1
$\begingroup$

I am trying to figure out the correct expression for the noncentrality parameter $\lambda_{ws}$ for the within-subjects effect in a one-way Repeated-Measures ANOVA with $k$ trials/groups. Comparing the calculation of $\lambda_{ws}$ from G*Power with my own attempts (based on the literature) I noticed an inconsistency. I'm not sure where things are going wrong, so I was hoping someone here could shed some light.

I started from a paper by Potvin and Schutz (2000, p. 348), who provided the following formula:

$$\lambda_{ws} = n\frac{\sum_{j = 1}^k (\mu_j-\mu)^2}{\sigma_{wg}^2(1-\bar{\rho})},$$

where $\mu_j$ is the mean for trial/group $j$, $\bar{\rho}$ is the average correlation between scores in the trials/groups, and $\sigma_{wg}^2$ is the within-trial (within-group) variance (assumed to be constant across trials/groups). Thus, $\lambda_{ws}$ is comparable to $\lambda$ from a 'regular' (between-subjects) one-way ANOVA, but it is just multiplied by a factor $c = \frac{1}{1-\bar{\rho}}$:

$$\lambda = nf^² = n\frac{\eta^2}{1-\eta^2} = n\frac{\sum_{j = 1}^k (\mu_j-\mu)^2}{\sigma_{wg}^2}= \frac{1}{c}\lambda_{ws} = (1-\bar{\rho})\lambda_{ws} \Rightarrow \lambda_{ws}=\frac{\lambda}{(1-\bar{\rho})}.$$

However, when I tried to calculate $\lambda_{ws}$ using G*Power, it seemed to be using a different formula. I couldn't find that formula anywhere in the official manual, but I came across this tutorial document. On page 34, it notes that:

$$\lambda_{gpower} = \frac{knf^2}{(1-\bar{\rho})} = k\lambda_{ws},$$

so the two are off by a factor $k$. My question therefore is: which one is correct? My hunch is that the G*Power formula is off: it appears to 'double-count' the 'treatment' sums of squares in the numerator (first it sums over the groups, and then it multiplies the result by $k$ again). This would have made sense to me if there were a within-between-interaction, and if $k$ represented the number of levels of the between-subjects factor, but that's not the case here.

I'm just not sure, though.

EDIT

Thanks to @dipetkov's answer below I was able to see for myself that the G*Power and Potvin & Schutz formulas are identical (I will maintain $k$ to denote the number of trials; @dipetkov used $q$):

$$\lambda_{gpower}= \frac{Nkf^2}{1-\bar{\rho}}= \frac{npkf^2}{(1-\bar{\rho})} = \frac{nkf^2}{(1-\bar{\rho})}$$ $$=\frac{nk}{(1-\bar{\rho})}\frac{\eta^2}{(1-\eta^2)}$$ $$= \frac{nk}{(1-\bar{\rho})}\frac{\sum_{j = 1}^k n_j(\mu_j-\mu)^2}{\sum_{j = 1}^k \sum_{i=1}^n (y_{ij}-\mu_j)^2}$$ $$= \frac{nk}{(1-\bar{\rho})}\frac{\sum_{j = 1}^k n_j(\mu_j-\mu)^2}{\sum_{j = 1}^k n_j\sigma_{wg(j)}^2}$$ $$= \frac{nk}{(1-\bar{\rho})}\frac{n\sum_{j = 1}^k (\mu_j-\mu)^2}{kn\sigma_{wg}^2}$$ $$= \frac{n}{(1-\bar{\rho})}\frac{\sum_{j = 1}^k (\mu_j-\mu)^2}{\sigma_{wg}^2}$$ $$= \lambda_{ws}$$

$\endgroup$

1 Answer 1

1
$\begingroup$

TL;DR The G*Power formula is correct and the (Potvin and Schutz, 2000) formula is also correct. The missing step is how to adapt a formula for two-way ANOVA to the one-way layout.


You seem to mix the notation used by two different sources and the power formulas for one- and two-way repeated measures ANOVAs.

In the (Potvin and Schutz, 2000) paper [1], $n$ is the group sample size (number of subjects in each group) in a balanced design. In the G*Power tutorial [2], $N$ is the total sample size (total number of subjects). The relationship between the group sample size and the total size is simple: $N = pn$ where $p$ in the number of levels of the between-subjects factor A.

Note: (Potvin and Schutz, 2000) denote the number of A levels by $p$; the G*Power tutorial denotes it by $a$ and you denote it by $k$. I'll use the notation in the paper.

On pages 32—36 the G*Power tutorial presents results for a two-way $A_p \times B_q$ repeated measures (RM) ANOVA. You seem to ignore this fact as you are interested in a one-way repeated measures design. To an extent, you sweep the difference between the between-subjects and within-subject factors by referring to "the mean for trial/group $j$", implying that groups and trials are interchangeable.

In your question you reference the formula for the within-subject effect B. Let's show that the G*Power formula is equivalent to the (Potvin and Schutz, 2000) formula.

$$ \begin{aligned} \eta_{\text{G*Power}} = \frac{Nqf^2}{1-\rho} = \frac{npq\sum_{j=1}^q(\mu_j-\mu)^2/(q\sigma^2)}{1-\rho} = \frac{np\sum_{j=1}^q(\mu_j-\mu)^2}{\sigma^2(1-\rho)} = \lambda_B \end{aligned} $$

For completeness, let's also look at the formula for the between-subjects effect A. Aside: In the tutorial there is a typo: the effect size $f^2$ should be in the numerator, not the denominator.

$$ \begin{aligned} \eta_{\text{G*Power}} = \frac{Nqf^2}{1-(1-q)\rho} = \frac{npq\sum_{i=1}^p(\mu_i-\mu)^2/(p\sigma^2)}{1-(1-q)\rho} = \frac{nq\sum_{i=1}^p(\mu_i-\mu)^2}{\sigma^2(1-(1-q)\rho)} = \lambda_A \end{aligned} $$

So can you use G*Power to calculate sample size for a one-way repeated measures ANOVA? Yes: It corresponds to a two-way layout with one level for the A factor, ie, $p=1$. The noncentrality parameter is $\lambda = nqf^2/(1-\rho)$ just as given in your question.

[1] P. Potvin and R. Schutz. Statistical power for the two-factor repeated measures ANOVA. Behavior Research Methods, 32:347–356, 2012.
[2] G*Power tutorial by StatPower. Available online.

$\endgroup$
1
  • $\begingroup$ Wasn't too long, and I did read it! Thanks so much for your help. I think I get it now (I also edited the initial post). $\endgroup$ Jul 26 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.