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I have used bootstrapping (percentile method) to calculate the confidence interval for the estimated mean of a set. I have now divided my data into two groups (a and b), and I want to test if the mean value in each group is different. I have considered the wilcoxon rank sum test, but would prefer a method similar to the bootstrap below.

library(boot)

set.seed(1)
value <- runif(100)
grp <- sample(c("a","b"), 100, replace = TRUE)

df <- data.frame(grp, value)

bootstrap_data <- function (data, func = mean, R = 10000) {
    boot_func <- function(x, d) {
        return(func(x[d]))
    }

    set.seed(1)
    boot(data, boot_func, R = R)
}

bootstrap_ci <- function (data, func = mean, R = 10000) {
    boot_data <- bootstrap_data(data, func, R)
    boot.ci(boot_data, type="perc")
}

bootstrap_ci(value)
# 95%   ( 0.4648,  0.5700 )
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  • $\begingroup$ In what sense do you mean "equivalent"? Note that randomization tests are operating under the restriction that the null is true, while bootstrap intervals are not. $\endgroup$
    – Glen_b
    Sep 18, 2016 at 6:13
  • $\begingroup$ By equivalent i mean: A test that has the same relation to a bootstrap CI as a t-test has to a CI based on the t-distribution. $\endgroup$
    – JohannesNE
    Sep 18, 2016 at 14:48

1 Answer 1

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You could bootstrap the difference of means:

library(boot)

set.seed(1)

grp <- sample(c("a","b"), 100, replace = TRUE)

#some data with an actual difference in means
value <- rnorm(100, mean = as.integer(factor(df$grp)))

df <- data.frame(grp, value)

b <- boot(df, function(DF, i) {
  DF <- DF[i,]
  mean(DF[DF$grp == "a", "value"]) -
    mean(DF[DF$grp == "b", "value"])
}, R = 1e4, strata = as.integer(factor(df$grp)))

#bootstrap t-value
b$t0/sd(b$t)
#[1] -6.027335

#compare with t from normality assumption
t.test(value ~ grp, data = df)$statistic
#        t 
#-5.896552 
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