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  • 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。 

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