我进行了不显着的 t 检验。这是数据和结果。
Var1 <- c(12.0, 12.3, 14.1, 6.2, 2.9, 5.0, 16.2, 2.3, 4.8, 5.9, 15.0, 12.0, 11.1)
Var2 <- c(11.2, 15.1, 16, 7.2, 3.1, 1.2, 5.2, 4.1, 3.1, 11.6, 2.1, 6.5, 9.1)
data <- data.frame(Var1, Var2)
testdata <- data %>%
pivot_longer(cols = c(Var1,Var2)) %>%
mutate(group = ifelse(name == 'Var1', 0, ifelse(name == 'Var2', 1, NA)))
t.test(testdata$value ~ testdata$group, mu = 1, var.eq = F)
Welch Two Sample t-test
data: testdata$value by testdata$group
t = 0.45553, df = 23.995, p-value = 0.6528
alternative hypothesis: true difference in means is not equal to 1
95 percent confidence interval:
-2.069071 5.807532
sample estimates:
mean in group 0 mean in group 1
9.215385 7.346154
当我引入异常值时（我将 Var1 的第一个值 12.0 更改为 120.0），它并没有像我预期的那样变得重要。
Var1 <- c(120.0, 12.3, 14.1, 6.2, 2.9, 5.0, 16.2, 2.3, 4.8, 5.9, 15.0, 12.0, 11.1)
Var2 <- c(11.2, 15.1, 16, 7.2, 3.1, 1.2, 5.2, 4.1, 3.1, 11.6, 2.1, 6.5, 9.1)
data <- data.frame(Var1, Var2)
testdata <- data %>%
pivot_longer(cols = c(Var1,Var2)) %>%
mutate(group = ifelse(name == 'Var1', 0, ifelse(name == 'Var2', 1, NA)))
t.test(testdata$value ~ testdata$group, mu = 1, var.eq = F)
Welch Two Sample t-test
data: testdata$value by testdata$group
t = 1.0491, df = 12.594, p-value = 0.3138
alternative hypothesis: true difference in means is not equal to 1
95 percent confidence interval:
-8.782548 29.136394
sample estimates:
mean in group 0 mean in group 1
17.523077 7.346154
t 检验如何纠正异常值以及其他导致 p 值不显着的情况。方差如何影响结果？
