R中列联表(4x2)的卡方检验
我在计算R中4x2列联表的卡方检验时遇到问题。我的脚本如下所示:R中列联表(4x2)的卡方检验,r,chi-squared,contingency,R,Chi Squared,Contingency,我在计算R中4x2列联表的卡方检验时遇到问题。我的脚本如下所示: # Read data read.table("Mortality_test.txt") # Assign a name to the data mortality<- read.table("Mortality_test.txt", ,col.names=c('treatment','dead'), header=TRUE, sep="\t", na.strings="NA", dec=",", strip.white=
# Read data
read.table("Mortality_test.txt")
# Assign a name to the data
mortality<- read.table("Mortality_test.txt", ,col.names=c('treatment','dead'), header=TRUE, sep="\t", na.strings="NA", dec=",", strip.white=TRUE)
table(mortality)
dead
treatment no yes
A 63 7
B 61 9
C 68 2
D 63 7
我现在的问题是,我想比较不同治疗(A、B、C、D)之间的死亡率是否有统计学差异。如果我没弄错的话,我可以在桌上做一个卡方检验。然而,我不确定下一步该采取哪一步 您有一个函数
chisq.test
,用于在一个接触表上执行chi测试
来,拿着你的桌子
dead <- read.table(text = "treatment no yes
A 63 7
B 61 9
C 68 2
D 63 7",header = T)
> dead
treatment no yes
1 A 40 15
2 B 61 9
3 C 68 2
4 D 63 7
这两种治疗方法之间没有区别。要查看另一个不同的示例,请执行以下操作:
dead <- read.table(text = "treatment no yes
A 55 12
B 61 9
C 68 2
D 63 7",header = T)
我们只需在
表()上应用summary()
,就可以方便地得到卡方检验
实例
注:“卡方近似可能不正确”是因为本例中只有32个观测值
有了你的数据,总结(表(死亡率))
应该是有效的。这很简单。非常感谢您的快速回答:-)
dead <- read.table(text = "treatment no yes
A 55 12
B 61 9
C 68 2
D 63 7",header = T)
Pearson's Chi-squared test
data: dead[, 2:3]
X-squared = 8.4334, df = 3, p-value = 0.03785
with(mtcars, table(cyl, gear))
# gear
# cyl 3 4 5
# 4 1 8 2
# 6 2 4 1
summary(with(mtcars, table(cyl, gear)))
# Number of cases in table: 32
# Number of factors: 2
# Test for independence of all factors:
# Chisq = 18.036, df = 4, p-value = 0.001214
# Chi-squared approximation may be incorrect