If a bootstrap confidence interval (CI) can be interpreted as a standard CI (e.g., the range of null hypothesis values that cannot be rejected) [also stated in this post]. Is it ok to derive a p-value from a bootstrap distribution like this? When the null hypothesis is $H_0: \theta=\theta_0$ and a bootstrap ($1-\alpha$)$\times 100\%$ CI is ($\theta_L$, $\theta_U$)$_{\alpha}$. The p-value is $\alpha$ corresponding with $\theta_U=\theta_0$ or $\theta_L=\theta_0$.

This post also describes examples of converting CIs to p-values, but I do not completely understand...

The following code derives a p-value from the percentile CI of the slope parameter of a linear regression model, and it can be applied to other types of CIs. If this is not ok, what is the appropriate way to compute a p-value, e.g., associated with the percentile CI? If the code below is ok, can it be described as a bootstrap hypothesis test (e.g., when describing it in a paper)?

# hypothestical data
x <- runif(20,10,50)
y <- rnorm(length(x),1+0.5*x,2)

model <- lm(y~x)

params    <- coef(model) 

nboot <- 2000
eboot <- rep(NA,nboot)
for(i in 1:nboot){
 booti <- sample(1:length(x),replace=T)
 eboot[i] <- coef(lm(y[booti]~x[booti]))[2]

# 95% CI for the slope
quantile(eboot,c(0.025,0.975))  # percentile CI
params[2]*2-quantile(eboot,c(0.975,0.025)) # basic CI

# null hypothesis
null <- 0.5

get.p <- function(x,null){
 if(null>quantile(eboot,0.5)) return(null-quantile(eboot,1-x/2))
 if(null<quantile(eboot,0.5)) return(null-quantile(eboot,x/2))

#x <- seq(0,2,length=100)
(p <- uniroot(get.p,null=null,c(0,1))$root)  # p-value

2 Answers 2


Just to expand a bit on @Maarten Buis answer, if you are testing the hypothesis $H_0: \theta=\theta_0$ within the framework of a linear model, it makes more sense to use the t-statistic as opposed to using just the coefficient of the model, which ignores the standard error. For example, you might end up with a coefficient > theta but with a standard error 2 or 3 times larger, and the approach will be blind to that. You can check out the tutorial by John Fox under Bootstrap Hypothesis Tests.

So using your example:

df = data.frame(
x = runif(20,10,50)
df$y = rnorm(length(df$x),1+0.5*df$x,2)

we need to define a function that calculates $\hat{\beta}-0.5$ and its t-statistic:

fun = function(mod){
d = deltaMethod(mod, "x-0.5")

Then boot and check the distribution:

bo <- car::Boot(fit, R=999, f=fun, labels=c("x-0.5","tstat"))
hist(bo, ci="none")

enter image description here

In this case, the method you highlight and the t-statistic will give very similar estimates:

sum(bo$t0[2] > bo$t[,2])/(nrow(bo$t)+1)
[1] 0.492
> sum(bo$t0[1] > bo$t[,1])/(nrow(bo$t)+1)
[1] 0.49

When the sample size is small or you have values that deviate, it will be useful to check how these two differ.

  • 1
    $\begingroup$ John Fox's tutorial given by you says that "Tests for individual coefficients equal to zero can be found by inverting a confidence interval: if the hypothesized value does not fall in a 95% confidence interval." And based on your answer, my understanding is that the approach I used in the question post is not necessarily wrong, but there is a better method (CI) than the basic CI for the example. $\endgroup$
    – quibble
    Commented Jun 8, 2020 at 7:24

Using the bootstrap to compute $p$-values is possible, but works differently: How to perform a bootstrap test to compare the means of two samples?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.