tipr: R tools for tipping point sensitivity analyses

Authors: Lucy D’Agostino McGowan
License: MIT

Installation

# install.packages(devtools)
devtools::install_github("lucymcgowan/tipr")

library("tipr")

Usage

After fitting your model, you can determine the unmeasured confounder needed to tip your analysis. This unmeasured confounder is determined by two quantities, the association between the exposure and the unmeasured confounder (if the unmeasured confounder is continuous, this is indicated with smd, if binary, with exposed_p and unexposed_p), and the association between the unmeasured confounder and outcome outcome_association. Using this 📦, we can fix one of these and solve for the other. Alternatively, we can fix both and solve for n, that is, how many unmeasured confounders of this magnitude would tip the analysis.

In this example, a model was fit and the exposure-outcome relationship was 1.5 (95% CI: 1.2, 1.8).

Continuous unmeasured confounder example

We are interested in a continuous unmeasured confounder, so we will use the tip_with_continuous() function.

Let’s assume the relationship between the unmeasured confounder and outcome is 1.5 (outcome_association = 1.5), let’s solve for the association between the exposure and unmeasured confounder needed to tip the analysis (in this case, we are solving for smd, the mean difference needed between the exposed and unexposed).

tip(1.2, outcome_association = 1.5)

## The observed effect (1.2) WOULD be tipped by 1 unmeasured confounder
## with the following specifications:
##   * estimated difference in scaled means between the unmeasured confounder
##     in the exposed population and unexposed population: 0.45
##   * estimated association between the unmeasured confounder and the outcome: 1.5

## # A tibble: 1 × 4
##   observed_effect   smd outcome_association n_unmeasured_confounders
##             <dbl> <dbl>               <dbl>                    <dbl>
## 1             1.2 0.450                 1.5                        1

A hypothetical unobserved continuous confounder that has an association of 1.5 with the outcome would need a scaled mean difference between exposure groups of 0.45 to tip this analysis at the 5% level, rendering it inconclusive.

Binary unmeasured confounder example

Now we are interested in the binary unmeasured confounder, so we will use the tip_with_binary() function.

Let’s assume the unmeasured confounder is prevalent in 25% of the exposed population (exposed_p = 0.25) and in 10% of the unexposed population (unexposed_p = 0.10) – let’s solve for the association between the unmeasured confounder and the outcome needed to tip the analysis (outcome_association).

tip_with_binary(1.2, exposed_p = 0.25, unexposed_p = 0.10)

## The observed effect (1.2) WOULD be tipped by 1 unmeasured confounder
## with the following specifications:
##   * estimated prevalence of the unmeasured confounder in the exposed population: 0.25
##   * estimated prevalence of the unmeasured confounder in the unexposed population: 0.1
##   * estimated association between the unmeasured confounder and the outcome: 2.54

## # A tibble: 1 × 5
##   observed_effect exposed_p unexposed_p outcome_association n_unmeasured_confou…
##             <dbl>     <dbl>       <dbl>               <dbl>                <dbl>
## 1             1.2      0.25         0.1                2.54                    1

A hypothetical unobserved binary confounder that is prevalent in 10% of the unexposed population and 25% of the exposed population would need to have an association with the outcome of 2.5 to tip this analysis at the 5% level, rendering it inconclusive.

Many unmeasured confounders

Suppose we are concerned that there are many small, independent, continuous, unmeasured confounders present.

tip(1.2, smd = 0.25, outcome_association = 1.05)

## The observed effect (1.2) WOULD be tipped by 15 unmeasured confounders
## with the following specifications:
##   * estimated difference in scaled means between the unmeasured confounder
##     in the exposed population and unexposed population: 0.25
##   * estimated association between the unmeasured confounder and the outcome: 1.05

## # A tibble: 1 × 4
##   observed_effect   smd outcome_association n_unmeasured_confounders
##             <dbl> <dbl>               <dbl>                    <dbl>
## 1             1.2  0.25                1.05                     14.9

It would take about 15 independent unmeasured confounders with a scaled mean difference between exposure groups of 0.25 to and an association with the outcome of 1.05 tip the observed analysis at the 5% level, rendering it inconclusive.

Integration with broom

These functions were created to easily integrate with models tidied using the broom package. This is not necessary to use these functions, but a nice feature if you choose to do so. Here is an example of a logistic regression fit with glm and tidied with the tidy function broom that can be directly fed into the tip() function.

if (requireNamespace("broom", quietly = TRUE) &&  requireNamespace("dplyr", quietly = TRUE)) {
   glm(am ~ mpg, data = mtcars, family = "binomial") %>%
    broom::tidy(conf.int = TRUE, exponentiate = TRUE) %>%
    dplyr::filter(term == "mpg") %>%
    dplyr::pull(conf.low) %>%
    tip(outcome_association = 2.5)
}

## The observed effect (1.13) WOULD be tipped by 1 unmeasured confounder
## with the following specifications:
##   * estimated difference in scaled means between the unmeasured confounder
##     in the exposed population and unexposed population: 0.13
##   * estimated association between the unmeasured confounder and the outcome: 2.5

## # A tibble: 1 × 4
##   observed_effect   smd outcome_association n_unmeasured_confounders
##             <dbl> <dbl>               <dbl>                    <dbl>
## 1            1.13 0.133                 2.5                        1