Normal-reference-test with two-cumulant (2-c) matched $\chi^2$-approximation for two-sample problem proposed by Zhang et al. (2020)
Source:R/ZGZC2020.TS.2cNRT.R
ZGZC2020.TS.2cNRT.Rd
Zhang et al. (2020)'s test for testing equality of two-sample high-dimensional mean vectors with assuming that two covariance matrices are the same.
Value
A list of class "NRtest"
containing the results of the hypothesis test. See the help file for NRtest.object
for details.
Details
Suppose we have two independent high-dimensional samples: $$ \boldsymbol{y}_{i1},\ldots,\boldsymbol{y}_{in_i}, \;\operatorname{are \; i.i.d. \; with}\; \operatorname{E}(\boldsymbol{y}_{i1})=\boldsymbol{\mu}_i,\; \operatorname{Cov}(\boldsymbol{y}_{i1})=\boldsymbol{\Sigma},i=1,2. $$ The primary object is to test $$H_{0}: \boldsymbol{\mu}_1 = \boldsymbol{\mu}_2\; \operatorname{versus}\; H_{1}: \boldsymbol{\mu}_1 \neq \boldsymbol{\mu}_2.$$ Zhang et al.(2020) proposed the following test statistic: $$T_{ZGZC} = \frac{n_1n_2}{n} \|\bar{\boldsymbol{y}}_1 - \bar{\boldsymbol{y}}_2\|^2,$$ where \(\bar{\boldsymbol{y}}_{i},i=1,2\) are the sample mean vectors. They showed that under the null hypothesis, \(T_{ZGZC}\) and a chi-squared-type mixture have the same normal or non-normal limiting distribution.
Examples
library("HDNRA")
data("COVID19")
dim(COVID19)
#> [1] 87 20460
group1 <- as.matrix(COVID19[c(2:19, 82:87), ]) ## healthy group
group2 <- as.matrix(COVID19[-c(1:19, 82:87), ]) ## COVID-19 patients
ZGZC2020.TS.2cNRT(group1, group2)
#>
#> Results of Hypothesis Test
#> --------------------------
#>
#> Test name: Zhang et al. (2020)'s test
#>
#> Null Hypothesis: Difference between two mean vectors is 0
#>
#> Alternative Hypothesis: Difference between two mean vectors is not 0
#>
#> Data: group1 and group2
#>
#> Sample Sizes: n1 = 24
#> n2 = 62
#>
#> Sample Dimension: 20460
#>
#> Test Statistic: T[ZGZC] = 228972526332
#>
#> Approximation method to the 2-c matched chi^2-approximation
#> null distribution of T[ZGZC]:
#>
#> Approximation parameter(s): df = 2.605400e+00
#> beta = 2.965057e+10
#>
#> P-value: 0.03771277
#>