View source: R/mean1_1931Hotelling.R
| mean1.1931Hotelling | R Documentation | 
Given a multivariate sample X and hypothesized mean \mu_0, it tests
H_0 : \mu_x = \mu_0\quad vs\quad H_1 : \mu_x \neq \mu_0
using the procedure by Hotelling (1931).
mean1.1931Hotelling(X, mu0 = rep(0, ncol(X)))
X | 
 an   | 
mu0 | 
 a length-  | 
a (list) object of S3 class htest containing: 
a test statistic.
p-value under H_0.
alternative hypothesis.
name of the test.
name(s) of provided sample data.
hotelling_generalization_1931SHT
## CRAN-purpose small example
smallX = matrix(rnorm(10*3),ncol=3)
mean1.1931Hotelling(smallX) # run the test
## Not run: 
## empirical Type 1 error 
niter   = 1000
counter = rep(0,niter)  # record p-values
for (i in 1:niter){
  X = matrix(rnorm(50*5), ncol=5)
  counter[i] = ifelse(mean1.1931Hotelling(X)$p.value < 0.05, 1, 0)
}
## print the result
cat(paste("\n* Example for 'mean1.1931Hotelling'\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=""))
## End(Not run)
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