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Hypothesis Testing in Linear Models · Tutorial 333 of 975

Interpreting and Reporting General Linear Tests

A clear multivariate test report makes the restrictions, effective hypothesis rank, statistic, calibration, and limits of interpretation explicit.

Intermediate 12 min read

What You'll Learn

  • Express a multivariate linear hypothesis and distinguish its restrictions from the number of restriction rows.
  • Determine the effective hypothesis rank for full-rank and potentially redundant restrictions.
  • Report a test statistic together with its reference distribution and calibration.
  • Interpret an omnibus result without overclaiming about individual coefficients or responses.
  • Identify when follow-up contrasts and multiplicity control are needed.

What a Multivariate Test Report Must Say

A multivariate test can answer a focused scientific question while still being easy to misreport. A small p-value for an overall hypothesis does not, by itself, say which response or coefficient differs. A large p-value does not establish that the coefficients are equal. A useful report therefore makes the hypothesis and the calibration visible, then limits its conclusion to what that test actually addresses.

Consider the multivariate linear model \(Y=XB+U\), where \(Y\) is an \(n\times q\) response matrix, \(X\) is an \(n\times p\) design matrix, and \(B\) is a \(p\times q\) coefficient matrix. A general linear hypothesis can be written \(LBA=C\), where \(L\) selects or combines predictor-space coefficients, \(A\) selects or combines response-space directions, and \(C\) specifies the hypothesized value. The dimensions are \(L:s\times p\), \(A:q\times t\), and \(C:s\times t\). The null hypothesis is the set of coefficient matrices satisfying this equation.

Definition: The effective hypothesis rank is the number of linearly independent scalar restrictions imposed by the null hypothesis on the identifiable coefficient space. When \(X\) has full column rank and the restrictions are consistent, it is the rank of the linear map \(B\mapsto LBA\). If \(L\) and \(A\) have full respective ranks, that number is \(\operatorname{rank}(L)\operatorname{rank}(A)\). With nonidentifiable coefficients, only restrictions on estimable combinations count.

The number of rows of \(L\), or the number of entries in \(C\), is not automatically the effective rank. Some restrictions may repeat information already contained in others. The effective rank matters because it describes the dimension of the tested hypothesis and, for example, determines the degrees of freedom of an asymptotic Wald reference distribution.

Why the Effective Rank Is a Rank

For full-column-rank \(X\), all entries of \(B\) are identifiable. Using column-wise vectorization, the identity \(\operatorname{vec}(LBA)=(A^\mathsf{T}\otimes L)\operatorname{vec}(B)\) converts the matrix restrictions into a linear system. This gives a direct way to count independent restrictions.

Theorem: If \(X\) has full column rank, the restrictions \(LBA=C\) are consistent, and \(L\) and \(A\) have the dimensions given above, their effective hypothesis rank is \(\operatorname{rank}(L)\operatorname{rank}(A)\).

Proof. Vectorization writes the restriction map as multiplication by \(A^\mathsf{T}\otimes L\). For any matrices \(G\) and \(H\), \(\operatorname{rank}(G\otimes H)=\operatorname{rank}(G)\operatorname{rank}(H)\). To see this, write each matrix in a rank factorization: \(G=P_G Q_G\), \(H=P_H Q_H\), where the columns of each \(P\) and rows of each \(Q\) are linearly independent. Then

$$ G\otimes H=(P_G\otimes P_H)(Q_G\otimes Q_H). $$

The first factor has full column rank and the second has full row rank, so their product has rank equal to the inner dimension, namely \(\operatorname{rank}(G)\operatorname{rank}(H)\). Applying this to \(G=A^\mathsf{T}\) and \(H=L\) gives the result. Since \(X\) has full column rank, this restriction map acts on an identifiable coefficient space. \(\square\)

If \(X\) is rank deficient, this product is not by itself enough: the restriction map must be considered only on the identifiable coefficient space. A report should then make clear that the tested restrictions are estimable. Redundant rows also need careful handling: removing a repeated equation preserves the hypothesis only when its right-hand side obeys the same linear dependence.

Result: Replacing an independent set of restriction equations by any invertible linear recombination leaves the null hypothesis unchanged and preserves its rank. Consequently, the report should describe the restriction, not rely on an arbitrary count of how many rows a particular matrix representation contains.

Proof. Write the restrictions as \(D\beta=d\), where \(\beta=\operatorname{vec}(B)\). If \(S\) is invertible, \(D\beta=d\) holds exactly when \(SD\beta=Sd\) holds, because multiplication by \(S\) is reversible. Also, \(\operatorname{rank}(SD)=\operatorname{rank}(D)\), since an invertible transformation preserves linear dependence among rows. Thus the hypothesis and its effective rank are unchanged. \(\square\)

Worked Examples: From Restrictions to a Report

Worked Example: One Predictor Effect Across Two Responses

Suppose a full-rank design has three predictors and two responses. The scientific question is whether the coefficient of predictor 2 is zero for both responses. This is \(LBA=0\), with \(L=(0,1,0)\) and \(A=I_2\). The ranks are \(\operatorname{rank}(L)=1\) and \(\operatorname{rank}(A)=2\), so the effective rank is \(1\times2=2\). The null comprises two scalar restrictions, one for each response.

Assume an asymptotic Wald analysis reports \(W=8.00\), calibrated against a chi-square distribution with 2 degrees of freedom. For 2 degrees of freedom, the upper-tail probability is \(\exp(-W/2)\). Thus \(p=\exp(-4)\approx0.0183\). A concise report is: “The joint null that predictor 2 has zero coefficients for both responses was rejected (Wald \(\chi^2_2=8.00\), \(p=0.018\), asymptotic calibration).”

This supports evidence against the joint null. It does not establish that both response-specific coefficients are nonzero, nor does it identify which one contributes most to the result. Those are separate questions.

Worked Example: Counting Redundant Restrictions

There are four predictors and three responses. Let the rows of \(L\) be \((0,1,0,0)\), \((0,0,1,0)\), and \((0,1,1,0)\), and let \(A=I_3\), \(C=0\). The third row of \(L\) is the sum of the first two, so \(\operatorname{rank}(L)=2\), not 3. Since \(\operatorname{rank}(A)=3\), the effective rank is \(2\times3=6\). Although the displayed matrix equation contains nine entries, only six independent scalar restrictions are imposed.

Suppose the asymptotic Wald statistic for this hypothesis is \(W=9.60\). Its reference distribution is \(\chi^2_6\). For six degrees of freedom, the upper-tail probability at \(x\) is \(\exp(-x/2)\{1+x/2+(x/2)^2/2\}\). Substituting \(x=9.60\) gives \(\exp(-4.8)(1+4.8+11.52)\approx0.143\). The report is: “The joint restrictions were not rejected (Wald \(\chi^2_6=9.60\), \(p=0.143\), asymptotic calibration).”

It would be incorrect to use nine degrees of freedom just because the equation has nine entries. The third row adds no independent restriction. The result is also not evidence that all restrictions are exactly true; it indicates that this test did not find sufficient evidence against their joint null.

Worked Example: A Single Response-Space Contrast

Suppose there are three responses, and the question concerns the average of responses 1 and 2 minus response 3 for a particular predictor. Let \(L=(0,1,0)\), selecting that predictor, and \(A=(1/2,1/2,-1)^\mathsf{T}\). The tested value is \(LBA=0\). Both \(L\) and \(A\) have rank 1, so the effective rank is 1. This test concerns one specified response contrast, not all three response-specific coefficients separately.

Assume the asymptotic Wald output is \(W=4.84\), with a \(\chi^2_1\) reference distribution. Since \(4.84=2.2^2\), the corresponding two-sided normal-tail probability is \(2\{1-\Phi(2.2)\}\approx0.0278\); this equals the upper-tail probability for \(\chi^2_1\). The report is: “The specified response contrast differed from zero for the predictor (Wald \(\chi^2_1=4.84\), \(p=0.028\), asymptotic calibration).”

The conclusion is about that weighted contrast. It does not establish a difference for every response, or imply that any one response-specific coefficient is individually significant.

Choosing and Reporting the Calibration

A test statistic has no interpretable p-value without a reference distribution and calibration. State both, including degrees of freedom where applicable. For an asymptotic Wald test, report the chi-square statistic, its degrees of freedom (the effective restriction rank under the usual regularity conditions), and that the calibration is asymptotic. The examples above use this form of report.

Classical multivariate analyses may instead report Wilks’ lambda, Pillai’s trace, the Hotelling–Lawley trace, or Roy’s largest root, together with the reference calibration used by the software. These statistics are different summaries of the same hypothesis and error information; their reported p-values need not agree exactly. Their finite-sample reference distributions or approximate \(F\) calibrations depend on the design, ranks, and assumptions. A careful report identifies the statistic actually used and states whether its calibration is exact under the stated model or approximate. Do not label an approximate \(F\) calibration as exact.

For classical finite-sample multivariate calibrations, the model assumptions must justify the chosen method; a common setting assumes independent multivariate normal errors with a common response covariance matrix and appropriate full-rank design and hypothesis matrices. An asymptotic Wald calibration instead requires the relevant regularity conditions, including estimability and a consistent covariance estimate for the unrestricted estimator. The test name alone does not establish that its calibration is justified.

A complete report ordinarily gives the scientific question in context, the tested restrictions, the effective rank or the degrees of freedom that reflect it, the chosen statistic and observed value, the reference calibration, and the p-value. It should also state the scope of the conclusion. If estimates or confidence intervals are important to interpretation, report them separately; a test p-value is not an effect-size estimate.

Common Reporting Pitfalls

  • Counting equations instead of independent restrictions. A matrix equation can contain redundant rows or entries. Check the rank of the restriction map on the identifiable coefficient space.
  • Reporting only a p-value. Without the hypothesis, statistic, and reference calibration, a reader cannot tell what was tested or how the p-value was obtained.
  • Turning an omnibus rejection into a coefficient-level claim. A joint test can reject even when it does not identify a particular response or coefficient. Follow-up tests or contrasts must be specified and their multiplicity addressed.
  • Treating nonrejection as proof of equality. A nonsignificant result is not an equivalence conclusion. Evidence for practical similarity requires a suitable equivalence hypothesis and design.
  • Calling an approximation exact. Name the calibration accurately and state the assumptions that support it.

These distinctions matter because a general linear test summarizes evidence about a specified collection of restrictions. It is not a substitute for interpreting the estimated coefficients, checking the model assumptions, or designing follow-up comparisons. Keeping the hypothesis and calibration explicit makes the report reproducible and keeps the conclusion within the evidence.

Check Your Understanding

Use the restriction matrices and calibration information, not just the number of displayed equations.

  1. In a full-rank model, if \(\operatorname{rank}(L)=2\) and \(\operatorname{rank}(A)=1\), what is the effective hypothesis rank?
  2. Why can a restriction equation with nine entries have effective rank six?
  3. What information should accompany a reported asymptotic Wald chi-square statistic?
  4. A joint test rejects the null that three response coefficients are all zero. What can you conclude, and what can you not conclude about the individual coefficients?
  5. Why is failure to reject a zero-effect null not, by itself, evidence that the effect is practically negligible?
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