Wald Tests for Logistic Coefficients

Why Test an Individual Logistic Coefficient?

A logistic regression coefficient describes how a predictor is associated with the outcome on the log-odds scale, conditional on the other predictors in the model. A Wald test asks whether the data provide evidence that this coefficient differs from zero. It is one way to assess an individual predictor’s conditional association with a binary outcome.

The test does not determine whether the association is practically important, establish causation, or show that the model is correctly specified. Its conclusion depends on the fitted model, the estimated coefficient, and the uncertainty in that estimate. In particular, a large coefficient is not automatically persuasive evidence if its standard error is also large.

Definition: For a logistic regression coefficient $\beta_j$, an individual two-sided Wald test has null hypothesis $H_0:\beta_j=0$ and alternative hypothesis $H_A:\beta_j\ne 0$. The test statistic is the estimated coefficient divided by its estimated standard error: $$ Z_W=\frac{\hat{\beta}_j}{\operatorname{SE}(\hat{\beta}_j)}. $$ Under the null hypothesis and standard large-sample conditions, $Z_W$ is approximately standard normal. The two-sided p-value is $2[1-\Phi(|Z_W|)]$, where $\Phi$ is the standard normal cumulative distribution function.

Model, Hypotheses, and Calculation

For independent Bernoulli outcomes $Y_i$, a logistic regression model specifies $P(Y_i=1\mid \mathbf{x}_i)=p_i$ and $\log\{p_i/(1-p_i)\}=\beta_0+\sum_{k=1}^q\beta_kx_{ik}$. Here $\beta_j$ is the coefficient for predictor $x_j$. Testing $\beta_j=0$ asks whether that predictor has a log-odds association with the outcome after accounting for the other predictors included in this model.

The usual Wald calculation uses the estimated coefficient $\hat{\beta}_j$ and its estimated standard error. The standard error is the square root of the corresponding diagonal element of the estimated covariance matrix of the coefficient estimates. It measures sampling uncertainty in the coefficient estimate, not variability among individual outcomes.

For a single coefficient, the same test can be written using a squared statistic. Under $H_0$, $Z_W^2$ is approximately distributed as a chi-square random variable with one degree of freedom. The two-sided p-value from that chi-square calculation is the same as the two-sided standard-normal p-value. The sign of $Z_W$ indicates the direction of the estimated association; the squared statistic does not retain that sign.

Why the Wald Formula Works

The justification is the large-sample behavior of the maximum likelihood estimate. Under a correctly specified model and regularity conditions, the estimate is approximately normal around the true coefficient, with variance estimated from the fitted model. Under the null, the true coefficient is zero. Standardizing the difference between the estimate and zero therefore gives an approximately standard normal statistic.

\hat{\beta}_j \ \dot{\sim}\ N\!\left(\beta_j,\operatorname{SE}(\hat{\beta}_j)^2\right) \quad\Longrightarrow\quad \left.\frac{\hat{\beta}_j-\beta_j}{\operatorname{SE}(\hat{\beta}_j)}\right|_{H_0:\beta_j=0} \ \dot{\sim}\ N(0,1).

The symbol $\dot{\sim}$ indicates an approximation that improves under suitable large-sample conditions. The p-value follows by adding the probabilities in the two tails at least as far from zero as the observed statistic:

P(|Z|\ge |Z_W|)=2\{1-\Phi(|Z_W|)\}.

This argument also explains the limits of the method: the normal approximation can be poor when the sample contains little information about a coefficient or when the estimate is unstable. A finite, reliable coefficient estimate and a sensible standard error are essential to interpreting the test.

Worked Tests and Interpretations

Worked Example: Testing a Rating Predictor

In a hypothetical model for whether a policy receives a specified claim outcome, an insurer includes a numerical vehicle-age rating along with other predictors. The fitted coefficient for the rating is $0.36$, with standard error $0.15$. Test for a conditional association at the 5% significance level. For a one-unit increase in the rating, the other included predictors are held fixed.

Step 1: State the hypotheses. The null is $H_0:\beta_j=0$, and the two-sided alternative is $H_A:\beta_j\ne0$. The question is whether the rating contributes evidence of an association after adjustment for the other model predictors.

Step 2: Calculate the Wald statistic. $Z_W=0.36/0.15=2.4$. The positive sign indicates a positive estimated coefficient.

Step 3: Calculate the p-value. The standard normal probability above $2.4$ is approximately $0.0082$. Thus the two-sided p-value is $2(0.0082)=0.0164$.

Step 4: Make the decision and interpret. Since $0.0164<0.05$, reject $H_0$ at the 5% level. The data provide evidence of a conditional association between vehicle-age rating and the modeled outcome, given the other predictors. This is not evidence by itself that changing the rating causes a change in outcome.

Method and mistake avoided. This is an individual two-sided Wald z test. It uses the coefficient divided by its standard error; it does not compare the unstandardized coefficient $0.36$ directly with a standard-normal cutoff.

Worked Example: A Negative Estimate With Limited Evidence

A hypothetical logistic model predicts whether a member makes a particular type of service request. The coefficient for annual account tenure, measured in years, is $-0.21$, with standard error $0.14$. Test $H_0:\beta_j=0$ against a two-sided alternative at the 5% level.

Step 1: Standardize the estimate. $Z_W=-0.21/0.14=-1.5$. The estimate is negative, but its magnitude must be considered relative to its uncertainty.

Step 2: Find the two-sided p-value. For a standard normal variable, $P(Z\le-1.5)$ is approximately $0.0668$. By symmetry, the two-sided p-value is $2(0.0668)=0.1336$.

Step 3: Interpret the result. Because $0.1336>0.05$, do not reject $H_0$ at the 5% level. This analysis does not provide sufficient evidence of a conditional association at that level. It does not prove that the true coefficient is zero; the estimate remains uncertain.

Method and mistake avoided. The sign of the statistic does not make this a one-sided test. For the stated two-sided alternative, both tails are counted, so the p-value is twice the lower-tail probability at $-1.5$.

Worked Example: Getting the Standard Error From a Covariance Estimate

For a hypothetical logistic model of whether an electronic equipment lease requires a service visit, the coefficient estimate for monthly operating hours is $0.45$. The estimated covariance matrix has diagonal entry $0.0324$ for this coefficient. Test for a conditional association at both the 5% and 1% levels.

Step 1: Recover the standard error. The diagonal entry is an estimated variance, so $\operatorname{SE}(\hat{\beta}_j)=\sqrt{0.0324}=0.18$.

Step 2: Calculate the statistic. $Z_W=0.45/0.18=2.5$. The one-degree-of-freedom Wald chi-square statistic is $2.5^2=6.25$.

Step 3: Obtain the two-sided p-value. The standard-normal upper-tail probability beyond $2.5$ is approximately $0.0062$. Therefore $p=2(0.0062)=0.0124$.

Step 4: Decide at each level. Since $0.0124<0.05$, reject the null at 5%. Since $0.0124>0.01$, do not reject at 1%. The evidence is significant at the 5% level but not at the 1% level.

Method and mistake avoided. This is a Wald test using the square root of the estimated variance. Using $0.0324$ itself as the standard error would understate the standard error and produce an incorrect statistic.

Interpreting the Result Carefully

A significant Wald test supports evidence that the coefficient differs from zero in the specified model. For a numerical predictor measured in its stated units, the coefficient is a change in log odds for a one-unit increase, conditional on the other predictors. Equivalently, the corresponding odds are multiplied by $\exp(\beta_j)$ for that increase. The test is about whether the coefficient is zero; it is not a direct test of a change in probability.

A non-significant result is not proof of no association. It may reflect a small association, substantial uncertainty, or limited information. Conversely, a small p-value does not measure the size or practical value of an association. Always report the estimate and its uncertainty alongside the test conclusion when those quantities are available.

The usual approximation assumes independent observations, a correctly specified logistic model, and conditions under which maximum likelihood estimates are approximately normal and have a reliable estimated covariance matrix. The data must contain enough information to estimate the coefficient. When estimates are unstable or the normal approximation is poor, a Wald p-value can be unreliable. The test is also conditional on the predictors included: changing the model can change both the coefficient and its standard error.

Common Exam Trap

Common mistake: A candidate sees $\hat{\beta}_j=0.36$, compares $0.36$ with $1.96$, and concludes that the coefficient is not significant because $0.36<1.96$. This compares quantities on different scales: $0.36$ is an estimated coefficient, while $1.96$ is a standard-normal cutoff for a standardized statistic.

Correction: Divide by the standard error first. With standard error $0.15$, $Z_W=0.36/0.15=2.4$, which exceeds $1.96$ in absolute value. The two-sided p-value is $0.0164$, so the null is rejected at the 5% level.

Another frequent error is to report a one-tail probability for a two-sided alternative. First identify the alternative hypothesis. If it is $\beta_j\ne0$, calculate $2[1-\Phi(|Z_W|)]$; do not use only the tail in the direction of the estimate.

Key takeaway: For an individual logistic coefficient, divide the estimate by its standard error, use the stated alternative to obtain the p-value, and interpret the result as evidence about conditional association in the fitted model. Neither statistical significance nor a non-significant result, by itself, establishes practical importance or absence of an effect.

Check Your Understanding

For each problem, use a two-sided Wald test unless the question states otherwise. Give p-values to four decimals.

A hypothetical logistic model for a roadside-assistance outcome estimates the coefficient for annual mileage at $0.30$, with standard error $0.12$. Calculate the Wald statistic and p-value, and state the decision at 5%. (Answer: $Z_W=2.50$, $p=0.0124$; reject $H_0$ at 5%.)
In a model for a hypothetical equipment inspection outcome, a coefficient estimate is $-0.18$, with standard error $0.12$. Calculate the statistic and p-value, then state whether there is evidence of a conditional association at 5%. (Answer: $Z_W=-1.50$, $p=0.1336$; do not reject $H_0$ at 5%.)
A fitted model for a hypothetical travel-insurance outcome gives an estimated coefficient of $0.28$ for a numerical predictor and standard error $0.14$. Test at both the 5% and 1% levels. (Answer: $Z_W=2.00$, $p=0.0455$; reject at 5%, but not at 1%.)
For a coefficient in a hypothetical logistic model of a device-repair outcome, the estimate is $0.50$ and the estimated variance is $0.0625$. Calculate the standard error, Wald statistic, and p-value. (Answer: $\operatorname{SE}=0.25$, $Z_W=2.00$, $p=0.0455$.)
A hypothetical logistic model for a benefit-claim outcome estimates a coefficient of $-0.36$, with standard error $0.12$. A candidate reports only the lower-tail probability as the two-sided p-value. Correct the p-value and state the 1% decision. (Answer: $Z_W=-3.00$; two-sided $p=0.0027$, not $0.0013$; reject $H_0$ at 1%.)

What You'll Learn