Why Test Polynomial Terms?
A straight-line regression can miss systematic curvature. Adding powers of a predictor gives a model more flexibility, but a smaller residual sum of squares is guaranteed whenever terms are added. The relevant question is whether the reduction is large enough, relative to the remaining variation, to support the more complex model.
A partial F-test answers this question when one candidate regression is nested inside another. For polynomial regression, it can assess whether a quadratic term adds evidence of curvature, whether several higher-order terms contribute jointly, or whether a cubic term is useful after a quadratic term is already included. The test compares models for the same response and same observations; it is not a comparison of unrelated fits.
The Partial F-Test for Nested Regressions
Suppose the full model has \(p_F\) estimated regression coefficients, including the intercept, and is fit to \(n\) observations. Let \(\mathrm{SSE}_R\) and \(\mathrm{SSE}_F\) be the residual sums of squares from the reduced and full models. If the full model adds \(q\) coefficients, the partial F-statistic is
The numerator measures the decrease in residual sum of squares per restriction released. The denominator is the full model’s residual mean square. Under the null hypothesis that all \(q\) added coefficients are zero, the statistic has an \(F\) distribution with \(q\) and \(n-p_F\) degrees of freedom, provided the regression assumptions below hold.
For a polynomial comparison, the coefficients being tested are the powers included in the full model but absent from the reduced one. A linear-versus-quadratic test has \(q=1\). A linear-versus-cubic test has \(q=2\), because it tests the quadratic and cubic coefficients together. A quadratic-versus-quartic test also has \(q=2\), testing the cubic and quartic coefficients jointly.
Why the Formula Has This Form
Under the null hypothesis, the added terms do not improve the mean model, apart from random variation. In a normal linear model, the reduction \(\mathrm{SSE}_R-\mathrm{SSE}_F\), divided by \(\sigma^2\), has a chi-square distribution with \(q\) degrees of freedom. The full-model residual sum of squares divided by \(\sigma^2\) has a chi-square distribution with \(n-p_F\) degrees of freedom. These two quantities are independent because they arise from orthogonal parts of the normal-error regression decomposition.
Dividing each chi-square quantity by its degrees of freedom gives independent estimates of the same error variance under the null. Their ratio therefore has an \(F\) distribution:
A large value indicates that the reduction in residual variation is large relative to the full model’s remaining residual variation. At significance level \(\alpha\), reject the null if the statistic exceeds the upper-tail critical value \(F_{q,n-p_F;1-\alpha}\), or equivalently if the p-value is less than \(\alpha\).
Worked Comparisons
Worked Example: Testing a Quadratic Term
A hypothetical insurer models annual processing cost \(Y\), in dollars, against policy duration \(x\), in years. For \(n=20\) policies, the fitted linear model has \(\mathrm{SSE}_R=240\) (in squared response units), and the quadratic model has \(\mathrm{SSE}_F=180\). The quadratic fit is reported as \(\hat Y=42+1.8x+0.15x^2\). At the 5% level, test whether adding the quadratic term improves the model. Use \(F_{1,17;0.95}=4.451\).
Method and hypotheses. Use the partial F-test for the nested linear and quadratic models. The null hypothesis is \(H_0:\beta_2=0\), and the alternative is \(H_A:\beta_2\ne0\). The quadratic model estimates an intercept and two slopes, so \(p_F=3\); there is \(q=1\) added coefficient and \(n-p_F=20-3=17\) residual degrees of freedom.
Numerator. The reduction in SSE is \(240-180=60\). Dividing by the one added coefficient gives \(60/1=60\).
Denominator. The full-model residual mean square is \(180/17=10.5882\).
Test statistic and decision. Thus \(F=60/10.5882=5.6667\). Since \(5.6667>4.451\), reject \(H_0\) at the 5% level. The data provide evidence that the quadratic term improves the linear fit under the model assumptions.
Interpretation and trap avoided. The coefficient \(0.15\) is the coefficient of \(x^2\), not a constant increase in cost for each additional year. The fitted slope at duration \(x\) is \(1.8+2(0.15)x=1.8+0.30x\). A common error is to use the reduced-model SSE in the denominator, calculating \(60/(240/17)=4.25\). The denominator must use the full model’s residual mean square, so the correct statistic is \(5.6667\).
Worked Example: Testing Curvature Terms Jointly
A hypothetical analyst models the quarterly change in a reserve indicator \(Y\) against a centered economic index \(z\). For \(n=24\) quarters, the linear model has \(\mathrm{SSE}_R=210\), while the cubic model has \(\mathrm{SSE}_F=126\). The cubic fit is \(\hat Y=5.2+0.7z-0.08z^2+0.004z^3\). At the 5% level, test whether the polynomial terms beyond the straight line contribute jointly. Use \(F_{2,20;0.95}=3.493\).
Method and hypotheses. Compare the linear model with the cubic model using a partial F-test. The null is \(H_0:\beta_2=\beta_3=0\); the alternative is that at least one of these coefficients is nonzero. Two coefficients are added, so \(q=2\). The full model has four coefficients, including its intercept, giving \(n-p_F=24-4=20\) residual degrees of freedom.
Numerator. The SSE reduction is \(210-126=84\), and the reduction per restriction is \(84/2=42\).
Denominator. The full-model residual mean square is \(126/20=6.3\).
Test statistic and decision. The statistic is \(F=42/6.3=6.6667\). Since \(6.6667>3.493\), reject the joint null at the 5% level. The quadratic and cubic terms, considered together, improve the model relative to the straight line.
Interpretation and trap avoided. This test does not say that each coefficient is individually significant; it tests the pair of restrictions together. Nor is \(-0.08\), the coefficient of \(z^2\), a constant linear effect. The fitted slope is \(0.7-0.16z+0.012z^2\), which varies with \(z\). The common trap is to substitute separate coefficient tests for the specified joint test; the partial F-test directly answers whether the two added terms contribute as a group.
Worked Example: Does a Cubic Term Improve a Quadratic Fit?
A hypothetical pension analyst relates a projected adjustment \(Y\) to a centered valuation index \(x\). With \(n=18\) observations, the quadratic model has \(\mathrm{SSE}_R=96\), and the cubic model has \(\mathrm{SSE}_F=88\). The cubic model has four coefficients including the intercept. Test at the 5% level whether the cubic term adds evidence beyond the quadratic model. Use \(F_{1,14;0.95}=4.600\).
Method and hypotheses. Compare the nested quadratic and cubic models. The null is \(H_0:\beta_3=0\), conditional on retaining the intercept, \(x\), and \(x^2\) terms. There is one added coefficient, so \(q=1\). The full-model residual degrees of freedom are \(18-4=14\).
Numerator. The SSE reduction is \(96-88=8\), giving \(8/1=8\) per restriction.
Denominator. The cubic model’s residual mean square is \(88/14=6.2857\).
Test statistic and decision. Therefore \(F=8/6.2857=1.2727\). Since \(1.2727<4.600\), do not reject \(H_0\) at the 5% level. This comparison does not provide sufficient evidence that the cubic term improves the quadratic model.
Interpretation and trap avoided. The result does not establish that the cubic coefficient is exactly zero, and it does not test whether the quadratic term should be removed. It only addresses the cubic term conditional on the quadratic model. A common mistake is to conclude that the entire polynomial fit is unsupported; that conclusion does not follow from this comparison.
Interpreting Polynomial Coefficients and Test Results
A polynomial model is linear in its coefficients, even though its fitted mean can curve as a function of the predictor. For a quadratic model,
Thus \(\beta_1\) is the slope at \(x=0\), not necessarily a slope that applies throughout the predictor range. The quadratic coefficient describes how the slope changes with \(x\); it is not itself a linear effect. In a cubic model the slope is \(\beta_1+2\beta_2x+3\beta_3x^2\), so interpreting \(\beta_3\) as a constant change in \(Y\) per unit of \(x\) is also incorrect.
The null hypothesis must match the model comparison. A linear-versus-cubic test asks whether \(\beta_2=\beta_3=0\) jointly. A quadratic-versus-cubic test asks only whether \(\beta_3=0\). These are different scientific questions and generally produce different test statistics. State the reduced model, full model, restrictions, and degrees of freedom before calculating.
Check that both models use the same response, observations, and error scale, and that the reduced model is obtained by setting full-model coefficients to zero.
Set \(q\) equal to the number of coefficients added, and count \(p_F\) in the full model, including the intercept.
Use the full-model residual mean square in the denominator, compare the statistic with the appropriate \(F\) reference distribution, and describe the tested terms jointly when appropriate.
Common Exam Traps
- Using the wrong residual mean square. The denominator is \(\mathrm{SSE}_F/(n-p_F)\), not the reduced model’s residual mean square.
- Miscounting the degrees of freedom. The numerator degrees of freedom equal the number of restrictions, while the denominator degrees of freedom are \(n-p_F\).
- Testing terms separately when the question is joint. A linear-versus-cubic test concerns both the quadratic and cubic coefficients together.
- Calling a higher-order coefficient a linear effect. The marginal slope depends on \(x\) whenever powers of \(x\) are present.
- Overstating a failure to reject. It means the comparison did not provide sufficient evidence at the chosen significance level; it does not prove that the added coefficient is zero.
A partial F-test evaluates improvement in fit under a specified regression model and its assumptions. It does not, by itself, establish that the polynomial is the best predictive choice, that the model’s diagnostics are satisfactory, or that a detected curve has a causal interpretation. Its contribution is narrower: it provides a principled test of whether specified nested polynomial terms reduce residual variation enough to justify the added parameters.
Check Your Understanding
For each comparison, identify the tested terms and use the supplied critical value to make the decision.
- A linear model and a quadratic model use \(n=16\) observations. Their SSE values are 150 and 110, respectively. The full model has three coefficients. Test at 5%, using \(F_{1,13;0.95}=4.667\). (Answer: \(F=4.7273\); reject the null that the quadratic coefficient is zero.)
- For \(n=25\), a linear model has \(\mathrm{SSE}=250\), and a cubic model has \(\mathrm{SSE}=160\). The cubic model has four coefficients. Test the added terms jointly at 5%, using \(F_{2,21;0.95}=3.467\). (Answer: \(F=5.9063\); reject \(H_0:\beta_2=\beta_3=0\).)
- A quadratic model has \(\mathrm{SSE}=132\), and a cubic model has \(\mathrm{SSE}=126\), based on \(n=22\). The full model has four coefficients. At 5%, use \(F_{1,18;0.95}=4.414\). (Answer: \(F=0.8571\); do not reject \(H_0:\beta_3=0\); this does not test the quadratic term.)
- For \(n=28\), the linear and quartic models have SSE values 280 and 196. The quartic model has five coefficients. Test the cubic and quartic terms jointly at 5%, using \(F_{3,23;0.95}=3.028\). (Answer: \(F=3.2857\); reject the joint null that \(\beta_2=\beta_3=\beta_4=0\).)
- A quadratic model is compared with a quintic model fitted to the same \(n=40\) observations. How many restrictions are tested, and which coefficients are in the null hypothesis? (Answer: \(q=3\); test \(H_0:\beta_3=\beta_4=\beta_5=0\), with numerator degrees of freedom 3.)