Why Test a Group of Added Predictors?
A multiple regression model may already include several predictors when an analyst considers adding a block of new variables. For example, a hypothetical insurer might model annual claim cost using policyholder age and coverage amount, then ask whether vehicle age and annual mileage contribute additional information. Testing each new coefficient separately does not answer the same question as testing whether the new variables, considered together, improve the model.
A partial F test addresses that joint question by comparing the residual variation from two nested models fitted to the same response observations. The reduced model contains the predictors already under consideration; the full model contains those predictors plus the added block. The test measures the reduction in residual sum of squares relative to the full model’s estimate of error variance.
Hypotheses, Statistic, and Conditions
Suppose there are \(n\) observations. Write the full model as a regression with an intercept and \(p_F\) predictor coefficients, and let the reduced model have \(p_R\) predictor coefficients. These counts exclude the intercept. If the full model adds \(q=p_F-p_R\) coefficients, then the joint test is
The notation labels the added coefficients as a block; it does not require that they appear last in a particular software output. More generally, the null hypothesis can specify a set of \(q\) independent linear restrictions on the full-model coefficients. The central question is whether the added terms provide evidence of an improvement conditional on the predictors already in the model.
The standard exact test assumes a correctly specified linear mean model, independent errors with mean zero and common variance \(\sigma^2\), normally distributed errors, and full rank for the fitted model matrices. Both models must use the same response and the same observations. In particular, a comparison is not valid if adding predictors causes missing-data deletion in one model but not the other. The \(F\) reference distribution is exact under normal errors; when that assumption is not appropriate, this exact small-sample test need not apply.
Why the Partial F Formula Works
Adding terms cannot increase the least-squares residual sum of squares when both models use the same observations: the full model has all the reduced model’s possible fits, plus additional ones. Thus \(SSE_R-SSE_F\) measures the reduction in residual variation attributable to the added block. Dividing by \(q\) gives that reduction per restriction, while \(SSE_F/(n-p_F-1)\) estimates the error variance using the full model.
The reason the ratio has an \(F\) distribution under the null is that, under the stated normal linear-model assumptions, the scaled reduction in sum of squares and the scaled full-model residual sum of squares are independent chi-square quantities. Under \(H_0\), both are scaled by the same \(\sigma^2\):
Dividing each quantity by its degrees of freedom gives two independent estimates of the same variance under \(H_0\). Their ratio is therefore \(F\)-distributed with the degrees of freedom shown above. If the added terms improve the fit more than would typically be expected from random error, the observed statistic will be large. Consequently, the p-value is the upper-tail probability \(P(F_{q,n-p_F-1}\geq F_{\text{observed}})\).
Applying the Test
Use the residual sums of squares from the models, count the added coefficients, and use the full model’s residual degrees of freedom. At a chosen significance level \(\alpha\), reject \(H_0\) when the p-value is less than \(\alpha\), or equivalently when the test statistic exceeds the appropriate upper-tail critical value. A rejection supports the conclusion that the block contributes statistically detectable information, given the reduced-model predictors and the assumptions. It does not establish that every added coefficient matters, that the effect is large in practical terms, or that the predictors cause the response.
Worked Example: Adding Two Regional Risk Measures
A hypothetical analyst models annual claim cost for 24 regions. The reduced model uses two predictors, and the full model adds two more regional measures. The reduced-model residual sum of squares is \(SSE_R=58\); the full-model value is \(SSE_F=38\). At \(\alpha=0.05\), test whether the two added predictors improve the model.
Method and hypotheses. This is a partial F test of the added block. The null says both added coefficients are zero; the alternative says at least one is nonzero. The number of restrictions is \(q=2\).
Degrees of freedom. The full model has four predictor coefficients plus an intercept, so its residual degrees of freedom are \(24-4-1=19\). The reference distribution is \(F_{2,19}\).
Calculate the statistic. The reduction in residual sum of squares is \(58-38=20\), or \(20/2=10\) per added coefficient. The full-model mean square error is \(38/19=2\). Therefore,
For \(F_{2,19}\), the upper-tail p-value is approximately \(0.0180\). Since \(0.0180<0.05\), reject \(H_0\). There is evidence that the two regional measures, considered jointly and conditional on the original predictors, improve the model. The test does not establish that both measures individually have nonzero coefficients.
Check. The full model has the smaller residual sum of squares, \(38<58\), and the statistic uses the full-model error estimate \(38/19\). The common mistake of using the reduced-model error estimate in the denominator is avoided.
Worked Example: A Three-Predictor Block That Falls Short
A hypothetical portfolio study uses 30 portfolio groups to model average annual claim cost. The reduced model has two predictors, while the full model adds three characteristics. The reported residual sums of squares are \(SSE_R=96\) and \(SSE_F=72\). Test the added block at the 5% level.
Method and hypotheses. The question concerns a joint test of the three added coefficients, so \(q=3\). The null sets all three to zero, and the alternative says at least one differs from zero.
Degrees of freedom. The full model has five predictor coefficients and an intercept. Its residual degrees of freedom are \(30-5-1=24\). The reference distribution is \(F_{3,24}\).
Calculate the statistic. The reduction is \(96-72=24\), giving \(24/3=8\) per added coefficient. The full-model mean square error is \(72/24=3\). Hence,
The 5% upper-tail critical value for \(F_{3,24}\) is approximately \(3.01\). Because \(2.667<3.01\), do not reject \(H_0\) at the 5% level. The observed reduction is not large enough, relative to the full-model error estimate, to conclude that the three predictors jointly improve the model at this significance level. This is not proof that all three coefficients are zero.
Check. The denominator uses \(72/24=3\), and there are three numerator degrees of freedom because three coefficients are tested. Dividing the reduction by the full model’s number of parameters instead of by the three restrictions would use the wrong numerator degrees of freedom.
Worked Example: Evidence for Two Added Service Variables
For 32 hypothetical service regions, an analyst fits a reduced model with two predictors and a full model that adds two service variables. The residual sums of squares are \(SSE_R=90\) and \(SSE_F=60\). Test the added block at \(\alpha=0.01\).
Method and hypotheses. The partial F test evaluates whether both added coefficients are zero. Here \(q=2\), and the alternative is that at least one is nonzero.
Degrees of freedom. There are four predictor coefficients in the full model, in addition to its intercept. The denominator degrees of freedom are \(32-4-1=27\), so the reference distribution is \(F_{2,27}\).
Calculate the statistic. The reduction is \(90-60=30\), which is \(30/2=15\) per added coefficient. The full-model mean square error is \(60/27=20/9\). Therefore,
For \(F_{2,27}\), the upper-tail p-value is approximately \(0.0042\). Since \(0.0042<0.01\), reject \(H_0\). The two added service variables jointly provide statistically significant improvement at the 1% level, conditional on the original predictors.
Check. The statistic is positive because the full model has lower SSE, and the p-value is below the stated significance level. The conclusion concerns the block, not the separate contribution of each variable.
Common Exam Traps
- Using the wrong variance estimate. The denominator is the full-model \(SSE_F\) divided by its residual degrees of freedom. It is not the reduced-model mean square error.
- Using the wrong numerator degrees of freedom. The numerator degrees of freedom equal the number of independent added restrictions, \(q\), not the number of observations or the total number of full-model predictors.
- Comparing models on different observations. The SSE difference has the intended nested-model interpretation only when both fits use the same response observations.
- Reversing the p-value tail. Evidence against the null corresponds to a large reduction in SSE and a large \(F\) statistic, so the p-value is an upper-tail probability.
- Overstating the result. A significant joint test does not show that every added predictor is individually significant, large in practical terms, or causal.
Check Your Understanding
For each problem, identify the test degrees of freedom, calculate the partial F statistic, and use the stated decision rule. Each scenario is hypothetical.
- A reduced model for 20 regions has two predictors. A full model adds one predictor and has \(SSE_R=50\) and \(SSE_F=42\). At the 5% level, use the \(F_{1,16}\) critical value \(4.49\). (Answer: \(F=3.05\), degrees of freedom \((1,16)\); do not reject at 5%.)
- For 28 hypothetical portfolios, the full model has five predictors, \(SSE_R=110\), and \(SSE_F=66\). It adds two predictors to the reduced model. Using a 5% critical value of \(3.44\) for \(F_{2,22}\), decide whether the block improves the fit. (Answer: \(F=7.33\), degrees of freedom \((2,22)\); reject at 5%.)
- A full model for 40 hypothetical service areas has six predictors, with \(SSE_F=100\). It adds three predictors to a reduced model with \(SSE_R=160\). Calculate the statistic and decide at 5%, using a critical value of \(2.89\) for \(F_{3,33}\). (Answer: \(F=6.60\), degrees of freedom \((3,33)\); reject at 5%.)
- Two models use the same response but not the same observations: the reduced model uses 120 complete cases, while the full model uses 113 because an added predictor is missing for some cases. Can their reported SSE values be inserted directly into the usual partial F formula? (Answer: No; refit or otherwise compare the models on the same observations before using the usual partial F test.)