多项T检验以找到主要效果的功能

如何解决多项T检验以找到主要效果的功能

df <- data.frame (rating1  = c(1,5,2,4,5),rating2  = c(2,1,2),rating3  = c(0,0),race = c("black","asian","white","black","white"),gender = c("male","female","male","female")
              ) 

我想对组平均值（例如，等级1中的亚洲人平均值）和每个等级的总体平均值（例如，等级1）进行t检验。下面是我对亚洲人的等级为1的代码。

asian_df <- df %>% 
   filter(race == "asian")
t.test(asian_df$rating1,mean(df$rating1)) 

然后在等级2的黑人中奔跑

   black_df <- df %>% 
       filter(race == "black")
    t.test(black_df$rating2,mean(df$rating2))

我该如何编写一个使每个小组的t检验自动化的函数？到目前为止，我必须手动将变量名称更改为实质上针对每个种族，每个性别和每个等级（等级1至等级3）运行。谢谢！

解决方法

执行多次t检验会增加I型错误的风险，您将需要adjust for multiple comparisons才能使结果有效/有意义。您可以通过遍历变量来运行t检验，例如

library(tidyverse)
df <- data.frame (rating1  = c(5,8,7,9,6,5,5),rating2  = c(2,4,3,1,1),rating3  = c(0,2,race = c("asian","asian","black","white","black"),gender = c("male","female","male","male")
)

for (rac in unique(df$race)){
tmp_df <- df %>% 
    filter(race == rac)
print(rac)
print(t.test(tmp_df$rating1,rep(mean(df$rating1),length(tmp_df$rating1))))
}
[1] "asian"

    Welch Two Sample t-test

data:  tmp_df$rating1 and rep(mean(df$rating1),length(tmp_df$rating1))
t = 0.19518,df = 3,p-value = 0.8577
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.550864  2.884198
sample estimates:
mean of x mean of y 
 7.250000  7.083333 

[1] "black"

    Welch Two Sample t-test

data:  tmp_df$rating1 and rep(mean(df$rating1),length(tmp_df$rating1))
t = -1.5149,df = 4,p-value = 0.2044
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.5022651  0.7355985
sample estimates:
mean of x mean of y 
 6.200000  7.083333 

[1] "white"

    Welch Two Sample t-test

data:  tmp_df$rating1 and rep(mean(df$rating1),length(tmp_df$rating1))
t = 3.75,df = 2,p-value = 0.06433
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1842176  2.6842176
sample estimates:
mean of x mean of y 
 8.333333  7.083333 


for (gend in unique(df$gender)){
  tmp_df <- df %>% 
    filter(gender == gend)
  print(gend)
  print(t.test(tmp_df$rating1,length(tmp_df$rating1))))
}
[1] "male"

    Welch Two Sample t-test

data:  tmp_df$rating1 and rep(mean(df$rating1),length(tmp_df$rating1))
t = -2.0979,df = 5,p-value = 0.09
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.4107761  0.2441094
sample estimates:
mean of x mean of y 
 6.000000  7.083333 

[1] "female"

    Welch Two Sample t-test

data:  tmp_df$rating1 and rep(mean(df$rating1),length(tmp_df$rating1))
t = 3.5251,p-value = 0.01683
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.2933469 1.8733198
sample estimates:
mean of x mean of y 
 8.166667  7.083333 

由于进行了多次测试（在本示例中为5次t检验），因此您出现假阳性的机率1 - (1 - 0.05)^5 = 22.62% Bonferroni correction，该方法基本上会获取所需的p值（在这种情况下，p

另一种方法是使用ANOVA比较所有条件下的均值，然后使用Tukey的HSD确定哪些组不同。 Tukey的HSD是一个事后测试，因此您无需考虑多个测试，并且结果是有效的。使这种方法适应您的问题可能是更好的解决方法，例如

anova_one_way <- aov(rating1 + rating2 + rating3 ~ race + gender,data = df)

summary(anova_one_way)

            Df Sum Sq Mean Sq F value  Pr(>F)   
race         2 266.70  133.35   14.01 0.00243 **
gender       1 140.08  140.08   14.72 0.00497 **
Residuals    8  76.13    9.52           
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


TukeyHSD(anova_one_way)

Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = rating1 + rating2 + rating3 ~ race + gender,data = df)

$race
                 diff        lwr       upr     p adj
black-asian -7.050000 -12.963253 -1.136747 0.0224905
white-asian  4.416667  -2.315868 11.149201 0.2076254
white-black 11.466667   5.029132 17.904201 0.0023910

$gender
                 diff       lwr       upr     p adj
male-female -3.416667 -7.523829 0.6904958 0.0913521