SHT (version 0.1.9)

meank.2019CPH: Test for Equality of Means by Cao, Park, and He (2019)

Description

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).

Usage

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

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.

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.

References

cao_test_2019SHT

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
meank.2019CPH(tinylist, method="h") # adopt one from 2017Hu

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=""))
}