r语言代写如何进行两组独立样本秩和检验

原文链接 

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))

如果您有任何疑问,请在下面发表评论。 

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