[英]Power for Hotelling's T^2 test
I am trying to estimate the performance of the hotelling's T^2 test.我试图估计hotelling 的T^2 测试的性能。 I simulate various test from assumptions of normality and also equal covariance matrix.
我根据正态性假设和等协方差矩阵模拟各种测试。 I succeed in computing the level of the test with the following formula.
我成功地用下面的公式计算了测试的水平。
#Generate two groups with equal multivariate vector of means but different covariances
## From the multivariate normal distributions
library(MASS)
set.seed(123)
X<-mvrnorm(50,rep(0,10),diag(rep(1,10)))
Y<-mvrnorm(50,rep(0,10),diag(rep(3,10)))
#Two sample Hotelling T^2 test
library(ICSNP)
HotellingsT2(X, Y)
# Performance of the test
## Significant level
n=10000 # testing 10,000 times
t1err=0
for (i in 1:n){
X<-mvrnorm(50,rep(0,10),diag(rep(1,10)))
Y<-mvrnorm(50,rep(0,10),diag(rep(3,10)))
if (((HotellingsT2(X, Y))$p.value)<=0.05) (t1err=t1err+1)
}
cat("The significance level in percentage is", (t1err/n)*100,"%")
Now my aim is to compute the power of the test.现在我的目标是计算测试的功效。 I notice there is no option for hotelling's T^2 in the power package.
我注意到在电源 package 中没有酒店的 T^2 选项。 So how can I compute the power of that test manually or any tips to compute a type II error of the test?
那么如何手动计算该测试的功效或计算测试 II 型错误的任何提示?
Here is a way of simulating Hotelling's T-squared power.这是一种模拟 Hotelling 的 T 平方幂的方法。 The question's code for the rate of Type I errors is an adaptation of the code in this Cross Validated post .
该问题的 I 类错误率代码是对这个 Cross Validated post中代码的改编。 I have also adapted that code, simplifying it a bit.
我还修改了该代码,稍微简化了一点。
The null of equal multivariate means is false, one is rep(0, 10)
and the other is rep(2, 10)
.等多元均值的 null 为假,一个是
rep(0, 10)
另一个是rep(2, 10)
。 Unlike in the question's code, the covariance matrices are equal.与问题的代码不同,协方差矩阵是相等的。
set.seed(123)
pval <- replicate(n, {
X <- mvrnorm(50, rep(0, 10), diag(rep(1, 10)))
Y <- mvrnorm(50, rep(2, 10), diag(rep(1, 10)))
HotellingsT2(X, Y)$p.value
})
t2err <- mean(pval > 0.05)
cat("The test power in percentage is", (1 - t2err)*100,"%\n")
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