Suppose we want to test the null hypothesis that a regression coefficient = 0 using bootstrap, and say we decide 0.05 to be the level of significance. Now, we can generate the sampling distribution for each coefficient using bootstrap. It is easy to check if 0 falls within 95% confidence interval, thus we can easily decide whether we can reject the null or not.

To get a p-value, we need to check what is the quantile value of 0 in the sampling distribution. (I am using a quantile based approach there are other methods to do it which can be found here Fox on Regression) After you get the quantile,Q, the p-value is 2*Q or 2*(1-Q) depending on whether Q > 0.5 or less than 0.5.

As an illustration of the approach, consider this

library(faraway) 

Build linear model

mdl <- lm(divorce ~ ., data = divusa) 

Bootstrap

bootTest <- sapply(1:1e4,function(x){   rows <- sample(nrow(divusa),nrow(divusa),replace = T)   mdl <- lm(divorce ~ ., data = divusa[rows,])   return(mdl$coefficients) }) 

Here is the model

summary(mdl)  Call: lm(formula = divorce ~ ., data = divusa)  Residuals:     Min      1Q  Median      3Q     Max  -2.9087 -0.9212 -0.0935  0.7447  3.4689   Coefficients:              Estimate Std. Error t value Pr(>|t|)     (Intercept) 380.14761   99.20371   3.832 0.000274 *** year         -0.20312    0.05333  -3.809 0.000297 *** unemployed   -0.04933    0.05378  -0.917 0.362171     femlab        0.80793    0.11487   7.033 1.09e-09 *** marriage      0.14977    0.02382   6.287 2.42e-08 *** birth        -0.11695    0.01470  -7.957 2.19e-11 *** military     -0.04276    0.01372  -3.117 0.002652 **  --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 1.513 on 70 degrees of freedom Multiple R-squared:  0.9344,    Adjusted R-squared:  0.9288  F-statistic: 166.2 on 6 and 70 DF,  p-value: < 2.2e-16 

Notice the p-values

Subset of regression coefficient generated using bootstrap

bootTest[,1:5]                     [,1]          [,2]         [,3]         [,4]         [,5] (Intercept) 335.70970574 372.525260160 569.85830341 338.70069977 344.69261238 year         -0.18107568  -0.201798080  -0.30579380  -0.18125215  -0.18328105 unemployed   -0.02916575   0.006828023   0.01197723  -0.05610887  -0.11463230 femlab        0.79078784   0.842924808   1.02607863   0.77527548   0.76472406 marriage      0.17372382   0.199571033   0.18782967   0.15289119   0.15693996 birth        -0.11613752  -0.118507758  -0.11998122  -0.11666450  -0.13344442 military     -0.04051730  -0.056277118  -0.04062756  -0.05167556  -0.07251748 

Generate p-values with bootstrap

pvals <- sapply(1:nrow(bootTest),function(x) {   distribution <- ecdf(bootTest[x,])   qt0 <- distribution(0)   if(qt0 < 0.5){         return(2*qt0)       } else {         return(2*(1-qt0))       } }) 

Comparing p-values from t-test and bootstrap

T test

summary(mdl)$coefficients[,4]  (Intercept)         year   unemployed       femlab     marriage        birth     military  2.744830e-04 2.966776e-04 3.621708e-01 1.085196e-09 2.419284e-08 2.191964e-11 2.652003e-03  

Bootstrap

> pvals [1] 0.0008 0.0008 0.2196 0.0000 0.0000 0.0000 0.0188 

The highly significant p-values with coefficients < 1e-8 are all 0 with bootstrap with 1e4 iterations. Furthermore, the ranking of p-values is comparable as well.