Power_ANOVA_FDR

Sample size and power calculations for 1-way ANOVA rolling in FDR

Background and setup

• 4-group exposure ($n_1 = 127$, $n_2 = 74$, $n_3 = 45$, $n_4 = 32$) - can be generalized to $G$ groups. Let $N$ be the total sample size.
• Continuous outcome (can assumed to be normally distributed and standardized)
• Comparisons via ANOVA
  • Assume group 1 has mean $\delta$ and remaining groups have mean 0. See below section on defining the effect size relative to Cohen's definition (1988) in the standard ANOVA with equal group sizes.
• 3000 proteins (outcomes) so want to control FDR
• Ideally, assumed N, 80% and produce minimal detectable effect sizes

Approach

• Write a function to calculate power (via simulations) for 1-way ANOVA with possibly unequal group sizes
• Suppose that of the $m=3000$ hypothesis tests performed, $\pi_0$ denotes the proportion that are truly null. The goal is to design a study so that we are powered at $(1-\beta)$ to declare significance for the $(1-\pi_0)\times m$ tests that have a signal, while controlling for the FDR, denoted $f$. Jung (2005 Bioinformatics) has shown that the p-value threshold ($\alpha$) used to declare significance is estimated by the formula below. We could the bespoke power function described above to calculate the minimum detectable effect sizes, given the group sample sizes and desired power based on value of $\alpha$.

$$\alpha = \displaystyle\frac{f(1-\pi_0)(1-\beta)}{\pi_0(1-f)}.$$

Defining effect size in these ANOVA power calculations

For ANOVA with equal group sizes $n_1=n_2=\dots=n_G$ and common variance $\sigma^2$. Let $\mu_W$ denoted the weighted mean of all groups. For the case of the 4 groups above, $$\mu_W=\displaystyle\frac{n_1}{N}\delta.$$ Additionally, define $\sigma_m^2$ as follows:

$$\sigma_m^2 = \displaystyle\sum_{i=1}^G \displaystyle\frac{n_i}{N}\left(\mu_i - \mu_W \right)^2,$$

which for the case of 4 groups is

$$\sigma_m^2 = \displaystyle\frac{n_1}{N}\left(\delta - \mu_W\right)^2 +\displaystyle\frac{n_2}{N}\mu_W^2 + \displaystyle\frac{n_3}{N}\mu_W^2 + \displaystyle\frac{n_4}{N}\mu_W^2 = \displaystyle\frac{n_1}{N}\left(\delta - \mu_W\right)^2 +\displaystyle\frac{N - n_1}{N}\mu_W^2= \delta^2\left[\displaystyle\frac{n(N - n_1)}{N^2}\right]$$

Cohen (1988) defined the effect size $f$ as

$$f = \sqrt{\displaystyle\frac{\sigma_m^2}{\sigma^2}}.$$

In our case, we assume $\sigma=1$ and thus $f = \sigma_m$. Note: this differs from what I intuitively assumed was the effect size which was just the difference in means of the group with the largest mean vs. the smallest mean scaled by the variance, i.e., $\delta$.

So for a given effect size $f$, in the setting described above, this specified that the mean of the first group, given by $\delta$ is

$$\delta = f \displaystyle\frac{N}{\sqrt{n_1(N-n_1)}}.$$