Function for multiple T-test to find main effect - r

Question

df <- data.frame (rating1 = c(1,5,2,4,5),
rating2 = c(2,1,2,4,2),
rating3 = c(0,2,1,2,0),
race = c("black", "asian", "white","black","white"),
gender = c("male","female","female","male","female")
)
I'd like to conduct t-test of group mean (e.g. mean of asians in rating1) and the overall mean of each rating (e.g. rating1). Below is my code for Asians in rating1.
asian_df <- df %>%
filter(race == "asian")
t.test(asian_df$rating1, mean(df$rating1))
Then for Blacks in rating 2, I'd run
black_df <- df %>%
filter(race == "black")
t.test(black_df$rating2, mean(df$rating2))
How can I write a function that automates the t-test for each group? So far I have to manually change the variable name to essentially run for each race, each gender and on each rating (rating 1 to rating 3). Thanks!

Answer 1

Performing multiple t-tests increases your risk of Type I error and you will need to adjust for multiple comparisons in order for your results to be valid/meaningful. You can run the t-tests by looping through your variables, e.g.
library(tidyverse)
df <- data.frame (rating1 = c(5,8,7,8,9,6,9,7,8,5,8,5),
rating2 = c(2,7,8,4,9,3,6,1,7,3,9,1),
rating3 = c(0,6,1,2,7,2,9,1,6,2,3,1),
race = c("asian", "asian", "asian","black","asian","black","white","black","white","black","white","black"),
gender = c("male","female","female","male","female","male","female","male","female","male","female","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$rating1,
rep(mean(df$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, df = 5, 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
Due to multiple testing (in this example, 5 t-tests) your chance of a false positive is 1 - (1 - 0.05)^5 = 22.62% <- very high. To account for this, you can apply the Bonferroni correction, which basically takes your required p-value (in this case, p < 0.05) and divides it by the number of tests (i.e. the new p-value required to reject the null is p < 0.01). When you apply this correction, even the 'best' t-test result (gender; p-value = 0.01683) is not statistically significant.
An alternative approach would be to compare means in all conditions using ANOVA, then use Tukey's HSD to determine which groups are different. Tukey's HSD is a single post-hoc test, so you don't need to account for multiple testing, and your results are valid. Adapting this approach to your problem might be a better way to go e.g.
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