Authors: Lucy
D’Agostino McGowan
License: MIT
Install the CRAN version
install.packages("tipr")Or install the development version from GitHub:
# install.packages(devtools)
devtools::install_github("lucymcgowan/tipr")library(tipr)After fitting your model, you can determine the unmeasured confounder
needed to tip your analysis. This unmeasured confounder is determined by
two quantities, the relationship between the exposure and the unmeasured
confounder (if the unmeasured confounder is continuous, this is
indicated with exposure_confounder_effect, if binary, with
exposed_confounder_prev and
unexposed_confounder_prev), and the relationship between
the unmeasured confounder and outcome
confounder_outcome_effect. 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.
This package comes with a few example data sets. For this example, we
will use exdata_rr. This data set was simulated such that
there are two confounders, one that was “measured” (and thus usable in
the main analysis, this is called measured_confounder) and
one that is “unmeasured” (we have access to it because this is simulated
data, but ordinarily we would not, this variable is called
.unmeasured_confounder).
Using the example data exdata_rr, we can estimate the
exposure-outcome relationship using the measured confounder as
follows:
mod <- glm(outcome ~ exposure + measured_confounder, data = exdata_rr,
family = poisson)
mod %>%
broom::tidy(exponentiate = TRUE, conf.int = TRUE)## # A tibble: 3 Ă— 7
## term estimate std.error statistic p.value conf.low conf.high
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0366 0.151 -21.9 2.56e-106 0.0269 0.0486
## 2 exposure 1.49 0.166 2.43 1.52e- 2 1.09 2.08
## 3 measured_confounder 2.43 0.0754 11.7 7.51e- 32 2.09 2.81We see the above example, the exposure-outcome relationship is 1.5 (95% CI: 1.09, 2.08).
We are interested in a continuous unmeasured confounder, so we will
use the tip_with_continuous() function.
Let’s assume the unmeasured confounder is normally distributed with a
mean of 0.5 in the exposed group and 0 in the unexposed (and unit
variance in both), resulting in a mean difference of 0.5
(exposure_confounder_effect = 0.5), let’s solve for the
relationship between the unmeasured confounder and outcome needed to tip
the analysis (in this case, we are solving for
confounder_outcome_effect).
tip(effect_observed = 1.5, exposure_confounder_effect = 0.5)## The observed effect (1.5) 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: 0.5
## * estimated relationship between the unmeasured confounder and the outcome: 2.25
## # A tibble: 1 Ă— 5
## effect_adjusted effect_observed exposure_confounder_effect confounder_outcome…
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.5 0.5 2.25
## # … with 1 more variable: n_unmeasured_confounders <dbl>A hypothetical unobserved continuous confounder a scaled mean
difference between exposure groups of 0.5 would need a
relationship of at least 2.25 with the outcome to tip this analysis at
the point estimate.
tip(effect_observed = 1.09, exposure_confounder_effect = 0.5)## The observed effect (1.09) 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: 0.5
## * estimated relationship between the unmeasured confounder and the outcome: 1.19
## # A tibble: 1 Ă— 5
## effect_adjusted effect_observed exposure_confounder_effect confounder_outcome…
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.5 1.19
## # … with 1 more variable: n_unmeasured_confounders <dbl>A hypothetical unobserved continuous confounder a scaled mean
difference between exposure groups of 0.5 would need a
relationship of at least 1.19 with the outcome to tip this analysis at
the 5% level, rendering it inconclusive.
Because this is simulated data, we can see what the true unmeasured confounder looked like. First we will calculate the difference in scaled means.
exdata_rr %>%
dplyr::group_by(exposure) %>%
dplyr::summarise(m = mean(.unmeasured_confounder / sd(.unmeasured_confounder))) %>%
tidyr::pivot_wider(names_from = exposure,
values_from = m,
names_prefix = "u_") %>%
dplyr::summarise(estimate = u_1 - u_0)## # A tibble: 1 Ă— 1
## estimate
## <dbl>
## 1 0.494Now we can refit the above model with this unmeasured confounder included. According to our tipping point result, as long as the relative risk of the unmeasured confounder and outcome in the model is greater than 2.25, the result that we observed will be “tipped” (the point estimate will cross the null).
mod_true <- glm(
outcome ~ exposure + measured_confounder + .unmeasured_confounder,
data = exdata_rr,
family = poisson)
mod_true %>%
broom::tidy(exponentiate = TRUE, conf.int = TRUE)## # A tibble: 4 Ă— 7
## term estimate std.error statistic p.value conf.low conf.high
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.0245 0.163 -22.7 1.49e-114 0.0176 0.0334
## 2 exposure 0.921 0.172 -0.477 6.34e- 1 0.660 1.30
## 3 measured_confounder 2.44 0.0746 11.9 6.95e- 33 2.11 2.82
## 4 .unmeasured_confoun… 2.42 0.0742 11.9 1.35e- 32 2.09 2.80Notice here the .unmeasured_confounder effect is 2.42
(which is greater than the 2.25 we calculated that would be needed to
render our result null) and, as expected, the point estimate for the
exposure has crossed the null (and now is less than 1).
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_confounder_prev = 0.25) and in
10% of the unexposed population
(unexposed_confounder_prev = 0.10) – let’s solve for the
relationship between the unmeasured confounder and the outcome needed to
tip the analysis (confounder_outcome_effect).
tip_with_binary(effect_observed = 1.09,
exposed_confounder_prev = 0.25,
unexposed_confounder_prev = 0.10)## The observed effect (1.09) 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 relationship between the unmeasured confounder and the outcome: 1.64
## # A tibble: 1 Ă— 6
## effect_adjusted effect_observed exposed_confounder_prev unexposed_confounder_…
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.25 0.1
## # … with 2 more variables: confounder_outcome_effect <dbl>,
## # n_unmeasured_confounders <dbl>A hypothetical unobserved binary confounder that is prevalent in 10% of the unexposed population and 25% of the exposed population would need to have a relationship with the outcome of 1.64 to tip this analysis at the 5% level, rendering it inconclusive.
Suppose we are concerned that there are many small, independent, continuous, unmeasured confounders present.
tip(effect_observed = 1.09,
exposure_confounder_effect = 0.25,
confounder_outcome_effect = 1.05)## The observed effect (1.09) WOULD be tipped by 7 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 relationship between the unmeasured confounder and the outcome: 1.05
## # A tibble: 1 Ă— 5
## effect_adjusted effect_observed exposure_confounder_effect confounder_outcome…
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.25 1.05
## # … with 1 more variable: n_unmeasured_confounders <dbl>It would take about 7 independent standardized Normal
unmeasured confounders with a mean difference between exposure groups of
0.25 and a relationship with the outcome of 1.05 tip the observed
analysis at the 5% level, rendering it inconclusive.
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(outcome ~ exposure + measured_confounder, data = exdata_rr,
family = poisson) %>%
broom::tidy(conf.int = TRUE, exponentiate = TRUE) %>%
dplyr::filter(term == "exposure") %>%
dplyr::pull(conf.low) %>%
tip(confounder_outcome_effect = 2.5)
}## The observed effect (1.09) 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: 0.09
## * estimated relationship between the unmeasured confounder and the outcome: 2.5
## # A tibble: 1 Ă— 5
## effect_adjusted effect_observed exposure_confounder_effect confounder_outcome…
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.0907 2.5
## # … with 1 more variable: n_unmeasured_confounders <dbl>Please note that the tipr project is released with a Contributor Code of Conduct. By contributing to this project, you agree to abide by its terms.