R 中的统计测试：fisher 测试和二元逻辑变量
Statistical testing in R: fisher test and logical variables as binary
我应该在变量“亚麻酸大于 0.33”和变量“区域等于撒丁岛”之间选择和使用正确的统计检验。这是两个二进制变量，所以我正在尝试使用 fisher.test，但是如何将它们更改为 table 的逻辑变量？
library(dslabs)
dt <- as.data.table(olive)
这是您要找的吗？
table(dt[,.(Region = region == "Sardinia", Linolelic = linolenic > 0.33)])
# Linolelic
#Region FALSE TRUE
# FALSE 220 254
# TRUE 86 12
然后顺手：
fisher.test(table(dt[,.(Region = region == "Sardinia", Linolelic = linolenic > 0.33)]))
# Fisher's Exact Test for Count Data
#
#data: table(dt[, .(Region = region == "Sardinia", Linolelic = linolenic > 0.33)])
#p-value = 4.58e-15
#alternative hypothesis: true odds ratio is not equal to 1
#95 percent confidence interval:
# 0.05870264 0.23036344
#sample estimates:
#odds ratio
# 0.1212736
我应该在变量“亚麻酸大于 0.33”和变量“区域等于撒丁岛”之间选择和使用正确的统计检验。这是两个二进制变量，所以我正在尝试使用 fisher.test，但是如何将它们更改为 table 的逻辑变量？
library(dslabs)
dt <- as.data.table(olive)
这是您要找的吗？
table(dt[,.(Region = region == "Sardinia", Linolelic = linolenic > 0.33)])
# Linolelic
#Region FALSE TRUE
# FALSE 220 254
# TRUE 86 12
然后顺手：
fisher.test(table(dt[,.(Region = region == "Sardinia", Linolelic = linolenic > 0.33)]))
# Fisher's Exact Test for Count Data
#
#data: table(dt[, .(Region = region == "Sardinia", Linolelic = linolenic > 0.33)])
#p-value = 4.58e-15
#alternative hypothesis: true odds ratio is not equal to 1
#95 percent confidence interval:
# 0.05870264 0.23036344
#sample estimates:
#odds ratio
# 0.1212736