r语言中如何进行两组独立样本秩和检验

标签：ack   ssi   dep   percent   func   pdf   pre   data   sample   

安装所需的包

wants <- c("coin")
has   <- wants %in% rownames(installed.packages())
if(any(!has)) install.packages(wants[!has])>

一个样本

 测试

set.seed(123)
medH0 <- 30
DV    <- sample(0:100, 20, replace=TRUE)
DV    <- DV[DV != medH0]
N     <- length(DV)
(obs  <- sum(DV > medH0))

[1] 15

(pGreater <- 1-pbinom(obs-1, N, 0.5))

[1] 0.02069

(pTwoSided <- 2 * pGreater)

[1] 0.04139

 威尔科克森排检验

IQ    <- c(99, 131, 118, 112, 128, 136, 120, 107, 134, 122)
medH0 <- 110

wilcox.test(IQ, alternative="greater", mu=medH0, conf.int=TRUE)

1.  
    
2.  
   Wilcoxon signed rank test
3.  
    
4.  
   data: IQ
5.  
   V = 48, p-value = 0.01855
6.  
   alternative hypothesis: true location is greater than 110
7.  
   95 percent confidence interval:
8.  
   113.5 Inf
9.  
   sample estimates:
10.  
    (pseudo)median
11.  
    121

两个独立样本

 测试

Nj  <- c(20, 30)
DVa <- rnorm(Nj[1], mean= 95, sd=15)
DVb <- rnorm(Nj[2], mean=100, sd=15)
wIndDf <- data.frame(DV=c(DVa, DVb),
                     IV=factor(rep(1:2, Nj), labels=LETTERS[1:2]))

查看每组中低于或高于组合数据中位数的个案数。

library(coin)
median_test(DV ~ IV, distribution="exact", data=wIndDf)

1.  
    
2.  
   Exact Median Test
3.  
    
4.  
   data: DV by IV (A, B)
5.  
   Z = 1.143, p-value = 0.3868
6.  
   alternative hypothesis: true mu is not equal to 0

Wilcoxon秩和检验（曼 - 惠特尼检疫）

wilcox.test(DV ~ IV, alternative="less", conf.int=TRUE, data=wIndDf)

1.  
    
2.  
   Wilcoxon rank sum test
3.  
    
4.  
   data: DV by IV
5.  
   W = 202, p-value = 0.02647
6.  
   alternative hypothesis: true location shift is less than 0
7.  
   95 percent confidence interval:
8.  
   -Inf -1.771
9.  
   sample estimates:
10.  
    difference in location
11.  
    -9.761

1.  
   library(coin)
2.  
   wilcox_test(DV ~ IV, alternative="less", conf.int=TRUE,
3.  
   distribution="exact", data=wIndDf)

1.  
    
2.  
   Exact Wilcoxon Mann-Whitney Rank Sum Test
3.  
    
4.  
   data: DV by IV (A, B)
5.  
   Z = -1.941, p-value = 0.02647
6.  
   alternative hypothesis: true mu is less than 0
7.  
   95 percent confidence interval:
8.  
   -Inf -1.771
9.  
   sample estimates:
10.  
    difference in location
11.  
    -9.761

两个依赖样本

 测试

N      <- 20
DVpre  <- rnorm(N, mean= 95, sd=15)
DVpost <- rnorm(N, mean=100, sd=15)
wDepDf <- data.frame(id=factor(rep(1:N, times=2)),
                     DV=c(DVpre, DVpost),
                     IV=factor(rep(0:1, each=N), labels=c("pre", "post")))

medH0  <- 0
DVdiff <- aggregate(DV ~ id, FUN=diff, data=wDepDf)
(obs   <- sum(DVdiff$DV < medH0))

[1] 7

(pLess <- pbinom(obs, N, 0.5))

[1] 0.1316

排名威尔科克森检验

wilcoxsign_test(DV ~ IV | id, alternative="greater",
                distribution="exact", data=wDepDf)

1.  
    
2.  
   Exact Wilcoxon-Signed-Rank Test
3.  
    
4.  
   data: y by x (neg, pos)
5.  
   stratified by block
6.  
   Z = 2.128, p-value = 0.01638
7.  
   alternative hypothesis: true mu is greater than 0

分离（自动）加载的包 

try(detach(package:coin))
try(detach(package:modeltools))
try(detach(package:survival))
try(detach(package:mvtnorm))
try(detach(package:splines))
try(detach(package:stats4))

r语言中如何进行两组独立样本秩和检验

标签：ack   ssi   dep   percent   func   pdf   pre   data   sample   

原文地址：https://www.cnblogs.com/tecdat/p/9361796.html