What Does a Single Added Term Test?
A regression coefficient represents the contribution of one model term after accounting for the other terms already included. A single-term deviance test asks whether the data provide evidence that this coefficient differs from zero. It compares two models fitted to the same observations: a reduced model without the term and a full model that adds it. The comparison concerns the improvement in maximized likelihood, not merely whether the full model’s fitted values look different.
Here, “single added term” means a term controlled by one free regression coefficient. The test is therefore for one scalar restriction. A factor represented by several coefficients, or a collection of terms tested together, requires a different degrees-of-freedom calculation and is outside this tutorial’s scope.
The test presumes that the full and reduced models use the same response observations and the same likelihood specification. The added coefficient must be identifiable: its predictor contribution cannot be exactly represented by the other terms. Under the null, the reduced model is nested in the full model.
The Statistic and Its Degrees of Freedom
Write \(\ell_F\) and \(\ell_R\) for the maximized log-likelihoods under the full and reduced models. Write \(D_F\) and \(D_R\) for their unscaled likelihood deviances, using the same observations and likelihood. The likelihood-ratio statistic is
The full model includes the reduced model as a special case, so maximizing over the full model’s larger parameter space cannot produce a smaller maximized log-likelihood. Thus \(G^2\geq 0\). Larger values indicate a greater improvement in fit from allowing the added coefficient to vary.
The identity follows directly from the definition of likelihood deviance. If \(\ell_{\mathrm{sat}}\) is the maximized log-likelihood in the saturated model, then \(D_M=2(\ell_{\mathrm{sat}}-\ell_M)\) for model \(M\). Therefore
The saturated-model term cancels. This algebra also shows why the comparison must use the same observations and likelihood: otherwise the two saturated-model terms need not be the same, and the difference would not simplify to the likelihood-ratio statistic.
For the chi-square calibration, the regularity conditions of Wilks’ theorem are needed. In particular, the null parameter must be an interior point of a regular, identifiable model; the log-likelihood must satisfy the usual smoothness and information conditions; the true parameter must be represented by the model; and the number of parameters is fixed as the sample size grows. The added coefficient must not be a boundary parameter or unidentified under the null. Under these conditions, with one independent restriction,
This is the single-coefficient specialization of Wilks’ theorem for nested models. The p-value is the upper-tail probability \(P(\chi^2_1\geq G^2_{\mathrm{obs}})\). It measures how unusual a statistic at least this large would be under the null model and the stated asymptotic approximation. It is not the probability that the null hypothesis is true.
Worked Tests
Worked Example: A Binary Predictor in a Grouped-Binomial Model
Two groups each contain 20 individuals. There are 8 successes in group 1 and 14 in group 2. Consider a grouped-binomial logistic regression with an intercept and a group indicator in the full model; the reduced model has only an intercept. Test whether the group coefficient is zero.
Under the null, the fitted success proportion is pooled across groups: \(\hat p_R=(8+14)/(20+20)=22/40=0.55\). In the full model, the fitted proportions are \(\hat p_1=8/20=0.40\) and \(\hat p_2=14/20=0.70\). The binomial coefficients in the log-likelihood are the same in the two models, so they cancel in the likelihood difference. The statistic is
One coefficient was added, so the reference distribution is \(\chi^2_1\). The upper-tail probability is \(P(\chi^2_1\geq 3.6961)\approx 0.0546\). At the 0.05 significance level, the test does not reject \(H_0\). The observed proportions differ, but this test does not find evidence at that threshold that the group coefficient differs from zero.
Worked Example: A Poisson Rate Ratio
Suppose two groups each contribute 10 units of exposure, with observed event counts 2 and 8. Use a Poisson log-linear model with log exposure as an offset. The reduced model has a common rate; the full model adds a group indicator. Test whether the group coefficient is zero.
The reduced model’s common rate estimate is \((2+8)/(10+10)=0.5\) events per exposure unit. The full model estimates rates \(2/10=0.2\) and \(8/10=0.8\). For Poisson observations, the terms involving the observed counts but not the fitted means cancel in the likelihood ratio. Hence
The added group coefficient gives one degree of freedom. Thus \(P(\chi^2_1\geq 3.8560)\approx 0.0496\). At the 0.05 level, this test rejects the common-rate null, though the result is close to the cutoff. The fitted rate ratio is \(0.8/0.2=4\); the likelihood-ratio test evaluates evidence against a ratio of one, rather than testing whether the ratio is greater than one alone.
Worked Example: Testing a Logistic Slope
In a grouped-binomial logistic regression, suppose a binary predictor \(x\) has 10 individuals at each value. There is 1 success at \(x=0\) and 6 successes at \(x=1\). The full model is \(\operatorname{logit}(p)=\beta_0+\beta_1x\); the reduced model sets \(\beta_1=0\). Test \(H_0:\beta_1=0\).
The reduced estimate is the pooled proportion, \(7/20=0.35\). The full estimates are \(0.10\) and \(0.60\). As in the first example, the binomial coefficients cancel. The likelihood-ratio statistic is
Using one degree of freedom gives \(P(\chi^2_1\geq 5.9360)\approx 0.0148\). The test rejects at the 0.05 level. In the full model, the estimated slope is \(\log(0.60/0.40)-\log(0.10/0.90)=\log(13.5)\approx 2.603\). The test’s null is that this slope equals zero; the likelihood-ratio statistic itself does not encode a one-sided alternative.
Interpreting the Test Carefully
The likelihood-ratio test is a two-sided test of the equality constraint \(\beta_j=0\): departures in either direction can increase the statistic. The sign of the fitted coefficient supplies the direction of the estimated association, but it does not change the chi-square reference distribution. A small p-value is evidence against the reduced model under the assumed model and calibration; it does not establish that the added term is causally important or practically large.
Because the chi-square result is asymptotic, its calibration can be inaccurate in small samples, with sparse outcomes, or when the regularity assumptions fail. The examples illustrate the calculation, not a guarantee that the approximation is accurate for every data set. In particular, if the coefficient is on a boundary or is not identified under the null, the ordinary \(\chi^2_1\) reference need not apply.
A frequent mistake is to subtract the deviances in the wrong order or to compare the full-model statistic directly with a one-degree-of-freedom distribution. The correct statistic is \(D_R-D_F\), which is nonnegative, and the degrees of freedom are the difference in model dimensions: one for one added coefficient. Another mistake is to infer that a coefficient is significant merely because its full-model estimate is nonzero. The test uses the likelihood improvement and its reference distribution, not that observation alone.
Check Your Understanding
For each question, distinguish the model comparison, the test statistic, and its calibration.
- What is the null hypothesis when a full model adds one identifiable regression coefficient to a reduced model?
- Why does the statistic use the reduced-model deviance minus the full-model deviance?
- If exactly one coefficient is added, what is the asymptotic reference distribution under the null?
- A comparison gives \(D_R=18.4\) and \(D_F=15.9\). Calculate the likelihood-ratio statistic and state which tail probability defines the p-value.
- Name one regularity condition under which the usual chi-square calibration may fail.