Why Compare an F-Test With a t-Test?
A partial F-test asks whether restrictions imposed by a smaller regression model worsen its fit enough to count as evidence against those restrictions. When the smaller model imposes just one restriction on the larger model, there is also a familiar coefficient t-test for that restriction. The two calculations are not merely similar: the F statistic is exactly the square of the t statistic.
This identity links two common ways of expressing the same evidence. The t statistic retains a sign, indicating whether an estimate is above or below its null value. Squaring removes that sign, so the F-test is nondirectional. As a result, the matching t-test is two-sided, not one-sided.
Models and Test Statistics
Let the full linear regression model have \(n\) observations and \(p\) coefficients, including any intercept. Write its design matrix as \(X\), its coefficient vector as \(\boldsymbol{\beta}\), and its outcome vector as \(\mathbf{y}\). Suppose the reduced model imposes one linear restriction on the full model:
Here \(\mathbf{c}\) is a specified vector of constants and \(r\) is a specified value. For example, testing whether one coefficient equals zero is the restriction \(\beta_j=0\). The full and reduced models must use the same outcome and observations, and the restriction must reduce the number of freely estimated coefficients by exactly one.
For the usual exact regression tests, assume the full model is correctly specified with a full-rank design matrix and errors that, conditional on the predictors, are independent normal random variables with mean zero and common variance \(\sigma^2\). Under \(H_0\), the t statistic has a \(t\) distribution with \(df\) degrees of freedom, and the partial F statistic has an \(F\) distribution with 1 and \(df\) degrees of freedom. These assumptions justify the reference distributions; a numerical identity between statistics does not itself verify the assumptions.
Why the Statistics Are Identical
The key fact is that restricting a least-squares fit by one linear condition increases its residual sum of squares by a particular squared, standardized distance. The following derivation gives that increase directly.
For a full-rank design, \(\hat{\boldsymbol{\beta}}=(X^{\mathsf T}X)^{-1}X^{\mathsf T}\mathbf{y}\). Define \(A=X^{\mathsf T}X\), \(v=\mathbf{c}^{\mathsf T}A^{-1}\mathbf{c}\), and \(d=\mathbf{c}^{\mathsf T}\hat{\boldsymbol{\beta}}-r\).
The least-squares estimate subject to \(\mathbf{c}^{\mathsf T}\boldsymbol{\beta}=r\) is \(\hat{\boldsymbol{\beta}}_R=\hat{\boldsymbol{\beta}}-A^{-1}\mathbf{c}\,d/v\). It satisfies the restriction because \(\mathbf{c}^{\mathsf T}\hat{\boldsymbol{\beta}}_R=\mathbf{c}^{\mathsf T}\hat{\boldsymbol{\beta}}-v d/v=r\).
The full-model residual is orthogonal to every column of \(X\), so changing the coefficient vector by \(\boldsymbol{\delta}=\hat{\boldsymbol{\beta}}_R-\hat{\boldsymbol{\beta}}\) increases SSE by \(\boldsymbol{\delta}^{\mathsf T}A\boldsymbol{\delta}\). Substituting \(\boldsymbol{\delta}=-A^{-1}\mathbf{c}\,d/v\) gives \(SSE_R-SSE_F=d^2/v\).
Since the estimated variance of \(\mathbf{c}^{\mathsf T}\hat{\boldsymbol{\beta}}\) is \(MSE_Fv\), the squared t statistic is \(d^2/(MSE_Fv)\). The partial F statistic, using the one-restriction SSE increase just derived, is the same quantity:
This is an algebraic identity for the observed statistics. Under the null model and the stated error assumptions, it also means that the two-sided t-test and the one-restriction F-test produce the same p-value. In distributional terms, if \(T\sim t_{df}\), then \(T^2\sim F_{1,df}\).
Worked Examples
Worked Example: Checking the Identity From a Hypothetical Data Set
Consider these hypothetical observations, with outcome \(y\) and predictors \(x\) and \(z\). The reduced model includes an intercept and \(x\); the full model also includes \(z\). The test asks whether the coefficient of \(z\) is zero.
| Observation | \(x\) | \(z\) | \(y\) |
|---|---|---|---|
| 1 | -1 | -1 | 6 |
| 2 | -1 | 1 | 12 |
| 3 | 0 | -1 | 5 |
| 4 | 0 | 1 | 11 |
| 5 | 1 | 0 | 13 |
| 6 | 1 | 0 | 13 |
For these values, \(x\) and \(z\) each sum to zero, their cross-product sum is zero, and their sums of squares are both 4. The outcome can be written \(y=10+2x+3z+e\), where the residual vector \(e=(1,1,-2,-2,1,1)\) is orthogonal to the intercept, \(x\), and \(z\). Thus the full fit has coefficient estimates 10, 2, and 3, and \(SSE_F=\sum e_i^2=12\). It has \(df=6-3=3\), so \(MSE_F=12/3=4\).
In the reduced model, the omitted \(3z\) component remains in the residuals along with \(e\). Because \(z\) is orthogonal to the intercept, \(x\), and \(e\), the reduced residual sum of squares is \(SSE_R=3^2(4)+12=48\). Therefore,
The standard error of the full-model estimate for the coefficient of \(z\) is \(\sqrt{MSE_F/4}=\sqrt{4/4}=1\), so \(t=(3-0)/1=3\) and \(t^2=9\). Both tests use 1 and 3 degrees of freedom in their reference distributions; their p-values match (about 0.0577 for the two-sided t-test and the F-test). For this hypothetical data set, the tests do not reject at the 0.05 level. The arithmetic verifies the identity, but the data alone do not establish normality or constant error variance; those assumptions must be justified separately.
Worked Example: Testing a Nonzero Coefficient Value
Suppose a hypothetical full-model summary has \(n=24\), \(p=4\), \(SSE_F=180\), and an estimated coefficient of 2.4 with standard error 0.8. Test \(H_0:\beta_j=1.6\) by comparing the full model with the model that fixes this coefficient at 1.6. Assume the models use the same 24 observations and that the regression assumptions for inference are reasonable.
The full-model residual degrees of freedom are \(df=24-4=20\), and \(MSE_F=180/20=9\). The coefficient’s t statistic is
The coefficient variance factor is \(0.8^2/9=0.07111\). The one-restriction increase in SSE is therefore \((2.4-1.6)^2/0.07111=9\), so the restricted model has \(SSE_R=180+9=189\). Its partial F statistic is
The two-sided t-test and the F-test have the same p-value, about 0.329 with 20 denominator degrees of freedom. This example shows that the identity applies to a specified coefficient value, not only to a test that the coefficient equals zero.
Worked Example: Why the Matching t-Test Is Two-Sided
In another hypothetical full-model output, a coefficient estimate is 1.7 and its standard error is 0.5. Suppose the null value is 2.8 and the full model has 12 residual degrees of freedom. Then \(t=(1.7-2.8)/0.5=-2.2\), so \(t^2=4.84\). If the full-model \(MSE\) is 5, the SSE increase from imposing the null restriction is \(t^2MSE=4.84(5)=24.2\), and the partial F statistic is \(24.2/5=4.84\).
The F statistic records the size of the discrepancy but not its direction. Its p-value matches the two-sided t-test p-value (about 0.048), not a one-sided test chosen to look only for coefficients below 2.8. A one-sided alternative uses the sign of \(t\) and a different tail probability; squaring \(t\) discards precisely that directional information.
When the Identity Is Useful—and a Common Pitfall
The identity is useful as a consistency check: for a single coefficient restriction, software that reports both a coefficient t-test and a partial F-test should give \(F=t^2\), apart from rounding. It also helps select a concise result. A single coefficient can be reported with its estimate, standard error, t statistic, degrees of freedom, and p-value; the corresponding one-restriction F statistic adds no different test evidence.
The restriction count matters. If a reduced model imposes two or more independent restrictions, the partial F statistic averages the SSE increase across those restrictions, and it is not generally the square of one t statistic. Likewise, comparisons require the same outcome and observations; changing the sample as well as the model can make an SSE difference unsuitable for this nested-model test.
- Wrong reasoning: “An F-test is always the square of a t-test.”
- Correction: This exact identity applies when the nested models differ by one restriction and the t-test concerns that same restriction, using the full model’s residual variance estimate.
- Wrong reasoning: “Because \(F=t^2\), the F-test gives the direction of the coefficient effect.”
- Correction: Squaring removes the sign. Use the signed t statistic and a prespecified one-sided alternative if the scientific question is directional.
Check Your Understanding
Use the definitions and derivation above to answer these questions.
- A full model has \(SSE_F=150\) and 25 residual degrees of freedom. A reduced model imposes one restriction and has \(SSE_R=156\). Calculate the partial F statistic.
- If the corresponding coefficient estimate is 4.0 and its standard error is 2.0 under a null value of zero, verify that the t statistic gives the same test statistic.
- Why does the p-value from the F-test match a two-sided t-test rather than a one-sided t-test?
- Does the identity imply that a partial F-test of three restrictions equals the square of any one coefficient’s t statistic? Explain.
- What assumptions support the usual exact t and F reference distributions in this regression setting?