma^2/n]=\sigma^2/n$.

Parallel RCT evaluating the efficacy of treatment $A$ only with sample size $N=4n$

Assumptions:

  • 1:1 assignment of treatment $A$ to control
  • $2n$ subjects assigned each trial arm (total sample size $N=4n$).
  • Normally distributed outcome modeled by $y = \beta_0 + \beta_A I(A=1) + \epsilon$, where $\epsilon$ are i.i.d. $N(0,\sigma^2)$.

Variance for estimate of $\beta_A$, which targets the efficacy of treatment $A$

Note that $\beta_A = E(y|A=1) - E(y|A=0)$. Each expectation can be estimated by the mean of the outcomes ($y$) within each trial arm, i.e., a mean of $2n$ values with resulting variance $\sigma^2/{2n}$.

It follows that the variance of the estimated $\beta_A$ is $\sigma^2/n$.

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