克鲁斯卡尔·沃利斯在R

我尝试过执行Kruskal-Wallis测试，但我不确定数据的顺序是否正确，因此，我不知道结果是否正确

table(datos$ï..meses)

    abr-17 dic-15 dic-16 dic-17 ene-15 ene-16 ene-17 ene-18 feb-15 feb-16 mar-15 
        10     10     10     12     10     10     13     20     15      2     10 
    mar-17 nov-15 
         4     12

    kruskal.test(fajerae ~ ï..meses, data = matriz)

#Results# 
Kruskal-Wallis rank sum test
data:  fajerae by ï..meses 
Kruskal-Wallis chi-squared = 24.493, df = 12, p-value = 0.01742

我仍然不知道如何验证它是否正确，我不知道您是否可以帮助我识别错误

多谢各位

数据：
共有13个样本和138次观察，按月份划分

             ï..meses fajerae
            1     ene-15       0
            2     ene-15       1
            3     ene-15       0
            4     ene-15       0
            5     ene-15       0
            6     ene-15       0
            7     ene-15       1
            8     ene-15       1
            9     ene-15       0
            10    ene-15       1
            11    feb-15       0
            12    feb-15       0
            13    feb-15       0
            14    feb-15       1
            15    feb-15       0
            16    feb-15       1
            17    feb-15       1
            18    feb-15       0
            19    feb-15       0
            20    feb-15       3
            21    feb-15       2
            22    feb-15       2
            23    feb-15       0
            24    feb-15       0
            25    feb-15       1
            26    mar-15       1
            27    mar-15       0
            28    mar-15       3
            29    mar-15       3
            30    mar-15       1
            31    mar-15       4
            32    mar-15       2
            33    mar-15       5
            34    mar-15       0
            35    mar-15       1

这里我做了Kruskal-Wallis测试

#kruskal-wallis test
  
setwd("~/")

datos <- read.csv(file.choose(), header = TRUE)
datos


#estructura de los datos
str(datos)
attach(datos)
names(datos)
class(datos)
factor(datos)

但在表中，它是正确的

table(datos$ï..meses)

    abr-17 dic-15 dic-16 dic-17 ene-15 ene-16 ene-17 ene-18 feb-15 feb-16 mar-15 
        10     10     10     12     10     10     13     20     15      2     10 
    mar-17 nov-15 
         4     12

    kruskal.test(fajerae ~ ï..meses, data = matriz)

#Results# 
Kruskal-Wallis rank sum test
data:  fajerae by ï..meses 
Kruskal-Wallis chi-squared = 24.493, df = 12, p-value = 0.01742

我做了同方差测试

# Homocedasticidad: la varianza debe de ser constante entre todos los grupos.
 
 leveneTest(fajerae ~ ï..meses, data = matriz, center = "median")
 
 #Results#
 Levene's Test for Homogeneity of Variance (center = "median")
       Df F value  Pr(>F)  
  group  12  1.8447 0.04784 *
        125                  
  ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
  Warning message:
  In leveneTest.default(y = y, group = group, ...) : group coerced to factor.

还有Kruskal-Wallis测试，它给了我一个结果，但我不知道如何检查它是否正确

table(datos$ï..meses)

    abr-17 dic-15 dic-16 dic-17 ene-15 ene-16 ene-17 ene-18 feb-15 feb-16 mar-15 
        10     10     10     12     10     10     13     20     15      2     10 
    mar-17 nov-15 
         4     12

    kruskal.test(fajerae ~ ï..meses, data = matriz)

#Results# 
Kruskal-Wallis rank sum test
data:  fajerae by ï..meses 
Kruskal-Wallis chi-squared = 24.493, df = 12, p-value = 0.01742

---

问题中的代码可以简化

1。加载要使用的非基本软件包

library(car)

2.使用

str

检查数据结构

str(datos)
#'data.frame':  35 obs. of  2 variables:
# $ ï..meses  : chr  "ene-15" "ene-15" "ene-15" "ene-15" ...
# $ fajerae   : int  0 1 0 0 0 0 1 1 0 1 ...

第一个名字，

ï..meses

，由于口音和无意义的点而相当难看，让它更漂亮

names(datos)[1] <- "meses"

但如果你想要一个因素，那很简单

datos$meses <- factor(datos$meses)

4.通过

mese组测定同质性

leveneTest(fajerae ~ meses, data = datos, center = "median")
#Levene's Test for Homogeneity of Variance (center = "median")
#      Df F value  Pr(>F)  
#group  2  3.4584 0.04367 *
#      32                  
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

5.最后是Kuskall-Wallis测试

#kruskal-wallis test
  
setwd("~/")

datos <- read.csv(file.choose(), header = TRUE)
datos


#estructura de los datos
str(datos)
attach(datos)
names(datos)
class(datos)
factor(datos)

如果您试图确定各组是否具有相同的中位数，为什么不先选中一个框并绘制胡须图查看您的数据

boxplot(fajerae ~ meses, data = datos)


各组似乎有不同的位置，通过非参数测试确认

kruskal.test(fajerae ~ meses, data = datos)
#
#   Kruskal-Wallis rank sum test
#
#data:  fajerae by meses
#Kruskal-Wallis chi-squared = 6.8633, df = 2, p-value = 0.03233
#

你问测试是否正确。是的，从某种意义上说，base R函数经过了多年的精心编码和无数用户的无数次测试。这是否是预期的是一个不同的问题，只有你能回答它

数据采用
dput
格式。

datos <-
structure(list(meses = c("ene-15", "ene-15", "ene-15", "ene-15", 
"ene-15", "ene-15", "ene-15", "ene-15", "ene-15", "ene-15", "feb-15", 
"feb-15", "feb-15", "feb-15", "feb-15", "feb-15", "feb-15", "feb-15", 
"feb-15", "feb-15", "feb-15", "feb-15", "feb-15", "feb-15", "feb-15", 
"mar-15", "mar-15", "mar-15", "mar-15", "mar-15", "mar-15", "mar-15", 
"mar-15", "mar-15", "mar-15"), fajerae = c(0L, 1L, 0L, 0L, 0L, 
0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 3L, 2L, 
2L, 0L, 0L, 1L, 1L, 0L, 3L, 3L, 1L, 4L, 2L, 5L, 0L, 1L)), 
class = "data.frame", row.names = c("1", "2", "3", "4", "5", 
"6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", 
"18", "19", "20", "21", "22", "23", "24", "25", "26", "27", "28", 
"29", "30", "31", "32", "33", "34", "35"))

datos