25  Hypothesis Tests

The marginaleffects package can conduct linear or non-linear hypothesis tests on the coefficients of any supported model class, or on the quantities generated by any of the other functions of the package: predictions(), comparisons(), or slopes().

There are two main entry points for hypothesis tests:

  1. marginaleffects functions: Use the hypothesis argument.
  2. Other R objects and models: Use the hypotheses() function.

Both the hypothesis argument and the hypotheses() function accept several input types, which allow a lot of flexibility in the specification of hypothesis tests:

  1. Numeric: Value of a null hypothesis.
  2. Formula: Equation of a (non-)linear hypothesis test.
  3. Function: Tests on arbitrary transformations or aggregations.
  4. Matrix: Contrast matrices and vectors.

This vignette shows how to use each of these strategies to conduct hypothesis tests on model coefficients or on the quantities estimated by the marginaleffects package. After reading it, you will be able to specify custom hypothesis tests and contrasts to assess statements like:

26 Numeric (null hypothesis)

The simplest way to modify a hypothesis test is to change the null hypothesis. By default, all functions in the marginaleffects package assume that the null is 0. This can be changed with the hypothesis argument.

26.1 Coefficients

Consider a simple logistic regression model:

library(marginaleffects)
mod <- glm(am ~ hp + drat, data = mtcars, family = binomial)

By default, the summary() function will report the results of hypothesis tests where the null is set to 0:

summary(mod)
#> 
#> Call:
#> glm(formula = am ~ hp + drat, family = binomial, data = mtcars)
#> 
#> Coefficients:
#>               Estimate Std. Error z value Pr(>|z|)  
#> (Intercept) -29.076080  12.416916  -2.342   0.0192 *
#> hp            0.010793   0.009328   1.157   0.2473  
#> drat          7.309781   3.046597   2.399   0.0164 *
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#> 
#> (Dispersion parameter for binomial family taken to be 1)
#> 
#>     Null deviance: 43.230  on 31  degrees of freedom
#> Residual deviance: 20.144  on 29  degrees of freedom
#> AIC: 26.144
#> 
#> Number of Fisher Scoring iterations: 7

Using hypotheses(), we can easily change the null hypothesis:

hypotheses(mod, hypothesis = 6)
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
(Intercept) -29.0761 12.41692 -2.82 0.00473 7.7 -53.41279 -4.7394
hp 0.0108 0.00933 -642.04 < 0.001 Inf -0.00749 0.0291
drat 7.3098 3.04660 0.43 0.66726 0.6 1.33856 13.2810

26.2 Predictions

Changing the value of the null is particularly important in the context of predictions, where the 0 baseline may not be particularly meaningful. For example, here we compute the predicted outcome for a hypothetical unit where all regressors are fixed to their sample means:

predictions(mod, newdata = "mean")
Estimate Pr(>|z|) S 2.5 % 97.5 % hp drat
Type: invlink(link)
0.231 0.135 2.9 0.0584 0.592 147 3.6

The Z statistic and p value reported above assume that the null hypothesis equals zero. We can change the null with the hypothesis argument:

predictions(mod, newdata = "mean", hypothesis = .5)
Estimate Pr(>|z|) S 2.5 % 97.5 % hp drat
Type: invlink(link)
0.231 0.0343 4.9 0.0584 0.592 147 3.6

26.3 Comparisons and slopes

When computing different quantities of interest like risk ratios, it can make sense to set the null hypothesis to 1 rather than 0:

avg_comparisons(
    mod,
    variables = "hp",
    comparison = "ratio",
    hypothesis = 1) |>
    print(digits = 5)
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
1.0027 0.0021743 1.2515 0.21074 2.2 0.99846 1.007

27 Equations

The hypotheses() function emulates the equation-based syntax of the well-established car::deltaMethod and car::linearHypothesis functions. However, marginaleffects supports more models, requires fewer dependencies, and offers some convenience features like the vcov argument for robust standard errors.

The syntax we illustrate in this section takes the form a string equation, ex: "am = 2 * vs"

27.1 Coefficients

Let’s start by estimating a simple model:

library(marginaleffects)
mod <- lm(mpg ~ hp + wt + factor(cyl), data = mtcars)

When the FUN and hypothesis arguments of hypotheses() equal NULL (the default), the function returns a data.frame of raw estimates:

Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
(Intercept) 35.8460 2.041 17.56 <0.001 227.0 31.8457 39.846319
hp -0.0231 0.012 -1.93 0.0531 4.2 -0.0465 0.000306
wt -3.1814 0.720 -4.42 <0.001 16.6 -4.5918 -1.771012
factor(cyl)6 -3.3590 1.402 -2.40 0.0166 5.9 -6.1062 -0.611803
factor(cyl)8 -3.1859 2.170 -1.47 0.1422 2.8 -7.4399 1.068169

Test of equality between coefficients:

hypotheses(mod, "hp = wt")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
3.16 0.72 4.39 <0.001 16.4 1.75 4.57

Non-linear function of coefficients

hypotheses(mod, "exp(hp + wt) = 0.1")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-0.0594 0.0292 -2.04 0.0418 4.6 -0.117 -0.0022

The vcov argument behaves in the same was as in all the other marginaleffects functions, allowing us to easily compute robust standard errors:

hypotheses(mod, "hp = wt", vcov = "HC3")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
3.16 0.805 3.92 <0.001 13.5 1.58 4.74

We can use shortcuts like b1, b2, ... to identify the position of each parameter in the output of FUN. For example, b2=b3 is equivalent to hp=wt because those term names appear in the 2nd and 3rd row when we call hypotheses(mod).

hypotheses(mod, "b2 = b3")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
3.16 0.72 4.39 <0.001 16.4 1.75 4.57 
hypotheses(mod, hypothesis = "b* / b3 = 1")
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
b1 / b3 = 1 -12.26735 2.07340 -5.9165 <0.001 28.2 -16.33 -8.204
b2 / b3 = 1 -0.99273 0.00413 -240.5537 <0.001 Inf -1.00 -0.985
b3 / b3 = 1 0.00000 NA NA NA NA NA NA
b4 / b3 = 1 0.05583 0.58287 0.0958 0.924 0.1 -1.09 1.198
b5 / b3 = 1 0.00141 0.82981 0.0017 0.999 0.0 -1.62 1.628

Term names with special characters must be enclosed in backticks:

hypotheses(mod, "`factor(cyl)6` = `factor(cyl)8`")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-0.173 1.65 -0.105 0.917 0.1 -3.41 3.07

27.2 Predictions

Now consider the case of adjusted predictions:

mod <- lm(mpg ~ am + vs, data = mtcars)

p <- predictions(
    mod,
    newdata = datagrid(am = 0:1, vs = 0:1))
p
am vs Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
0 0 14.6 0.926 15.8 <0.001 183.4 12.8 16.4
0 1 21.5 1.130 19.0 <0.001 266.3 19.3 23.7
1 0 20.7 1.183 17.5 <0.001 224.5 18.3 23.0
1 1 27.6 1.130 24.4 <0.001 435.0 25.4 29.8

Since there is no term column in the output of the predictions function, we must use parameter identifiers like b1, b2, etc. to determine which estimates we want to compare:

hypotheses(p, hypothesis = "b1 = b2")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-6.93 1.26 -5.49 <0.001 24.6 -9.4 -4.46

Or directly:

predictions(
    mod,
    hypothesis = "b1 = b2",
    newdata = datagrid(am = 0:1, vs = 0:1))
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
-6.93 1.26 -5.49 <0.001 24.6 -9.4 -4.46 

p$estimate[1] - p$estimate[2]
#> [1] -6.929365

There are many more possibilities:

predictions(
    mod,
    hypothesis = "b1 + b2 = 30",
    newdata = datagrid(am = 0:1, vs = 0:1))
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
6.12 1.64 3.74 <0.001 12.4 2.91 9.32 

p$estimate[1] + p$estimate[2] - 30
#> [1] 6.118254

predictions(
    mod,
    hypothesis = "(b2 - b1) / (b3 - b2) = 0",
    newdata = datagrid(am = 0:1, vs = 0:1))
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
-8.03 17 -0.473 0.636 0.7 -41.3 25.2

27.3 Comparisons and slopes

The avg_comparisons() function allows us to answer questions of this form: On average, how does the expected outcome change when I change one regressor by some amount? Consider this:

mod <- lm(mpg ~ am + vs, data = mtcars)

cmp <- avg_comparisons(mod)
cmp
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
am 6.07 1.27 4.76 <0.001 19.0 3.57 8.57
vs 6.93 1.26 5.49 <0.001 24.6 4.46 9.40

This tells us that, on average, moving from 0 to 1 on the am changes the predicted outcome by about 6.1, and changing .vs from 0 to 1 changes the predicted outcome by about 6.9.

Is the difference between those two estimates statistically significant? In other words, Is the effect of am equal to the effect of vs? To answer this question, we use the hypothesis argument:

avg_comparisons(mod, hypothesis = "am = vs")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
-0.863 1.94 -0.445 0.656 0.6 -4.66 2.94

The hypothesis string can include any valid R expression, so we can run some silly non-linear tests:

avg_comparisons(mod, hypothesis = "exp(am) - 2 * vs = -400")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
817 550 1.49 0.137 2.9 -261 1896

Note that the p values and confidence intervals are calculated using the delta method and are thus based on the assumption that the hypotheses expression is approximately normally distributed. For (very) non-linear functions of the parameters, this is not realistic, and we get p values with incorrect error rates and confidence intervals with incorrect coverage probabilities. For such hypotheses, it’s better to calculate the confidence intervals using the bootstrap (see inferences for details):

While the confidence interval from the delta method is symmetric, equal to the estimate ± 1.96 times the standard error, the (perhaps) more reliable confidence interval from the bootstrap is highly skewed.

set.seed(1234)

avg_comparisons(mod, hypothesis = "exp(am) - 2 * vs = -400") |>
  inferences(method = "boot")
Estimate Std. Error 2.5 % 97.5 %
Type: response
817 1854 414 6990

The same approach can be taken to compare slopes:

mod <- lm(mpg ~ qsec * hp, data = mtcars)

avg_slopes(mod, hypothesis = "10 * hp - qsec = 0")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
0.262 0.353 0.742 0.458 1.1 -0.43 0.954

28 Formulas (Group-wise)

Since version 0.20.1.5 of marginaleffects, the hypothesis argument also accepts a formula to specify hypotheses to be tested on every row of the output, or within subsets. Consider this set of average predictions:

dat <- within(mtcars, {
    gear = factor(gear)
    cyl = factor(cyl)
})
dat <- sort_by(dat, ~ gear + cyl + am)
mod <- lm(mpg ~ am * wt * factor(cyl), data = dat)

avg_predictions(mod, by = "gear")
gear Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
3 16.1 0.609 26.5 <0.001 510.5 14.9 17.3
4 24.2 0.613 39.5 <0.001 Inf 23.0 25.4
5 22.2 0.837 26.5 <0.001 511.3 20.5 23.8

We can make “sequential”, “reference”, or “meandev” comparisons, either as differences (the default) or ratios:

avg_predictions(mod, 
    hypothesis = ~ sequential,
    by = "gear")
Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
(4) - (3) 8.08 0.835 9.68 <0.001 71.2 6.45 9.721
(5) - (4) -2.01 0.829 -2.43 0.0152 6.0 -3.64 -0.387 

avg_predictions(mod, 
    hypothesis = ~ meandev,
    by = "gear")
Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
(3) - Avg -4.72 0.563 -8.38 <0.001 54.1 -5.822 -3.62
(4) - Avg 3.37 0.434 7.75 <0.001 46.6 2.514 4.22
(5) - Avg 1.35 0.560 2.42 0.0156 6.0 0.257 2.45 

avg_predictions(mod, 
    hypothesis = ratio ~ sequential,
    by = "gear")
Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
(4) / (3) 1.502 0.0662 22.7 <0.001 375.6 1.372 1.632
(5) / (4) 0.917 0.0336 27.3 <0.001 543.2 0.851 0.983

We can also test hypotheses by subgroup. For example, in this case, we want to compare every estimate to the reference category (first estimate) in each subset of am:

avg_predictions(mod, 
    by = c("am", "gear")
)
am gear Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
0 3 16.1 0.609 26.5 <0.001 510.5 14.9 17.3
0 4 21.0 1.091 19.3 <0.001 272.8 18.9 23.2
1 4 25.8 0.740 34.8 <0.001 880.0 24.3 27.2
1 5 22.2 0.837 26.5 <0.001 511.3 20.5 23.8 

avg_predictions(mod, 
    hypothesis = ~ sequential | am,
    by = c("am", "gear")
)
Hypothesis am Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
(4) - (3) 0 4.93 1.190 4.14 <0.001 14.8 2.59 7.26
(5) - (4) 1 -3.59 0.815 -4.41 <0.001 16.5 -5.19 -1.99

The grouping component of the formula is particularly useful when conducting tests on contrasts:

avg_comparisons(mod, 
    variables = "cyl",
    by = "am")
Term Contrast am Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
cyl 6 - 4 0 -11.33 6.04 -1.875 0.0608 4.0 -23.2 0.512
cyl 8 - 4 0 -9.59 3.84 -2.495 0.0126 6.3 -17.1 -2.057
cyl 6 - 4 1 -4.39 4.94 -0.889 0.3739 1.4 -14.1 5.291
cyl 8 - 4 1 -7.85 8.46 -0.928 0.3535 1.5 -24.4 8.730 

avg_comparisons(mod, 
    variables = "cyl",
    by = "am", 
    hypothesis = ~ sequential | contrast)
Contrast Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
6 - 4 (cyl, 1) - (cyl, 0) 6.93 7.80 0.889 0.374 1.4 -8.36 22.2
8 - 4 (cyl, 1) - (cyl, 0) 1.74 9.29 0.188 0.851 0.2 -16.46 20.0 

avg_comparisons(mod, 
    variables = "cyl",
    by = "am", 
    hypothesis = ~ sequential | am)
Hypothesis am Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
(cyl, 8 - 4) - (cyl, 6 - 4) 0 1.74 4.79 0.363 0.717 0.5 -7.64 11.1
(cyl, 8 - 4) - (cyl, 6 - 4) 1 -3.45 9.65 -0.358 0.720 0.5 -22.37 15.5

It is also a powerful tool in categorical outcome models, to compare estimates for different outcome levels:

library("MASS")

mod <- polr(gear ~ cyl + hp, dat)

avg_comparisons(mod, variables = "cyl")
Group Contrast Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 6 - 4 0.2679 0.0787 3.403 < 0.001 10.6 0.1136 0.4221
3 8 - 4 0.8582 0.0432 19.886 < 0.001 289.9 0.7737 0.9428
4 6 - 4 0.0108 0.0474 0.228 0.81962 0.3 -0.0822 0.1038
4 8 - 4 -0.1971 0.0651 -3.028 0.00246 8.7 -0.3246 -0.0695
5 6 - 4 -0.2787 0.0717 -3.887 < 0.001 13.3 -0.4192 -0.1382
5 8 - 4 -0.6612 0.0694 -9.533 < 0.001 69.2 -0.7971 -0.5253

Is the “8 vs. 4” contrast equal to the “6 vs. 4” contrast, within each outcome level?

avg_comparisons(mod, 
    variables = "cyl",
    hypothesis = ~ sequential | group)
Group Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 (cyl, 8 - 4) - (cyl, 6 - 4) 0.590 0.0815 7.24 < 0.001 41.0 0.431 0.7502
4 (cyl, 8 - 4) - (cyl, 6 - 4) -0.208 0.0717 -2.90 0.00376 8.1 -0.348 -0.0673
5 (cyl, 8 - 4) - (cyl, 6 - 4) -0.383 0.0575 -6.65 < 0.001 35.0 -0.495 -0.2698

Is the “8 vs. 4” contrast for outcome level 3 equal to the “8 vs. 4” contrast for outcome level 4?

avg_comparisons(mod, 
    variables = "cyl",
    hypothesis = ~ sequential | contrast)
Contrast Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
6 - 4 (4, cyl) - (3, cyl) -0.257 0.1084 -2.37 0.01775 5.8 -0.470 -0.0445
6 - 4 (5, cyl) - (4, cyl) -0.289 0.0927 -3.12 0.00178 9.1 -0.471 -0.1079
8 - 4 (4, cyl) - (3, cyl) -1.055 0.0859 -12.28 < 0.001 112.7 -1.224 -0.8869
8 - 4 (5, cyl) - (4, cyl) -0.464 0.1274 -3.64 < 0.001 11.9 -0.714 -0.2145

29 Functions

Hypothesis tests can also be conducted using arbitrary R functions. This allows users to test hypotheses on complex aggregations or transformations. To achieve this, we define a custom function which accepts a fitted model or marginaleffects objects, and returns a data frame with at least two columns: term (or hypothesis) and estimate.

29.1 Predictions

When supplying a function to the hypothesis argument, that function must accept an argument x which is a data frame with columns rowid and estimate (and optional columns for other elements of newdata). That function must return a data frame with columns term (or hypothesis) and estimate.

In this example, we test if the mean predicted value is different from 2:

dat <- transform(mtcars, gear = factor(gear), cyl = factor(cyl))

mod <- lm(wt ~ mpg * hp * cyl, data = dat)

hyp <- function(x) {
    data.frame(
        hypothesis = "Avg(Ŷ) = 2",
        estimate = mean(x$estimate) - 2
    )
}
predictions(mod, hypothesis = hyp)
Hypothesis Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
Avg(Ŷ) = 2 1.22 0.0914 13.3 <0.001 132.0 1.04 1.4

In this ordinal logit model, the predictions() function returns one row per observation and per level of the outcome variable:

library(MASS)
library(dplyr)

mod <- polr(gear ~ cyl + hp, dat)

avg_predictions(mod)
Group Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 0.471 0.0584 8.05 <0.001 50.2 0.3561 0.585
4 0.366 0.0715 5.12 <0.001 21.7 0.2263 0.507
5 0.163 0.0478 3.41 <0.001 10.6 0.0692 0.257

We can use a function in the hypothesis argument to collapse the rows, displaying the average predicted values in groups 3-4 vs. 5:

fun <- function(x) {
    out <- x |> 
        mutate(term = ifelse(group %in% 3:4, "3 & 4", "5")) |>
        summarize(estimate = mean(estimate), .by = term)
    return(out)
}
avg_predictions(mod, hypothesis = fun)
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 & 4 0.419 0.0239 17.51 <0.001 225.5 0.3717 0.465
5 0.163 0.0478 3.41 <0.001 10.6 0.0692 0.257

And we can compare the two categories by doing:

fun <- function(x) {
    out <- x |> 
        mutate(term = ifelse(group %in% 3:4, "3 & 4", "5")) |>
        summarize(estimate = mean(estimate), .by = term) |>
        summarize(estimate = diff(estimate), term = "5 - (3 & 4)")
    return(out)
}
avg_predictions(mod, hypothesis = fun)
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
-0.256 0.0717 -3.56 <0.001 11.4 -0.396 -0.115

29.2 Comparisons and slopes

In the same ordinal logit model, we can estimate the average effect of an increase of 1 unit in hp on the expected probability of every level of the outcome:

avg_comparisons(mod, variables = "hp")
Group Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 -0.00774 0.002294 -3.38 <0.001 10.4 -0.01224 -0.00325
4 0.00258 0.002201 1.17 0.24 2.1 -0.00173 0.00690
5 0.00516 0.000882 5.85 <0.001 27.6 0.00343 0.00689

Compare estimate for different outcome levels:

fun <- function(x) {
    x |> 
    mutate(estimate = (estimate - lag(estimate)),
           group = sprintf("%s - %s", group, lag(group))) |>
    filter(!is.na(estimate))
}
avg_comparisons(mod, variables = "hp", hypothesis = fun)
Group Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
4 - 3 0.01033 0.00441 2.34 0.0191 5.7 0.00169 0.01897
5 - 4 0.00258 0.00245 1.05 0.2922 1.8 -0.00222 0.00737

Now suppose we want to compare the effect of hp for different levels of the outcome, but this time we do the computation within levels of cyl:

avg_comparisons(mod, variables = "hp", by = "cyl")
Group Term cyl Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 hp 4 -0.007862 0.003445 -2.282 0.02248 5.5 -1.46e-02 -0.00111
4 hp 4 -0.003024 0.003714 -0.814 0.41560 1.3 -1.03e-02 0.00426
5 hp 4 0.010886 0.002652 4.104 < 0.001 14.6 5.69e-03 0.01608
3 hp 6 -0.013282 0.005038 -2.637 0.00838 6.9 -2.32e-02 -0.00341
4 hp 6 0.008573 0.006431 1.333 0.18251 2.5 -4.03e-03 0.02118
5 hp 6 0.004709 0.001929 2.442 0.01462 6.1 9.29e-04 0.00849
3 hp 8 -0.004882 0.001457 -3.350 < 0.001 10.3 -7.74e-03 -0.00203
4 hp 8 0.003995 0.001627 2.455 0.01409 6.1 8.06e-04 0.00719
5 hp 8 0.000887 0.000431 2.059 0.03947 4.7 4.27e-05 0.00173 
fun <- function(x) {
    x |> 
    mutate(estimate = (estimate - lag(estimate)),
           group = sprintf("%s - %s", group, lag(group)), 
           .by = "cyl") |>
    filter(!is.na(estimate))
}
avg_comparisons(mod, variables = "hp", by = "cyl", hypothesis = fun)
Group Term cyl Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
4 - 3 hp 4 0.00484 0.00666 0.727 0.46719 1.1 -0.008205 0.017882
5 - 4 hp 4 0.01391 0.00546 2.548 0.01082 6.5 0.003211 0.024607
4 - 3 hp 6 0.02185 0.01139 1.919 0.05503 4.2 -0.000471 0.044180
5 - 4 hp 6 -0.00386 0.00805 -0.480 0.63117 0.7 -0.019638 0.011911
4 - 3 hp 8 0.00888 0.00306 2.902 0.00371 8.1 0.002882 0.014874
5 - 4 hp 8 -0.00311 0.00188 -1.651 0.09869 3.3 -0.006798 0.000581

29.3 Fitted models

The hypothesis argument can be used to compute standard errors for arbitrary functions of model parameters. This user-supplied function must accept a single model object, and return a data.frame with two columns named term and estimate.

Here, we test if the sum of the hp and mpg coefficients is equal to 2:

mod <- glm(am ~ hp + mpg, data = mtcars, family = binomial)

fun <- function(x) {
    b <- coef(x)
    out <- data.frame(
        term = "hp + mpg = 2",
        estimate = b["hp"] + b["mpg"] - 2,
        row.names = NULL
    )
    return(out)
}

hypotheses(mod, hypothesis = fun)
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-0.685 0.593 -1.16 0.248 2.0 -1.85 0.476

Test of equality between two two predictions:

fun <- function(x) {
    p <- predict(x, newdata = mtcars)
    out <- data.frame(term = "pred[2] = pred[3]", estimate = p[2] - p[3])
    return(out)
}
hypotheses(mod, hypothesis = fun)
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-1.33 0.616 -2.16 0.0305 5.0 -2.54 -0.125

We can also use more complex aggregation patterns. In this ordinal logistic regression model, we model the number of gears for each ar. If we compute fitted values with the predictions() function, we obtain one predicted probability for each individual car and for each level of the response variable:

library(MASS)
library(dplyr)

dat <- transform(mtcars, 
    gear = factor(gear),
    cyl = factor(cyl))
mod <- polr(gear ~ cyl + hp, dat)

predictions(mod)
Group Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: probs
3 0.3931 0.19125 2.06 0.03982 4.7 0.0183 0.768
3 0.3931 0.19125 2.06 0.03982 4.7 0.0183 0.768
3 0.0440 0.04256 1.03 0.30081 1.7 -0.0394 0.127
3 0.3931 0.19125 2.06 0.03982 4.7 0.0183 0.768
3 0.9963 0.00721 138.17 < 0.001 Inf 0.9822 1.010
5 0.6969 0.18931 3.68 < 0.001 12.1 0.3258 1.068
5 0.0555 0.06851 0.81 0.41775 1.3 -0.0788 0.190
5 0.8115 0.20626 3.93 < 0.001 13.5 0.4073 1.216
5 0.9111 0.16818 5.42 < 0.001 24.0 0.5815 1.241
5 0.6322 0.19648 3.22 0.00129 9.6 0.2471 1.017

There are three levels to the outcome: 3, 4, and 5. Imagine that, for each car in the dataset, we want to collapse categories of the output variable into two categories (“3 & 4” and “5”) by taking sums of predicted probabilities. Then, we want to take the average of those predicted probabilities for each level of cyl. To do so, we define a custom function, and pass it to the hypothesis argument of the hypotheses() function:

fun <- function(x) {
    predictions(x, vcov = FALSE) |>
        # label the new categories of outcome levels
        mutate(group = ifelse(group %in% c("3", "4"), "3 & 4", "5")) |>
        # sum of probabilities at the individual level
        summarize(estimate = sum(estimate), .by = c("rowid", "cyl", "group")) |>
        # average probabilities for each value of `cyl`
        summarize(estimate = mean(estimate), .by = c("cyl", "group")) |>
        # the `FUN` argument requires a `term` column
        rename(term = cyl)
}

hypotheses(mod, hypothesis = fun)
Group Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
3 & 4 6 0.8390 0.0651 12.89 <0.001 123.9 0.7115 0.967
3 & 4 4 0.7197 0.1099 6.55 <0.001 34.0 0.5044 0.935
3 & 4 8 0.9283 0.0174 53.45 <0.001 Inf 0.8943 0.962
5 6 0.1610 0.0651 2.47 0.0134 6.2 0.0334 0.289
5 4 0.2803 0.1099 2.55 0.0108 6.5 0.0649 0.496
5 8 0.0717 0.0174 4.13 <0.001 14.7 0.0377 0.106

Note that this workflow will not work for bayesian models or with bootstrap. However, with those models it is trivial to do the same kind of aggregation by calling get_draws() and operating directly on draws from the posterior distribution. See the vignette on bayesian analysis for examples with the get_draws() function.

30 Vectors and Matrices

The predictions() function can estimate marginal means. The hypothesis argument of that function offers a powerful mechanism to estimate custom contrasts between marginal means, by way of linear combination.

30.1 Simple contrast

Consider a simple example:

library(marginaleffects)
library(emmeans)
library(nnet)

dat <- mtcars
dat$carb <- factor(dat$carb)
dat$cyl <- factor(dat$cyl)
dat$am <- as.logical(dat$am)
dat <- sort_by(dat, ~carb)

mod <- lm(mpg ~ carb + cyl, dat)

mm <- predictions(mod,
    by = "carb",
    newdata = "balanced")
mm
carb Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
1 21.7 1.44 15.06 <0.001 167.8 18.8 24.5
2 21.3 1.23 17.29 <0.001 220.0 18.9 23.8
3 21.4 2.19 9.77 <0.001 72.5 17.1 25.7
4 18.9 1.21 15.59 <0.001 179.7 16.5 21.3
6 19.8 3.55 5.56 <0.001 25.2 12.8 26.7
8 20.1 3.51 5.73 <0.001 26.6 13.2 27.0

The contrast between marginal means for carb==1 and carb==2 is:

21.66232 - 21.34058 
#> [1] 0.32174

or

21.66232 + -(21.34058)
#> [1] 0.32174

or

sum(c(21.66232, 21.34058) * c(1, -1))
#> [1] 0.32174

or

c(21.66232, 21.34058) %*% c(1, -1)
#>         [,1]
#> [1,] 0.32174

The last two commands express the contrast of interest as a linear combination of marginal means.

In the predictions() function, we can supply a hypothesis argument to compute linear combinations of marginal means. This argument must be a numeric vector of the same length as the number of rows in the output. For example, in the previous there were six rows, and the two marginal means we want to compare are at in the first two positions:

lc <- c(1, -1, 0, 0, 0, 0)
predictions(mod,
    by = "carb",
    newdata = "balanced",
    hypothesis = lc)
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
custom 0.322 1.77 0.181 0.856 0.2 -3.15 3.8

30.2 Complex contrast

Of course, we can also estimate more complex contrasts:

lc <- c(0, -2, 1, 1, -1, 1)
predictions(mod,
    by = "carb",
    newdata = "balanced",
    hypothesis = lc)
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
custom -2.02 6.32 -0.32 0.749 0.4 -14.4 10.4

emmeans produces similar results:

library(emmeans)
em <- emmeans(mod, "carb")
lc <- data.frame(custom_contrast = c(0, -2, 1, 1, -1, 1))
contrast(em, method = lc)
#>  contrast        estimate   SE df t.ratio p.value
#>  custom_contrast    -2.02 6.32 24  -0.320  0.7516
#> 
#> Results are averaged over the levels of: cyl

30.3 Multiple contrasts

Users can also compute multiple linear combinations simultaneously by supplying a numeric matrix to hypotheses. This matrix must have the same number of rows as the output of slopes(), and each column represents a distinct set of weights for different linear combinations. The column names of the matrix become labels in the output. For example:

lc <- matrix(c(
    -2, 1, 1, 0, -1, 1,
    1, -1, 0, 0, 0, 0
    ), ncol = 2)
colnames(lc) <- c("Contrast A", "Contrast B")
lc
#>      Contrast A Contrast B
#> [1,]         -2          1
#> [2,]          1         -1
#> [3,]          1          0
#> [4,]          0          0
#> [5,]         -1          0
#> [6,]          1          0

predictions(mod,
    by = "carb",
    newdata = "balanced",
    hypothesis = lc)
Term Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
Contrast A -0.211 6.93 -0.0304 0.976 0.0 -13.79 13.4
Contrast B 0.322 1.77 0.1814 0.856 0.2 -3.15 3.8

31 Arbitrary quantities

marginaleffects can also compute uncertainty estimates for arbitrary quantities hosted in a data frame, as long as the user can supply a variance-covariance matrix. (Thanks to Kyle F Butts for this cool feature and example!)

Say you run a monte-carlo simulation and you want to perform hypothesis of various quantities returned from each simulation. The quantities are correlated within each draw:

# simulated means and medians
draw <- function(i) { 
  x <- rnorm(n = 10000, mean = 0, sd = 1)
  out <- data.frame(median = median(x), mean =  mean(x))
  return(out)
}
sims <- do.call("rbind", lapply(1:25, draw))

# average mean and average median 
coeftable <- data.frame(
  term = c("median", "mean"),
  estimate = c(mean(sims$median), mean(sims$mean))
)

# variance-covariance
vcov <- cov(sims)

# is the median equal to the mean?
hypotheses(
  coeftable,
  vcov = vcov,
  hypothesis = "median = mean"
)
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-0.00121 0.00572 -0.211 0.833 0.3 -0.0124 0.01

32 Joint hypotheses tests

The hypotheses() function can also test multiple hypotheses jointly. For example, consider this model:

model <- lm(mpg ~ as.factor(cyl) * hp, data = mtcars)
coef(model)
#>        (Intercept)    as.factor(cyl)6    as.factor(cyl)8                 hp as.factor(cyl)6:hp as.factor(cyl)8:hp 
#>        35.98302564       -15.30917451       -17.90295193        -0.11277589         0.10516262         0.09853177

We may want to test the null hypothesis that two of the coefficients are jointly (both) equal to zero.

hypotheses(model, joint = c("as.factor(cyl)6:hp", "as.factor(cyl)8:hp"))
F Pr(>|F|) Df 1 Df 2
2.11 0.142 2 26

The joint argument allows users to flexibly specify the parameters to be tested, using character vectors, integer indices, or Perl-compatible regular expressions. We can also specify the null hypothesis for each parameter individually using the hypothesis argument.

Naturally, the hypotheses function also works with marginaleffects objects.

# ## joint hypotheses: regular expression
hypotheses(model, joint = "cyl")
F Pr(>|F|) Df 1 Df 2
5.7 0.00197 4 26 

# joint hypotheses: integer indices
hypotheses(model, joint = 2:3)
F Pr(>|F|) Df 1 Df 2
6.12 0.00665 2 26 

# joint hypotheses: different null hypotheses
hypotheses(model, joint = 2:3, hypothesis = 1)
F Pr(>|F|) Df 1 Df 2
6.84 0.00411 2 26 
hypotheses(model, joint = 2:3, hypothesis = 1:2)
F Pr(>|F|) Df 1 Df 2
7.47 0.00273 2 26 

# joint hypotheses: marginaleffects object
cmp <- avg_comparisons(model)
hypotheses(cmp, joint = "cyl")
F Pr(>|F|) Df 1 Df 2
1.6 0.221 2 26

We can also combine multiple calls to hypotheses to execute a joint test on linear combinations of coefficients:

# fit model
mod <- lm(mpg ~ factor(carb), mtcars)

# hypothesis matrix for linear combinations
H <- matrix(0, nrow = length(coef(mod)), ncol = 2)
H[2:3, 1] <- H[4:6, 2] <- 1

# test individual linear combinations
hyp <- hypotheses(mod, hypothesis = H)
hyp
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
-12.0 4.92 -2.44 0.01477 6.1 -21.6 -2.35
-25.5 9.03 -2.83 0.00466 7.7 -43.2 -7.85 

# test joint hypotheses
#hypotheses(hyp, joint = TRUE, hypothesis = c(-10, -20))

33 More

33.1 Difference-in-Differences

Now we illustrate how to use the machinery described above to do pairwise comparisons between contrasts, a type of analysis often associated with a “Difference-in-Differences” research design.

First, we simulate data with two treatment groups and pre/post periods:

library(data.table)

N <- 1000
did <- data.table(
    id = 1:N,
    pre = rnorm(N),
    trt = sample(0:1, N, replace = TRUE))
did$post <- did$pre + did$trt * 0.3 + rnorm(N)
did <- melt(
    did,
    value.name = "y",
    variable.name = "time",
    id.vars = c("id", "trt"))
head(did)
#>       id   trt   time          y
#>    <int> <int> <fctr>      <num>
#> 1:     1     1    pre  0.2050066
#> 2:     2     1    pre -0.0211390
#> 3:     3     0    pre  0.8581367
#> 4:     4     1    pre  0.8285873
#> 5:     5     0    pre -0.8229379
#> 6:     6     1    pre -0.4390908

Then, we estimate a linear model with a multiple interaction between the time and the treatment indicators. We also compute contrasts at the mean for each treatment level:

did_model <- lm(y ~ time * trt, data = did)

comparisons(
    did_model,
    newdata = datagrid(trt = 0:1),
    variables = "time")
trt Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
0 -0.0664 0.0797 -0.834 0.404 1.3 -0.223 0.0897
1 0.3862 0.0762 5.066 <0.001 21.2 0.237 0.5357

Finally, we compute pairwise differences between contrasts. This is the Diff-in-Diff estimate:

comparisons(
    did_model,
    variables = "time",
    newdata = datagrid(trt = 0:1),
    hypothesis = "pairwise")
Estimate Std. Error z Pr(>|z|) S 2.5 % 97.5 %
Type: response
-0.453 0.11 -4.11 <0.001 14.6 -0.669 -0.237