Description Usage Arguments Details Value Author(s) References See Also Examples
Sample size calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.
1 2 3 4 5 6 7 | ssEQTL.ANOVA(
MAF,
typeI = 0.05,
nTests = 2e+05,
mypower = 0.8,
mystddev = 0.13,
deltaVec = c(0.13, 0.13))
|
MAF |
Minor allele frequency. |
typeI |
Type I error rate for testing if a SNP is associated to a gene probe. |
nTests |
integer. Number of tests in eQTL analysis. |
mypower |
Desired power for the eQTL analysis. |
mystddev |
Standard deviation of gene expression levels in one group of subjects. Assume all 3 groups of subjects (mutation homozygote, heterozygote, wild-type homozygote) have the same standard deviation of gene expression levels. |
deltaVec |
A vector having 2 elements. The first element is equal to mu_2 - mu_1 and the second elementis equalt to mu_3 - mu_2, where mu_1 is the mean gene expression level for the mutation homozygotes, mu_2 is the mean gene expression level for the heterozygotes, and mu_3 is the mean gene expression level for the wild-type gene expression level. |
The assumption of the ANOVA approach is that the association of a SNP to a gene probe is tested by using un-balanced one-way ANOVA (e.g. Lonsdale et al. 2013). According to SAS online document https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a0000000982.htm, the power calculation formula is
power = Pr(F >= F(1 - alpha, k - 1, N - k)| F ~ F(k - 1, N - k, lambda)),
where k = 3 is the number of groups of subjects, N is the total number of subjects, F_{1 - alpha}(k - 1, N - k) is the 100 * (1 - alpha)-th percentile of central F distribution with degrees of freedoms k - 1 and N - k, and F_{k - 1, N - k, lambda} is the non-central F distribution with degrees of freedoms k - 1 and N - k and non-central parameter (ncp) lambda. The ncp lambda is equal to
lambda = N * sum(wi * (mu_i - mu)^2, i = 1,.., k)/sigma^2,
where mu_i is the mean gene expression level for the i-th group of subjects, w_i is the weight for the i-th group of subjects, sigma^2 is the variance of the random errors in ANOVA (assuming each group has equal variance), and mu is the weighted mean gene expression level
mu = sum(w_i * mu_i, i = 1, ..., k).
The weights w_i are the sample proportions for the 3 groups of subjects. Hence, sum(w_i, i = 1, 2, 3) = 1.
sample size required for the eQTL analysis to achieve the desired power.
Xianjun Dong <XDONG@rics.bwh.harvard.edu>, Tzuu-Wang Chang <Chang.Tzuu-Wang@mgh.harvard.edu>, Scott T. Weiss <restw@channing.harvard.edu>, Weiliang Qiu <stwxq@channing.harvard.edu>
Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.
minEffectEQTL.ANOVA, powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA2
1 2 3 4 5 6 | ssEQTL.ANOVA(MAF = 0.1,
typeI = 0.05,
nTests = 200000,
mypower = 0.8,
mystddev = 0.13,
deltaVec = c(0.13, 0.13))
|
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.