Test for Equality of Means by Cao, Park, and He (2019)

Given univariate samples \(X_1~,\ldots,~X_k\), it tests $$H_0 : \mu_1 = \cdots \mu_k\quad vs\quad H_1 : \textrm{at least one equality does not hold}$$ using the procedure by Cao, Park, and He (2019).

meank.2019CPH(dlist, method = c("original", "Hu"))

Arguments

dlist

a list of length \(k\) where each element is a sample matrix of same dimension.

method

a method to be applied to estimate variance parameter. "original" for the estimator proposed in the paper, and "Hu" for the one used in 2017 paper by Hu et al. Case insensitive and initials can be used as well.

Value

a (list) object of S3 class htest containing:

statistic

a test statistic.

p.value

\(p\)-value under \(H_0\).

alternative

alternative hypothesis.

method

name of the test.

data.name

name(s) of provided sample data.

References

Cao M, Park J, He D (2019). “A test for the k sample Behrens–Fisher problem in high dimensional data.” Journal of Statistical Planning and Inference, 201, 86--102. ISSN 03783758.

Examples

## CRAN-purpose small example
tinylist = list()
for (i in 1:3){ # consider 3-sample case
  tinylist[[i]] = matrix(rnorm(10*3),ncol=3)
}
meank.2019CPH(tinylist, method="o") # newly-proposed variance estimator
#> 
#> 	Test for Equality of Means by Cao, Park, and He (2019)
#> 
#> data:  tinylist
#> T = 0.79157, p-value = 0.3134
#> alternative hypothesis: one of equalities does not hold.
#> 
meank.2019CPH(tinylist, method="h") # adopt one from 2017Hu
#> 
#> 	Test for Equality of Means by Cao, Park, and He (2019)
#> 
#> data:  tinylist
#> T = 0.79157, p-value = 0.3696
#> alternative hypothesis: one of equalities does not hold.
#> 

if (FALSE) {
## test when k=5 samples with (n,p) = (10,50)
## empirical Type 1 error 
niter   = 10000
counter = rep(0,niter)  # record p-values
for (i in 1:niter){
  mylist = list()
  for (j in 1:5){
     mylist[[j]] = matrix(rnorm(10*50),ncol=50)
  }
  
  counter[i] = ifelse(meank.2019CPH(mylist)$p.value < 0.05, 1, 0)
}

## print the result
cat(paste("\n* Example for 'meank.2019CPH'\n","*\n",
"* number of rejections   : ", sum(counter),"\n",
"* total number of trials : ", niter,"\n",
"* empirical Type 1 error : ",round(sum(counter/niter),5),"\n",sep=""))
}