I am running a glm model using bootstrap, I can extract the coefficient mean and the confidence intervals for all the factors in my model.

But how can I get the pvalue from there?


rate.lm <- glm(one ~ two + factor(three), data = data)

lmfit <- function(data, indices) {
  fit <- glm(one ~ two + factor(three), data = data[indices, ])

results <- boot(data= data, statistic = lmfit, R = 1000, strata=data$two)

for (i in 2:length(rate.lm$coefficients)) {
      bci <- boot.ci(results, type = "basic", index = i)
      print(sprintf("%s,%.4f,%.4f,%.4f", names(rate.lm$coefficients)[i], results$t0[i],
                    bci$basic[4], bci$basic[5]))

Which gives me back the coeff mean and the 95% confidence interval. In that case I know the variable two is significant but I need a pvalue absolutely for the rest of my analysis and I am unsure on how to get it. Thanks for your help!

[1] "two,0.0101,-0.0022,0.0229"
[1] "factor(three)2,-0.2029,-0.2185,-0.1888"
[1] "factor(three)3,-0.1752,-0.1914,-0.1595"
[1] "factor(three)4,-0.2093,-0.2259,-0.1947"
[1] "factor(three)5,-0.1745,-0.1910,-0.1588"
[1] "factor(three)6,-0.1988,-0.2139,-0.1849"
  • $\begingroup$ Try running predFun1<-function(.)(coef(summary(.))[,"Pr(>|z|)"]) , and then replace coef(fit) with predFun1(fit) in your function $\endgroup$
    – Tom
    Oct 20, 2020 at 22:39

1 Answer 1


John Fox's Bootstrapping Regression Models is a great resource to get started as it provides lots of examples.

To answer your question directly, you'll simply need to access the t0 and t values provided by the boot object. To wit, boot$t0 is a numeric (of length $M+1$ where $M$ is the number of covariates) with the coefficients returned by running the regression with the original data and boot$t is a a matrix of dimension $R \times (M+1)$ holding the coefficients returned from a regression run on a bootstrapped sample of the data in each row.

Therefore, in your example, the one-sided p-values can be calculated pretty simply as follows:

R <- 1e3 # number of bootstraps

extrema <- apply(X=results$t, MARGIN=1, FUN=">", results$t0)

pvals <- rowSums(extrema)/(R+1)


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