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View source: R/powerInteract2by2.R
Power calculation for interaction effect in 2x2 two-way ANOVA given effect sizes.
1 | powerInteract2by2(n, tauBetaSigma, alpha = 0.05, nTests = 1, verbose = FALSE)
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n |
integer. Number of subjects per group. |
tauBetaSigma |
Effect sizes ≤ft(τβ\right)_{ij}/σ, i=1, …, a, j=1,…, b, where a=b=2 and σ is
the standard deviation of random error. Rows are for factor 1 and columns are for factor 2. Note that ∑_{i=1}^a ≤ft(τβ\right)_{ij} = ∑_{j=1}^b ≤ft(τβ\right)_{ij}=0. We can get
≤ft(τβ\right)_{11}=θ,
≤ft(τβ\right)_{12}=-θ,
≤ft(τβ\right)_{21}=-θ,
≤ft(τβ\right)_{22}=θ. So
|
alpha |
family-wise type I error rate. |
nTests |
integer. For high-throughput omics study,
we perform two-way ANOVA for each of 'nTests' probes.
We use Bonferroni correction to control for family-wise type I error rate.
That is, for each probe, type I error rate would be |
verbose |
logical. Indicating if intermediate results should be printed out. |
We assume the following model:
y_{ijk}=μ+τ_i + β_j + ≤ft(τβ\right)_{ij} + ε_{ijk},
where i=1,…, a, j=1,…, b, k=1, …, n, ∑_{i=1}^{a}τ_i = 0, ∑_{j=1}^{b}β_j = 0, ∑_{i=1}^{a} ≤ft(τβ\right)_{ij} = 0, ∑_{j=1}^{b} ≤ft(τβ\right)_{ij} = 0, and ε_{ijk}\stackrel{i.i.d}{\sim} N≤ft(0, σ^2\right).
The group means are
μ_{ij} = μ+τ_i + β_j + ≤ft(τβ\right)_{ij}, i=1 …, a, j=1,…, b.
Note that μ = ∑_{i=1}^{a}∑_{j=1}^b μ_{ij} / (ab), τ_i = ∑_{j=1}^b μ_{ij}/b - μ, and β_j = ∑_{i=1}^a μ_{ij}/a - μ.
The null hypothesis H_0: all ≤ft(τβ\right)_{ij}, i=1, …, a, j=1,…, b are equal to zero. The alternative hypothesis H_a: at least one ≤ft(τβ\right)_{ij} is different from zero.
The F test statistic is
F=MS_{AB}/MS_{E}\stackrel{H_a}{\sim} F_{(a-1)(b-1), ab(n - 1), ncp},
where ncp
is the
non-centrality parameter of the F test statistic:
ncp=n∑_{i=1}^{a}∑_{j=1}^{b}≤ft[\frac{≤ft(τβ\right)_{ij}}{σ}\right]^2.
For the scenario a=b=2, we have ≤ft(τβ\right)_{11}=θ, ≤ft(τβ\right)_{12}=-θ, ≤ft(τβ\right)_{21}=-θ, ≤ft(τβ\right)_{22}=θ. Hence, the non-centrality parameter can be simplified to
ncp=4n≤ft(\frac{θ}{σ}\right)^2.
The power for testing the null hypothesis H_0 versus the alternative hypothesis H_a is
power=Pr≤ft(F > F_0 | H_a\right),
where the rejection region boundary F_0 satisfies:
Pr≤ft(F > F_0 | H_0\right) = α/nTests.
A list with 5 elements:
power |
the power of the two-way ANOVA test |
df1 |
the first degree of freedom of the F test statistic ( |
df2 |
the second degree of freedom of the F test statistic ( |
F0 |
the rejection region boundary |
ncp |
the non-centrality parameter |
Weiliang Qiu weiliang.qiu@gmail.com
Chow SC, Shao J, and Wang H. Sample size calculations in clinical research. 2nd edition. Chapman & Hall/CRC. 2008
Montgomery DC. Design and Analysis of Experiments. 8th edition. John Wiley & Sons. Inc.
1 2 3 4 5 6 | n = 25
tauBetaSigma = 0.3
# power = 0.8437275
res2 = powerInteract2by2(n = n, tauBetaSigma = tauBetaSigma,
alpha = 0.05, nTests = 1, verbose = TRUE)
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