Wald test

The Wald test is a statistical test, typically used to test whether an effect exists or not. In other words, it tests whether an independent variable has a statistically significant relationship with a dependent variable.

Suppose an economist, who has data on social class and shoe size, wonders whether social class is associated with shoe size. Say &theta; is the average increase in shoe size for upper class people compared to middle class people: then the Wald test can be used to test whether &theta; is 0 (in which case social class has no association with shoe size) or non-zero (shoe size varies between social classes). Or, for a medical example, suppose smoking multiplies the risk of lung cancer by some number R: then the Wald test can be used to test whether R = 1 (i.e. there is no effect of smoking) or is greater (or less) than 1 (i.e. smoking alters risk).

A Wald test can be used in a great variety of different models including models for dichotomous variables and models for continuous variables.

Mathematical details
Under the Wald statistical test, named after Abraham Wald, the maximum likelihood estimate $$\hat\theta$$ of the parameter(s) of interest $$\theta$$ is compared with the proposed value $$\theta_0$$, with the assumption that the difference between the two will be approximately normal. Typically the square of the difference is compared to a chi-squared distribution. In the univariate case, the Wald statistic is



\frac{ ( \widehat{ \theta}-\theta_0 )^2 }{\operatorname{var}(\hat \theta )} $$

which is compared against a chi-square distribution.

Alternatively, the difference can be compared to a normal distribution. In this case the test statistic is


 * $$\frac{\widehat{\theta}-\theta_0}{\operatorname{se}(\hat\theta)}$$

where $$\operatorname{se}(\widehat\theta)$$ is the standard error of the maximum likelihood estimate. A reasonable estimate of the standard error for the MLE can be given by $$ \frac{1}{\sqrt{I_n(MLE)}} $$, where $$ I_n $$ is the Fisher information of the parameter.

In the multivariate case, a test about several parameters at once is carried out using a variance matrix . A common use for this is to carry out a Wald test on a categorical variable by recoding it as several dichotomous variables.

Alternatives to the Wald test
The likelihood-ratio test can also be used to test whether an effect exists or not. Usually the Wald test and the likelihood ratio test give very similar conclusions (as they are asymptotically equivalent), but very rarely, they disagree enough to lead to different conclusions: the researcher finds him/herself asking, or being asked, why the p-value is significant when the confidence interval includes 0, or why the p-value is not significant when the confidence interval excludes 0. In this situation, first remember that statistical significance is always somewhat arbitrary, as it depends on an arbitrarily chosen significance level.

There are several reasons to prefer the likelihood ratio test above the Wald test . One is that the Wald test can give different answers to the same question, according to how the question is phrased . For example, asking whether R = 1 is the same as asking whether log R = 0; but the Wald statistic for R = 1 is not the same as the Wald statistic for log R = 0 (because there is in general no neat relationship between the standard errors of R and log R). Likelihood ratio tests will give exactly the same answer whether we work with R, log R or any other transformation of R. The other reason is that the Wald test uses two approximations (that we know the standard error, and that the distribution is chi-squared), whereas the likelihood ratio test uses one approximation (that the distribution is chi-squared).

Yet another alternative is the score test, which have the advantage that it can be formulated in situations where the variability is difficult to estimate; e.g. the Cochran-Mantel-Haenzel test is a score test.