|
We can easily do power analyses for single-factor and multi-factor experiments. In G*Power, you select F-Test (ANOVA), Global for F-Test (ANOVA), Special for
Note: For F-Test (ANOVA) (as well as for F-Test (MCR)), you can choose whether you want to perform power analyses for global (i.e., omnibus) tests or for special tests. Global test is the default option. This test refers to the H0 that all means in the design are equal (ANOVA) or that all regression coefficients (next to the additive constant) are zero (MCR). Random effects ANOVAs and mixed effects ANOVAs are not considered. We may add them at a later time, however. A discussion of how to do power analyses for repeated measures ANOVAs and MANOVAs can be found in the Other F-Tests section. For the ANOVA designs, we will use the effect size index f (Cohen, 1977). The relation of f to the noncentrality parameter lambda is given by
|
H0 is that the population means in k conditions are identical. More formally
|
H0 : |
k sum (mi - m)2 = 0 , i=1 |
|
H1: |
k sum (mi - m)2 = c, c > 0. i=1 |
|
where k is the number of conditions, mi is the mean in condition i, and m is the grand mean. You can compute the effect size index from group means and s which is assumed to be constant across groups by clicking on "Calc 'f'". We will spare you the formula behind that. (Just that much: You can save a lot of time when you use the "Calc 'f'" option.) |
|
We compare 10 groups, and we have reason to expect a "medium" effect size (f = .25). How many subjects do we need? |
|
Select: |
Type of Power Analysis: | |
|
|
Type of Test: |
F-Test (ANOVA), Global |
|
|
|
|
Input: |
.05 |
| |
|
|
.95 |
| |
|
|
.25 |
(Note that in G*Power you can compute f directly from the population means and the population standard deviation sigma; simply click "Calc f" after selecting "F-Test (ANOVA), Global".) | |
|
|
Groups: |
10 |
(This is a new item that pops up only when you do power analyses for ANOVAs.) |
|
Result: |
390 | |
|
|
0.9524 | |
|
|
F(9,380) = 1.9045 | |
|
|
24.1237 |
|
Thus, we need 39 subjects in each of the 10 groups. What if we had only 200 subjects available? Assuming that both alpha and beta are equally serious (i.e., the ratio q := beta/alpha = 1) which probably is the default in basic research, we can compute the following compromise power analysis: |
|
Select: |
Type of Power Analysis: | |
|
|
Type of Test: |
F-Test (ANOVA), Global Accuracy |
|
|
|
|
Input: |
200 |
| |
|
|
0.25 |
| |
|
|
1 |
|
|
Result: |
0.1592 | |
|
|
0.8408 | |
|
|
1.4762 | |
|
|
12.50000 |
|
In multi-factor designs, we want to determine separately the power for the main effects and for the interactions involved. For main effects, the H0, the interpretation of the effect size index f, and the procedure are basically the same as for single-factor designs. The major difference is that the numerator df (df = degrees of freedom) are reduced relative to a single-factor design because other factors have to be taken into account. Thus, the only new part is that you need to specify, as the "Groups", all cells of your multi-factor design, and as numerator df (the new item for this type of power analysis) you enter i - 1, where i represents the levels of the specific factor to be tested. Note that there may be considerable differences between the power analysis values as determined by G*Power and those determined according to the "approximations" suggested by Cohen (1977, p. 365). G*Power is correct, while Cohen's approximations systematically underestimate the power. This problem with Cohen's approximation method is described in more detail in the context of our description of the accuracy of the algorithms used in G*Power. |
|
Let us assume we have a 3 x 5 design in which Factor A has 3 levels and Factor B has 5 levels. We first want to compute a power analysis for main effect A: |
|
Select: |
Type of Power Analysis: | |
|
|
Type of Test: |
F-Test (ANOVA), Special |
|
|
|
|
Input: |
.05 |
| |
|
|
.25 |
| |
|
|
270 |
| |
|
|
Groups: |
15 |
(That is, all cells in your 3x5 design, thus 15 because there are 3 * 5 levels for Factor A and Factor B, respectively.) |
|
|
Numerator DF: |
2 |
(Factor A has 3 levels, thus the test of the main effect of Factor A has 3-1=2 df.) |
|
Result: |
0.9637 (Yeah!) | |
|
|
F(2,255) = 3.0312 | |
|
|
16.8750 |
|
Assume that your H0 states that there is no interaction between A and B. How do you perform a power analysis for this case? The number of Groups is again 3 * 5 = 15. The numerator df is (3-1) * (5-1) = 8. If you enter these values and leave the rest as it was for the main effect, then this is the result: |
|
Result: |
0.8396 | |
|
|
F(8,255) = 1.9748 | |
|
|
16.8750 |
|
To extend this further, assume that you have a 3 x 4 x 6 design with factors A, B, and C. You test the main and interaction effects of this design using the following values (assuming alpha = .05, effect size f = .25, and a total sample size of 288): |
|
|
|
|
Effects |
|
|
|
|
|
|
A |
|
|
|
|
|
|
B |
|
|
|
|
|
|
C |
|
|
|
|
|
|
A x B |
|
|
|
|
|
|
A x C |
|
|
|
|
|
|
B x C |
|
|
|
|
|
|
A x B x C |
|
|
|
|
|
|
|
So far, we have limited the discussion to post-hoc power analyses. However, in planning a multi-factor design, we want to know how many participants we need to recruit for our experiment. How do we proceed in that case? Essentially, our decisions involve the following steps: Step 1:
Step 2:
Let us return to our 3 x 5 design in which Factor A has 3 levels and Factor B has 5 levels. For simplicity, we assume that we want to detect effects of size f = .40 for the two main effects and the interaction given alpha = beta = .05. The relevant a priori power analyses suggest the following sample sizes: |
|
|
|
|
Effects |
|
|
|
|
|
|
A |
|
|
|
|
|
|
B |
|
|
|
|
|
|
A x B |
|
|
|
|
|
* Note that the total sample size values produced by G*Power are somewhat smaller. However, we use the next largest number that can be divided by 15 because our design has 15 cells and we wish to assure that the n's in all cells are equal.
|
|
In a Case 2 situation, we would need a total sample size of 165. Given that our assumptions about alpha and the effect size remained unchanged, a total sample size of 165 would imply power values > .99 for tests of the effects of Factors A and B. This result is a dream come true! Alternatively, let us assume a Case 1 situation in which Factor B is the most important factor from a theoretical point of view. What would the implications be of accepting 135 as the total sample size? Given that our assumptions about alpha and the effect size remained unchanged, a total sample size of 135 would imply power values of . 9890 and .9195 for tests of the effects of Factors A and the A x B interaction, respectively. This result is certainly acceptable and we may decide to use 135 as the total sample size. |
|
With planned comparisons, the H0 is that the contrasts among the means do not explain, in the dependent variable, any variance which has not already been accounted for by other sources of the effect. The effect size f is defined as
where R2p is the partial multiple correlation between the dependent variable and the variable(s) coding the contrast among the means. In G*Power, click "Calc F" after selecting "F-Test (ANOVA), Special" to calculate f from the partial multiple correlation (referred to as partial eta-square in G*Power). For a power analysis, it does not matter whether the contrasts are orthogonal or not. However, note that f does not only depend on the population means but also on the correlations among the contrast variables. |
|
Assume you have a Factor A with 4 levels. We want to determine whether the effect of A on our dependent variable Y is linear, but not quadratic or cubic. You can code A into 4-1=3 orthogonal contrast variables as follows. |
|
|
|
|
x1, linear |
-3 |
-1 |
1 |
3 |
|
|
|
x2, quadratic |
1 |
-1 |
-1 |
1 |
|
|
|
x3, cubic |
-1 |
3 |
-3 |
1 |
|
|
|
|
Assume that your H1 specifies that R2p = .20 for the linear contrast (x1). Thus, f = .1667. If you have 60 subjects in your experiment (i.e., 15 in each of the four groups), what are the alpha and beta error probabilities if both types of errors are equally important (i.e., the ratio q := beta/alpha = 1)? |
|
Select: |
Type of Power Analysis: | |
|
|
Type of Test: |
F-Test (ANOVA), Special |
|
|
|
|
Input: |
60 |
| |
|
|
0.2857 |
| |
|
|
1 |
|
|
Result: |
0.1888 | |
|
|
0.8112 | |
|
|
F(1,56) = 1.7700 | |
|
|
4.8975 |
|
In an analysis of covariance, we replace a dependent variable Y by a corrected dependent variable Y' which we arrive at by partialling out the linear relation between Y and a set X = (Xa, ... , Xq) of q covariates Xi (Cohen, 1977, p. 379).
where
In other words, covariate Xi differs in each of the populations we look at, but its relation to Y and, hence, its regression weight bi is the same in all of those populations. The analysis of covariance is essentially an analysis of variance of the Y' measures. However, we need to adjust the denominator df, which is why we need to select the "F-Test (ANOVA), Special" option. If k is the number of cells of your design, choose
In this way, the denominator df are reduced appropriately because G*Power assumes that
If the correlation between Y and the covariates is substantial, then the power of your statistical test is increased. This is so because the within-population standard deviation sigmaY' in the denominator of the F ratio is smaller than sigmaY. Specifically, where r is the (multiple) population correlation between Y' and Y, we find that
The numerator does not decrease correspondingly. It may even increase. ExampleAssume a 2 x 3 design. A covariate X has been partialled out of a dependent variable Y'. We want to detect 'large' effects (f = .40) according to Cohen's effect size conventions for Factor B which has 3 levels. We had 60 subjects, and we decide that alpha = .05. What is the power of the F-test in this situation?
|
|
Select: |
Type of Power Analysis: | |
|
|
Type of Test: |
F-Test (ANOVA), Special |
|
|
|
|
Input: |
.05 |
| |
|
|
.40 |
| |
|
|
60 |
| |
|
|
Groups: |
7 |
(That is, 2 * 3 + 1 = 7.) |
|
|
Numerator DF: |
2 |
(Factor B has 3 levels, thus the test of the main effect of Factor B has 3-1=2 df.) |
|
Result: |
0.7740 | |
|
|
F(2,53) = 3.1716 | |
|
|
9.6000 |
|
|
|
|
|
|
|
|
Please report suggestions for improvements to Axel Buchner, Franz Faul, or Edgar Erdfelder. |