ssEQTL.ANOVA: Sample Size Calculation for EQTL Analysis Based on...

In sterding/powerEQTL: Power and Sample Size Calculation for eQTL Analysis

Description

Sample size calculation for eQTL analysis that tests if a SNP is associated to a gene probe by using un-balanced one-way ANOVA.

Usage

ssEQTL.ANOVA(
  MAF,
  typeI = 0.05,
  nTests = 2e+05,
  mypower = 0.8,
  mystddev = 0.13,
  deltaVec = c(0.13, 0.13))

Arguments

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.

Details

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.

Value

sample size required for the eQTL analysis to achieve the desired power.

Author(s)

Xianjun Dong <[e-mail]>, Tzuu-Wang Chang <[e-mail]>, Scott T. Weiss <[e-mail]>, Weiliang Qiu <[e-mail]>

References

Lonsdale J and Thomas J, et al. The Genotype-Tissue Expression (GTEx) project. Nature Genetics, 45:580-585, 2013.

See Also

minEffectEQTL.ANOVA, powerEQTL.ANOVA, powerEQTL.ANOVA2, ssEQTL.ANOVA2

Examples

ssEQTL.ANOVA(MAF = 0.1,
       typeI = 0.05,
       nTests = 200000,
       mypower = 0.8,
       mystddev = 0.13,
       deltaVec = c(0.13, 0.13))

sterding/powerEQTL documentation built on May 30, 2019, 4:42 p.m.