CODEWITHSUNDEEP
rlinear-regressionmultivariate-testingbonferroni
T K
T K

Reputation: 415

Multiple Linear Regression Models with Bonferroni corrected p-values

Please see my sample data below:


dad <- data.frame(type = c("new", "new", "old", "new", "old", "new", "old", "old", "new", "new"),
                   outcome = c(68,76,57,67,89,98,99,120,99,67),
                  loan1=c(98000 ,56668,67000,87999,87945, 89768, 65738, 87547, 78937, 20000),
                  loan2 =c(56768,45679, 23453, 87673, 56749, 45783, 67836, 54673, 45379, 78483),
                  house=c("semi", "detached", "town", "semi", "town", "town", "semi", "detached", "semi", "town"))

I am trying to run separate multivariate linear regression models for each house and would like to correct for multiple testing by adjusting the p-values with Bonferroni correction. In my actual dataset I am running 7 backward stepwise selection models (initially starting with the same independent variables) to come to the final multivariate regression models. Every model has the same outcome as well. I am wondering if there is a way to get the adjusted p-value with the below code at all? Any help would be appreciated, thank you!

#Example 
full <- lm(outcome ~loan1+loan2,data = subset(dad, house=="semi"))
summary(full)
tab_model(full)

Upvotes: 0

Views: 1549

Answers (2)

GuedesBF
GuedesBF

Reputation: 9878

Try p.adjust(). Call it with the argument method = 'bonferroni'

library(purrr)

Unnadjusted p.values:

full |>
    summary() |>
    pluck(coefficients) |>
    (\(x) x[-1, 4])()

    loan1     loan2 
0.0814972 0.1389506

adjusted p.values:

full |>
    summary() |>
    pluck(coefficients) |>
    (\(x) x[-1, 4])() |>
    p.adjust(method = 'bonferroni')

    loan1     loan2 
0.1629944 0.2779012

Upvotes: 0

DaveArmstrong
DaveArmstrong

Reputation: 22034

The tab_model() function has an argument for adjusted p-values.

library(sjPlot)
dad <- data.frame(type = c("new", "new", "old", "new", "old", "new", "old", "old", "new", "new"),
                  outcome = c(68,76,57,67,89,98,99,120,99,67),
                  loan1=c(98000 ,56668,67000,87999,87945, 89768, 65738, 87547, 78937, 20000),
                  loan2 =c(56768,45679, 23453, 87673, 56749, 45783, 67836, 54673, 45379, 78483),
                  house=c("semi", "detached", "town", "semi", "town", "town", "semi", "detached", "semi", "town"))

full <- lapply(unique(dad$house), \(h){
  args <- list(
    formula = outcome ~ loan1 + loan2, 
    data= subset(dad, house == h))
  do.call(lm, args)
})
tab_model(full[[1]], p.adjust="bonferroni")
tab_model(full[[2]], p.adjust="bonferroni")
tab_model(full[[3]], p.adjust="bonferroni")
})

Upvotes: 1

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