# Wald tests of multiple-constraint null hypotheses

#### 2021-01-23

Version 0.5.0 of clubSandwich introduced a new syntax for Wald_test(), a function for conducting tests of multiple-constraint hypotheses. In previous versions, this function was poorly documented and, consequently, probably little used. This vignette will demonstrate the new syntax.

For purposes of illustration, I will use the STAR data (available in the AER package), which is drawn from a randomized trial evaluating the effects of elementary school class size on student achievement. The data consist of individual-level measures for students in each of several dozen schools. For purposes of illustration, I will look at effects on math performance in first grade. Treatment conditions (the variable called stark) were assigned at the classroom level, and consisted of either a) a regular-size class, b) a small-size class, or c) a regular-size class but with the addition of a teacher’s aide. In all of what follows, I will cluster standard errors by school in order to allow for generalization to a super-population of schools.

library(clubSandwich)
data(STAR, package = "AER")

# clean up a few variables
levels(STAR$stark) <- "aide" levels(STAR$schoolk) <- "urban"
STAR <- subset(STAR,
!is.na(schoolidk),
select = c(schoolidk, schoolk, stark, gender, ethnicity, math1, lunchk))
head(STAR)
##      schoolidk  schoolk   stark gender ethnicity math1   lunchk
## 1137        63    rural   small female      cauc   538 non-free
## 1143        20 suburban   small female      afam   592 non-free
## 1183        19    urban    aide   male      afam    NA     free
## 1277        69    rural regular   male      cauc   584 non-free
## 1292        79    rural   small   male      cauc   545     free
## 1308         5    rural regular   male      cauc   553     free

# 1 The Wald test function

The Wald_test() function can be used to conduct hypothesis tests that involve multiple constraints on the regression coefficients. Consider a linear model for an outcome $$Y_{ij}$$ regressed on a $$1 \times p$$ row vector of predictors $$\mathbf{x}_{ij}$$ (which might include a constant intercept term): $Y_{ij} = \mathbf{x}_{ij} \boldsymbol\beta + \epsilon_{ij}$ The regression coefficient vector is $$\boldsymbol\beta$$. In quite general terms, a set of constraints on the regression coefficient vector can be expressed in terms of a $$q \times p$$ matrix $$\mathbf{C}$$, where each row of $$\mathbf{C}$$ corresponds to one constraint. A joint null hypothesis is then $$H_0: \mathbf{C} \boldsymbol\beta = \mathbf{0}$$, where $$\mathbf{0}$$ is a $$q \times 1$$ vector of zeros.1

Wald-type test are based on the test statistic $Q = \left(\mathbf{C}\boldsymbol{\hat\beta}\right)' \left(\mathbf{C} \mathbf{V}^{CR} \mathbf{C}'\right)^{-1} \left(\mathbf{C}\boldsymbol{\hat\beta}\right),$ where $$\boldsymbol{\hat\beta}$$ is the estimated regression coefficient vector and $$\mathbf{V}^{CR}$$ is a cluster-robust variance matrix. If the number of clusters is sufficiently large, then the distribution of $$Q$$ under the null hypothesis is approximately $$\chi^2(q)$$. Tipton & Pustejovsky (2015) investigated a wide range of other approximations to the null distribution of $$Q$$, many of which are included as options in Wald_test(). Based on a large simulation, they (…er…we…) recommended a method called the “approximate Hotelling’s $$T^2$$-Z” test, or “AHZ.” This test approximates the distribution of $$Q / q$$ by a $$T^2$$ distribution, which is a multiple of an $$F$$ distribution, with numerator degrees of freedom $$q$$ and denominator degrees of freedom based on a generalization of the Satterthwaite approximation.

The Wald_test() function has three main arguments:

args(Wald_test)
## function (obj, constraints, vcov, test = "HTZ", tidy = FALSE,
##     ...)
## NULL
• The obj argument is used to specify the estimated regression model on which to perform the test,
• the constraints argument is a $$\mathbf{C}$$ matrix expressing the set of constraints to test, and
• the vcov argument is a cluster-robust variance matrix, which is used to construct the test statistic. (Alternately, vcov can be the type of cluster-robust variance matrix to construct, in which case it will be computed internally.)

By default, Wald_test() will use the HTZ small-sample approximation. Other options are available (via the test argument) but not recommended for routine use. The optional tidy argument will be demonstrated below.

## 1.1 Testing treatment effects

Returning to the STAR data, let’s suppose we want to examine differences in math performance across class sizes. This can be done with a simple linear regression model, while clustering the standard errors by schoolidk. The estimating equation is $\left(\text{Math}\right)_{ij} = \beta_0 + \beta_1 \left(\text{small}\right)_{ij} + \beta_2 \left(\text{aide}\right)_{ij} + e_{ij},$ which can be estimated in R as follows:

lm_trt <- lm(math1 ~ stark, data = STAR)
V_trt <- vcovCR(lm_trt, cluster = STAR$schoolidk, type = "CR2") coef_test(lm_trt, vcov = V_trt) ## Coef. Estimate SE t-stat d.f. p-val (Satt) Sig. ## 1 (Intercept) 531.727 2.78 191.506 59.9 <0.001 *** ## 2 starksmall 9.469 2.30 4.114 65.6 <0.001 *** ## 3 starkaide -0.483 1.86 -0.259 65.6 0.796 In this estimating equation, the coefficients $$\beta_1$$ and $$\beta_2$$ represent treatment effects, or differences in average math scores relative to the reference level of stark, which in this case is a regular-size class. The t-statistics and p-values reported by coef_test are separate tests of the null hypotheses that each of these coefficients are equal to zero, meaning that there is no difference between the specified treatment condition and the reference level. We might want to instead test the joint null hypothesis that there are no differences among any of the conditions. This null can be expressed by a set of multiple constraints on the parameters: $$\beta_1 = 0$$ and $$\beta_2 = 0$$. To test the null hypothesis that $$\beta_1 = \beta_2 = 0$$ based on the treatment effects model specification, we can use: C_trt <- matrix(c(0,0,1,0,0,1), 2, 3) C_trt ## [,1] [,2] [,3] ## [1,] 0 1 0 ## [2,] 0 0 1 Wald_test(lm_trt, constraints = C_trt, vcov = V_trt) ## test Fstat df_num df_denom p_val sig ## HTZ 10.2 2 65.3 <0.001 *** The result includes details about the form of test computed, the $$F$$-statistic, the numerator and denominator degrees of freedom used to compute the reference distribution, and the $$p$$-value corresponding to the specified null hypothesis. In this example, $$p = 0.000141$$, so we can rule out the null hypothesis that there are no differences in math performance across conditions. The representation of null hypotheses as arbitrary constraint matrices is useful for developing theory about how to test such hypotheses, but it is not all that helpful for actually running tests—constructing constraint matrices “by hand” is just too cumbersome of an exercise. Moreover, $$\mathbf{C}$$ matrices typically follow one of a small number of patterns. Two common use cases are a) constraining a set of $$q > 1$$ parameters to all be equal to zero and b) constraining a set of $$q + 1$$ parameters to be equal to a common value. The clubSandwich package now includes a set of helper functions to create constraint matrices for these common use cases. ## 1.2constrain_zero() To constrain a set of $$q$$ regression coefficients to all be equal to zero, the simplest form of the $$\mathbf{C}$$ matrix would consist of a set of $$q$$ rows, where a single entry in each row would be equal to 1 and the remaining entries would all be zero. For the lm_trt model, the C matrix would look like this: $\mathbf{C} = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right],$ so that $\mathbf{C}\boldsymbol\beta = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \\ \beta_2 \end{array} \right] = \left[\begin{array}{c} \beta_1 \\ \beta_2 \end{array} \right],$ which is set equal to $$\left[\begin{array}{c} 0 \\ 0 \end{array} \right]$$. The constrain_zero() function will create matrices like this automatically. The function takes two main arguments: args(constrain_zero) ## function (constraints, coefs, reg_ex = FALSE) ## NULL • The constraints argument is used to specify which coefficients in a regression model to set equal to zero. • The coefs argument is the set of estimated regression coefficients, for which to calculate the constraints. Constraints can be specified by position index, by name, or via a regular expression. To test the joint null hypothesis that average math performance is equal across the three treatment conditions, we need to constrain the second and third coefficients to zero: constrain_zero(2:3, coefs = coef(lm_trt)) ## [,1] [,2] [,3] ## [1,] 0 1 0 ## [2,] 0 0 1 Or equivalently: constrain_zero(c("starksmall","starkaide"), coefs = coef(lm_trt)) ## [,1] [,2] [,3] ## [1,] 0 1 0 ## [2,] 0 0 1 or constrain_zero("^stark", coefs = coef(lm_trt), reg_ex = TRUE) ## [,1] [,2] [,3] ## [1,] 0 1 0 ## [2,] 0 0 1 Note that if constraints is a regular expression, then the reg_ex argument needs to be set to TRUE. The result of constrain_zero() can then be fed into the Wald_test() function: C_trt <- constrain_zero(2:3, coefs = coef(lm_trt)) Wald_test(lm_trt, constraints = C_trt, vcov = V_trt) ## test Fstat df_num df_denom p_val sig ## HTZ 10.2 2 65.3 <0.001 *** To reduce redundancy in the syntax, we can also omit the coefs argument to constrain_zero, so long as we call it inside of Wald_test2: Wald_test(lm_trt, constraints = constrain_zero(2:3), vcov = V_trt) ## test Fstat df_num df_denom p_val sig ## HTZ 10.2 2 65.3 <0.001 *** ## 1.3constrain_equal() Another common type of constraints involve setting a set of $$q + 1$$ regression coefficients to be all equal to a common (but unknown) value ($$q + 1$$ because it takes $$q$$ constraints to do this). There are many equivalent ways to express such a set of constraints in terms of a $$\mathbf{C}$$ matrix. One fairly simple form consists of a set of $$q$$ rows, where the entry corresponding to one of the coefficients of interest is equal to -1 and the entry corresponding to another coefficient of interest is equal to 1. To see how this works, let’s look at a different way of parameterizing our simple model for the STAR data, by using separate intercepts for each treatment condition. The estimating equation would then be $\left(\text{Math}\right)_{ij} = \beta_0 \left(\text{regular}\right)_{ij} + \beta_1 \left(\text{small}\right)_{ij} + \beta_2 \left(\text{aide}\right)_{ij} + e_{ij}.$ This model can be estimated in R by dropping the intercept term: lm_sep <- lm(math1 ~ 0 + stark, data = STAR) V_sep <- vcovCR(lm_sep, cluster = STAR$schoolidk, type = "CR2")
coef_test(lm_sep, vcov = V_sep)
##          Coef. Estimate   SE t-stat d.f. p-val (Satt) Sig.
## 1 starkregular      532 2.78    192 59.9       <0.001  ***
## 2   starksmall      541 2.89    187 65.0       <0.001  ***
## 3    starkaide      531 2.72    195 64.3       <0.001  ***

In this parameterization, the coefficients $$\beta_0$$, $$\beta_1$$, and $$\beta_2$$ represent the average math performance levels of students in each of the treatment conditions. The t-tests and p-values now have a very different interpretation because they pertain to the null hypothesis that the average performance level for a given condition is equal to zero. With this separate-intercepts model, the joint null hypothesis that performance levels are equal across conditions amounts to constraining the intercepts to be equal to each other: $$\beta_0 = \beta_1$$ and $$\beta_0 = \beta_2$$ (note that we don’t need the constraint $$\beta_1 = \beta_2$$ because it is implied by the first two).

For the lm_sep model, which has separate intercepts $$\beta_0$$, $$\beta_1$$, and $$\beta_2$$, the C matrix would look like this: $\mathbf{C} = \left[\begin{array}{ccc} -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right],$ so that $\mathbf{C}\boldsymbol\beta = \left[\begin{array}{ccc} -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \\ \beta_2 \end{array} \right] = \left[\begin{array}{c} \beta_1 - \beta_0 \\ \beta_2 - \beta_0 \end{array} \right],$ which is set equal to $$\left[\begin{array}{c} 0 \\ 0 \end{array} \right]$$.

The constrain_equal() function will create matrices like this automatically, given a set of coefficients to constrain. The syntax is identical to constrain_zero():

args(constrain_equal)
## function (constraints, coefs, reg_ex = FALSE)
## NULL

To test the joint null hypothesis that average math performance is equal across the three treatment conditions, we can constrain all three coefficients of lm_sep to be equal:

constrain_equal(1:3, coefs = coef(lm_sep))
##      [,1] [,2] [,3]
## [1,]   -1    1    0
## [2,]   -1    0    1

Or equivalently:

constrain_equal(c("starkregular","starksmall","starkaide"), coefs = coef(lm_sep))
##      [,1] [,2] [,3]
## [1,]   -1    1    0
## [2,]   -1    0    1

or

constrain_equal("^stark", coefs = coef(lm_sep), reg_ex = TRUE)
##      [,1] [,2] [,3]
## [1,]   -1    1    0
## [2,]   -1    0    1

Just as with constrain_zero, if constraints is a regular expression, then the reg_ex argument needs to be set to TRUE.

This constraint matrix can then be fed into Wald_test():

C_sep <- constrain_equal("^stark", coefs = coef(lm_sep), reg_ex = TRUE)
Wald_test(lm_sep, constraints = C_sep, vcov = V_sep)
##  test Fstat df_num df_denom  p_val sig
##   HTZ  10.2      2     65.3 <0.001 ***

or equivalently:

Wald_test(lm_sep, constraints = constrain_equal(1:3), vcov = V_sep)
##  test Fstat df_num df_denom  p_val sig
##   HTZ  10.2      2     65.3 <0.001 ***

Note that these test results are exactly equal to the tests based on lm_trt with constrain_zero(). They’re algebraically equivalent—just different ways of parameterizing the same model and constraints.

# 2 Testing an interaction

Let’s now consider how these functions can be applied in a more complex model. Suppose that we are interested in understanding whether the effect of being in a small class is consistent across schools in different areas, where areas are categorized as urban, suburban, or rural. To answer this question, we need to test for an interaction between urbanicity and treatment condition. One estimating equation that would let us examine this question is: \begin{aligned} \left(\text{Math}\right)_{ij} &= \beta_0 + \beta_1 \left(\text{suburban}\right)_{ij} + \beta_2 \left(\text{rural}\right)_{ij} \\ & \quad + \beta_3 \left(\text{small}\right)_{ij} + \beta_4 \left(\text{aide}\right)_{ij} \\ & \quad\quad + \beta_5 \left(\text{small}\right)(\text{suburban})_{ij} + \beta_6 \left(\text{aide}\right)(\text{suburban})_{ij} \\ & \quad\quad\quad + \beta_{7} \left(\text{small}\right)(\text{rural})_{ij} + \beta_{8} \left(\text{aide}\right)(\text{rural})_{ij} \\ & \quad\quad\quad\quad + \mathbf{x}_{ij} \boldsymbol\gamma + e_{ij}, \end{aligned} where $$\mathbf{x}_{ij}$$ is a row vector of student characteristics, included just to make the regression look fancier. In this specification, $$\beta_3$$ and $$\beta_4$$ represent the effects of being in a small class or aide class, compared to being in a regular class, but only for the reference level of urbanicity—in this case, urban schools. The coefficients $$\beta_5, \beta_6, \beta_7, \beta_8$$ all represent interactions between treatment condition and urbanicity. Here’s the model, estimated in R:

lm_urbanicity <- lm(math1 ~ schoolk * stark + gender + ethnicity + lunchk, data = STAR)
V_urbanicity <- vcovCR(lm_urbanicity, cluster = STAR$schoolidk, type = "CR2") coef_test(lm_urbanicity, vcov = V_urbanicity) ## Coef. Estimate SE t-stat d.f. p-val (Satt) Sig. ## 1 (Intercept) 542.62 5.91 91.8599 21.70 <0.001 *** ## 2 schoolksuburban 2.77 6.76 0.4100 28.35 0.6849 ## 3 schoolkrural 1.03 6.38 0.1616 30.74 0.8727 ## 4 starksmall 9.42 4.56 2.0649 17.10 0.0544 . ## 5 starkaide -4.27 2.17 -1.9631 16.73 0.0665 . ## 6 genderfemale 2.14 1.20 1.7773 67.14 0.0800 . ## 7 ethnicityafam -16.79 4.19 -4.0026 34.94 <0.001 *** ## 8 ethnicityasian 13.19 11.02 1.1963 6.23 0.2751 ## 9 ethnicityhispanic 39.23 20.62 1.9028 1.01 0.3067 ## 10 ethnicityother 8.86 18.78 0.4720 3.02 0.6690 ## 11 lunchkfree -19.37 2.04 -9.4848 57.38 <0.001 *** ## 12 schoolksuburban:starksmall 3.03 6.39 0.4746 28.94 0.6386 ## 13 schoolkrural:starksmall -0.31 5.58 -0.0555 34.04 0.9560 ## 14 schoolksuburban:starkaide 5.10 3.72 1.3711 28.64 0.1810 ## 15 schoolkrural:starkaide 8.16 3.16 2.5857 34.30 0.0141 * With this specification, there are several different null hypotheses that we might want to test. For one, perhaps we want to see if there is any variation in treatment effects across different levels of urbanicity. This can be expressed in the null hypothesis that all four interaction terms are zero, or $$H_{0A}: \beta_5 = \beta_6 = \beta_7 = \beta_8 = 0$$. With Wald test: Wald_test(lm_urbanicity, constraints = constrain_zero("schoolk.+:stark", reg_ex = TRUE), vcov = V_urbanicity) ## test Fstat df_num df_denom p_val sig ## HTZ 1.96 4 37.5 0.121 Another possibility is that we might want to focus on variation in the effect of being in a small class or regular class, while ignoring whatever is going on in the aide class condition. Here, the null hypothesis would be simply $$H_{0B}: \beta_5 = \beta_6 = 0$$, tested as: Wald_test(lm_urbanicity, constraints = constrain_zero("schoolk.+:starksmall", reg_ex = TRUE), vcov = V_urbanicity) ## test Fstat df_num df_denom p_val sig ## HTZ 0.189 2 34.5 0.828 ## 2.1 Lists of constraints In models like the urbanicity-by-treatment interaction specification, we may need to run multiple tests on the same estimating equation. This can be accomplished with Wald_test by providing a list of constraints to the constraints argument. For example, we could test the hypotheses described above by creating a list of several constraint matrices and then passing it to Wald_test: C_list <- list( Any interaction = constrain_zero("schoolk.+:stark", coef(lm_urbanicity), reg_ex = TRUE), Small vs regular = constrain_zero("schoolk.+:starksmall", coef(lm_urbanicity), reg_ex = TRUE) ) Wald_test(lm_urbanicity, constraints = C_list, vcov = V_urbanicity) ##$Any interaction
##  test Fstat df_num df_denom p_val sig
##   HTZ  1.96      4     37.5 0.121
##
## $Small vs regular ## test Fstat df_num df_denom p_val sig ## HTZ 0.189 2 34.5 0.828 Setting the option tidy = TRUE will arrange the output of all the tests into a single data frame: Wald_test(lm_urbanicity, constraints = C_list, vcov = V_urbanicity, tidy = TRUE) ## hypothesis test Fstat df_num df_denom p_val sig ## Any interaction HTZ 1.960 4 37.5 0.121 ## Small vs regular HTZ 0.189 2 34.5 0.828 The list of constraints can also be created inside Wald_test, so that the coefs argument can be omitted from constrain_zero(): Wald_test( lm_urbanicity, constraints = list( Any interaction = constrain_zero("schoolk.+:stark", reg_ex = TRUE), Small vs regular = constrain_zero("schoolk.+:starksmall", reg_ex = TRUE) ), vcov = V_urbanicity, tidy = TRUE ) ## hypothesis test Fstat df_num df_denom p_val sig ## Any interaction HTZ 1.960 4 37.5 0.121 ## Small vs regular HTZ 0.189 2 34.5 0.828 # 3 Pairwise t-tests The clubSandwich package also provides a further convenience function, constrain_pairwise() that can be used in combination with Wald_test() to conduct pairwise comparisons among a set of regression coefficients. This function differs from the other two constrain_*() functions because it returns a list of constraint matrices, each of which corresponds to a single linear combination of covariates. Specifically, the constrain_pairwise() function provides a list of constraints that represent the differences between every possible pair among a specified set of coefficients. The syntax is very similar to the other constrain_*() functions. To demonstrate, consider the separate-intercepts specification of the simpler regression model: coef_test(lm_sep, vcov = V_sep) ## Coef. Estimate SE t-stat d.f. p-val (Satt) Sig. ## 1 starkregular 532 2.78 192 59.9 <0.001 *** ## 2 starksmall 541 2.89 187 65.0 <0.001 *** ## 3 starkaide 531 2.72 195 64.3 <0.001 *** This specification is nice because it lets us simply read off the average outcomes for each group. However, we will naturally also want to know about whether there are differences between the groups, so we’ll want to compare the small-class condition to the regular-size class condition, the aide condition to the regular-size class condition, and the small-class condition to the aide condition. Thus, we’ll want comparisons among all three coefficients: C_pairs <- constrain_pairwise(1:3, coefs = coef(lm_sep)) C_pairs ##$starksmall - starkregular
##      [,1] [,2] [,3]
## [1,]   -1    1    0
##
## $starkaide - starkregular ## [,1] [,2] [,3] ## [1,] -1 0 1 ## ##$starkaide - starksmall
##      [,1] [,2] [,3]
## [1,]    0   -1    1

Feeding these constraints into Wald_test() gives us significance tests for each pair:

Wald_test(lm_sep, constraints = C_pairs, vcov = V_sep, tidy = TRUE)
##                 hypothesis test   Fstat df_num df_denom  p_val sig
##  starksmall - starkregular  HTZ 16.9238      1     65.6 <0.001 ***
##   starkaide - starkregular  HTZ  0.0673      1     65.6  0.796
##     starkaide - starksmall  HTZ 17.8137      1     66.9 <0.001 ***

The first two of these tests are equivalent to the tests of the treatment effect coefficients in the other parameterization of the model. Indeed, the denominator degrees of freedom are identical to the results of coef_test(lm_trt, vcov = V_trt); the Fstats here are equal to the squared t-statistics from the first model:

t_stats <- coef_test(lm_trt, vcov = V_trt)$tstat[2:3] F_stats <- Wald_test(lm_sep, constraints = C_pairs, vcov = V_sep, tidy = TRUE)$Fstat[1:2]
all.equal(t_stats^2, F_stats)
##  TRUE

It is important to note that the p-values from the pairwise comparisons are not corrected for multiplicity.3 For now, please correct-your-own using p.adjust() or your preferred method.

Pairwise comparisons might also be of use in the model with treatment-by-urbanicity interactions. Here’s the model results again:

coef_test(lm_urbanicity, vcov = V_urbanicity)
##                         Coef. Estimate    SE  t-stat  d.f. p-val (Satt) Sig.
## 1                 (Intercept)   542.62  5.91 91.8599 21.70       <0.001  ***
## 2             schoolksuburban     2.77  6.76  0.4100 28.35       0.6849
## 3                schoolkrural     1.03  6.38  0.1616 30.74       0.8727
## 4                  starksmall     9.42  4.56  2.0649 17.10       0.0544    .
## 5                   starkaide    -4.27  2.17 -1.9631 16.73       0.0665    .
## 6                genderfemale     2.14  1.20  1.7773 67.14       0.0800    .
## 7               ethnicityafam   -16.79  4.19 -4.0026 34.94       <0.001  ***
## 8              ethnicityasian    13.19 11.02  1.1963  6.23       0.2751
## 9           ethnicityhispanic    39.23 20.62  1.9028  1.01       0.3067
## 10             ethnicityother     8.86 18.78  0.4720  3.02       0.6690
## 11                 lunchkfree   -19.37  2.04 -9.4848 57.38       <0.001  ***
## 12 schoolksuburban:starksmall     3.03  6.39  0.4746 28.94       0.6386
## 13    schoolkrural:starksmall    -0.31  5.58 -0.0555 34.04       0.9560
## 14  schoolksuburban:starkaide     5.10  3.72  1.3711 28.64       0.1810
## 15     schoolkrural:starkaide     8.16  3.16  2.5857 34.30       0.0141    *

Suppose that we are interested in the effect of small versus regular size classes, and in particular whether this effect varies across schools in different areas. The coefficients on schoolksuburban:starksmall and schoolkrural:starksmall already give us the differences in treatment effects between suburban schools versus urban schools and between rural schools versus urban schools. The difference between these coefficients gives us the difference in treatment effects between suburban schools and rural schools. We can look at all three of these contrasts using constrain_pairwise() by setting the option with_zero = TRUE:

Wald_test(lm_urbanicity,
constraints = constrain_pairwise(":starksmall", reg_ex = TRUE, with_zero = TRUE),
vcov = V_urbanicity,
tidy = TRUE)
##                                            hypothesis test   Fstat df_num
##                            schoolksuburban:starksmall  HTZ 0.22526      1
##                               schoolkrural:starksmall  HTZ 0.00308      1
##  schoolkrural:starksmall - schoolksuburban:starksmall  HTZ 0.36471      1
##  df_denom p_val sig
##      28.9 0.639
##      34.0 0.956
##      24.4 0.551

Again, the results of the first two tests are identical to the t-tests reported in coef_test().

# 4 Remark

All of the preceding examples were based on ordinary linear regression models with clustered standard errors. However, Wald_test() and its helper functions all work identically for all of the other models with supporting clubSandwich methods, including nlme::lme(), nlme::gls(), lme4::lmer(), rma.uni(), rma.mv(), and robu(), among others.

1. In Pustejovsky & Tipton (2018) we used a more general formulation of multiple-constraint null hypotheses, expressed as $$H_0: \mathbf{C} \boldsymbol\beta = \mathbf{d}$$ for some fixed $$q \times 1$$ vector $$\mathbf{d}$$. In practice, it’s often possible to modify the $$\mathbf{C}$$ matrix so that $$\mathbf{d}$$ can always be set to $$\mathbf{0}$$.↩︎

2. How does this work? If we omit the coefs argument, constrain_zero() acts as a functional, by returning a function equivalent to

function(coefs) constrain_zero(constraints, coefs = coefs)    

If this function is fed into the constraints argument of Wald_test(), Wald_test() recognizes that it is a function and evaluates the function with coef(obj). It’s a kinda-sorta hacky substitute for lazy evaluation. If you have suggestions for how to do this more elegantly, please send them my way.↩︎

3. Options to include multiplicity corrections (Bonferroni, Holm, Benjamini-Hochberg, etc.) might be included in a future release. Reach out if this is of interest to you.↩︎