Introduction to Type I/II/III Hypothesis Testing

Type I/II/III hypothesis testing originates from analysis of variance, however, Goodnight (1980) suggests viewing this issue as hypothesis testing of fixed effects in linear models, which avoid the “usual assumptions” and defined three types of estimable functions. Given that type I estimable functions are dependent on the specific order of used covariates, we focus on the type II and III testing here.

For MMRM, obtaining estimable functions is equivalent to obtaining the contrasts for linear hypothesis tests. For more details, see Goodnight (1980).

Contained Effect

Before we discuss the testing, it is important that we understand the concept of “contained effect”. The definition of “contained effect” is:

For effect $$E_2$$, it is said to contain $$E_1$$ if

1. the involved numerical covariate(s) are identical for these two effects
2. all the categorical covariate(s) associated with $$E_1$$ are also associated with $$E_2$$

For example, for the following model formula

$$Y = A + B + A*B$$

using $$E_A$$, $$E_B$$ and $$E_{A*B}$$ to denote the effect for $$A$$, $$B$$ and $$A*B$$ respectively, we have

1. If $$A$$ and $$B$$ are both categorical, then $$E_{A*B}$$ contains $$E_{A}$$ and $$E_{B}$$.
2. If $$A$$ and $$B$$ are both numerical, then $$E_{A*B}$$ do not contain $$E_{A}$$ or $$E_{B}$$.
3. If $$A$$ is categorical and $$B$$ is numerical, then $$E_{A*B}$$ contains $$E_{B}$$ but not $$E_{A}$$.

Type II Hypothesis Testing

For type II hypothesis testing, the contrasts can be obtained through the following steps:

Create a contrast matrix $$L$$, with rows equal to the number of parameters associated with the effect, columns equal to the number of all parameters.

1. All columns of $$L$$ associated with effect not containing $$E_1$$ (except $$E_1$$) are set to 0.
2. The submatrix of $$L$$ associated with $$E_1$$ is set to $$(X_1^\top M X_1)^{-}X_1^\top M X_1$$
3. Each of the remaining submatrices of L associated with an effect $$E_2$$ that contains $$E_1$$ is $$(X_1^\top M X_1)^{-}X_1^\top M X_2$$

where $$X_0$$ stands for columns of $$X$$ whose associated effect do not contain $$E_1$$, $$X_1$$ stands for columns of $$X$$ associated with $$E_1$$, $$X_2$$ stands for columns of $$X$$ whose associated effect contains $$E_1$$, $$M = I - X_0(X_0^\top X_0)^{-}X_0^\top$$, and $$Z^{-}$$ stands for the g2 inverse of $$Z$$.

Note: Here we do not allow for singularity in general, so the g2 inverse is just the usual inverse. Thus we can use an identity matrix as the submatrix in step 2.

Using fev_data to create our example, we have

library(mmrm)
fit <- mmrm(FEV1 ~ ARMCD + RACE + ARMCD * RACE + ar1(AVISIT | USUBJID), data = fev_data)

For this given example, we would like to test the effect of RACE, $$E_{RACE}$$. We initialize the contrast matrix with

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{matrix}$

Please note that this is a $$2\times 6$$ matrix, the rows is the number of coefficients that is associated with RACE, and the columns is the total number of coefficients 6.

Then following step 2, we filled in the identity matrix in the corresponding submatrix

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ \end{matrix}$

In the last step, for the last 2 columns, we fill the values with the calculated result.

x <- component(fit, "x_matrix")
x0 <- x[, c(1, 2)]
x1 <- x[, c(3, 4)]
x2 <- x[, c(5, 6)]
m <- diag(rep(1, nrow(x))) - x0 %*% solve(t(x0) %*% x0) %*% t(x0) # solve is used because the matrix is inversible
sub_mat <- solve(t(x1) %*% m %*% x1) %*% t(x1) %*% m %*% x2
sub_mat
##                               ARMCDTRT:RACEBlack or African American
## RACEBlack or African American                             0.42618692
## RACEWhite                                                -0.04372702
##                               ARMCDTRT:RACEWhite
## RACEBlack or African American         0.02751985
## RACEWhite                             0.58570969

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 1 & 0 & 0.42618692 & 0.02751985\\ 0 & 0 & 0 & 1 & -0.04372702 & 0.58570969\\ \end{matrix}$

Type III Hypothesis Testing

For type III hypothesis testing, we assume that the hypothesis tested should be the same for all designs with the same general form of estimable functions. It can be defined with the following steps:

Create a contrast matrix $$L$$, with rows equal to the number of parameters associated with the effect, columns equal to the number of all parameters.

1. All columns of $$L$$ associated with effect not containing $$E_1$$ (except $$E_1$$) are set to 0.
2. The submatrix of $$L$$ associated with $$E_1$$ is set to identity matrix.
3. For each of the remaining submatrices of L associated with an effect $$E_2$$ that contains $$E_1$$, equate the values.

Here equate the values means that, given that the coefficients in step 2 are all 1 (diagonal), in step 3 this 1 is divided by the total number of new levels.

Using the same example for type II testing, we also would like to test the effect of RACE, $$E_{RACE}$$. We initialize the contrast matrix with

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{matrix}$

Then following step 2, we filled in the identity matrix in the corresponding submatrix

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ \end{matrix}$

In the last step, for the last 2 columns, we fill the values to equate the coefficients in the sub matrix. We use the coefficients for RACE, and divide it by the additional levels (there is interaction between ARMCD and RACE, the additional covariate ARMCD has 2 levels) and we get 0.5 for the coefficients.

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 0 & 1 & 0 & 0.5 & 0\\ 0 & 0 & 0 & 1 & 0 & 0.5\\ \end{matrix}$

Similarly, if we are testing the effect of ARMCD we will have the following contrast because RACE has 3 levels.

$\begin{matrix} \mu & ARMCD & RACE & RACE & ARMCD:RACE & ARMCD:RACE\\ 0 & 1 & 0 & 0 & \frac{1}{3} & \frac{1}{3}\\ \end{matrix}$

Hypothesis Testing in SAS

In PROC MIXED we can specify HTYPE=1,2,3 to enable these hypothesis testings, and we can use option e1 e2 e3 to view the estimable functions.

Special Considerations

Reference Levels

For PROC MIXED, it is important that the categorical values have the correct order, especially for structured covariance (otherwise the correlation between visits can be messed up). We usually use CLASS variable(REF=) to define the reference level. However, SAS will put the reference level in the last, so if the visit variable is included in fixed effect, there can be issues with the comparison between SAS and R, as in R we still use the first level as our reference.

When we conduct the hypothesis testing, we will also face this issue. If reference level is different, the testing we have can be different, e.g. if we have 3 levels, $$l_1$$, $$l_2$$ and $$l_3$$. In theory we can test either $$l_2 - l_1$$, $$l_3 - l_1$$ (using $$l_1$$ as reference), or test $$l_1 - l_3$$, $$l_2 - l_3$$ (using $$l_3$$ as reference). But the result will be slightly different. However, if we have identical tests the result will be closer. The following example illustrate this.

Example of Reference Levels

Assume in SAS we have the following model

PROC MIXED DATA = fev cl method=reml;
CLASS AVISIT(ref = 'VIS4') ARMCD(ref = 'PBO') USUBJID;
MODEL FEV1 = ARMCD AVISIT ARMCD*AVISIT / ddfm=satterthwaite htype=3;
REPEATED AVISIT / subject=USUBJID type=ar(1);
RUN;

And in R we have the following model

fit <- mmrm(FEV1 ~ ARMCD + AVISIT + ARMCD * AVISIT + ar1(AVISIT | USUBJID), data = fev_data)
Anova(fit)

Note that for AVISIT there are 4 levels, VIS1, VIS2, VIS3 and VIS4. In SAS VIS4 is the reference, while in R VIS1 is the reference.

They both use the following matrix to test the effect of AVISIT (ignoring other part of contrasts that is not associated with AVISIT)

$\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix}$

But given that the reference level is different, we are testing different things. In SAS we are testing VIS1 - VIS4, VIS2 - VIS4 and VIS3 - VIS4, while in R we are testing VIS2 - VIS1, VIS3 - VIS1 and VIS4 - VIS1.

To correctly test VIS1 - VIS4, VIS2 - VIS4 and VIS3 - VIS4 in R, we need to update the contrast to the following matrix (ignoring the other part) to get a closer result.

$\begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & -1 & -1\\ \end{matrix}$

Why Model Covariate Order Changes My Testing in SAS?

For type II/III testing, it is true that the order of covariate should not affect the result. However, if there is an interaction term, the reference level can be changed. With different reference levels the result will be slightly different.

Why mmrm Gives More Covariates Than SAS?

For PROC MIXED, if a model is specified as follows

MODEL FEV1 = ARMCD SEX ARMCD * SEX ARMCD * FEV1_BL

Then it is equivalent to the following mmrm model

FEV1 ~ ARMCD * SEX + ARMCD * FEV1_BL - FEV1_BL

or

FEV1 ~ ARMCD * SEX + ARMCD:FEV1_BL

Because SAS will not include the covariate FEV1_BL by default, unless manually added. However in R we will add FEV1_BL by default. Please note that in this example we exclude the covariance structure part.

Goodnight JH (1980). “Tests of Hypotheses in Fixed Effects Linear Models.” Communications in Statistics-Theory and Methods, 9(2), 167–180.