Authors: Lucy D’Agostino McGowan, Malcolm Barrett
License: MIT
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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 × 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: 1.1, 2.1). Note, in practice when estimating the effect of an exposure on a binary outcome using a GLM with the Poisson distribution and log link function, it is important to use a sandwich estimator to appropriately estimate the variability (this can be done in R using the sandwich package), which in this case gives a very similar result (95% CI: 1.1, 2.0).
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).
## ℹ 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 × 5
## effect_adjusted effect_observed exposure_confounder_e…¹ confounder_outcome_e…²
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.5 0.5 2.25
## # ℹ abbreviated names: ¹exposure_confounder_effect, ²confounder_outcome_effect
## # ℹ 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.
## ℹ 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 × 5
## effect_adjusted effect_observed exposure_confounder_e…¹ confounder_outcome_e…²
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.5 1.19
## # ℹ abbreviated names: ¹exposure_confounder_effect, ²confounder_outcome_effect
## # ℹ 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 × 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 risk ratio 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 × 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)## ℹ 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 × 6
## effect_adjusted effect_observed exposed_confounder_prev unexposed_confounder…¹
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.25 0.1
## # ℹ abbreviated name: ¹unexposed_confounder_prev
## # ℹ 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.
## ℹ 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 × 5
## effect_adjusted effect_observed exposure_confounder_e…¹ confounder_outcome_e…²
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.25 1.05
## # ℹ abbreviated names: ¹exposure_confounder_effect, ²confounder_outcome_effect
## # ℹ 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)
}## ℹ 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 × 5
## effect_adjusted effect_observed exposure_confounder_e…¹ confounder_outcome_e…²
## <dbl> <dbl> <dbl> <dbl>
## 1 1 1.09 0.0907 2.5
## # ℹ abbreviated names: ¹exposure_confounder_effect, ²confounder_outcome_effect
## # ℹ 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.