Does it require fitting the alternative model?

Usually the statistic can be calculated from the null fit and derivatives of the larger model.

The common chi-square version is an asymptotic approximation.

Can it test several coefficients together?

Yes, with an appropriate joint statistic and degrees of freedom.

AI Glossary / Statistics and probability / Advanced

Score test

Definition

A score test is a likelihood-based hypothesis test that evaluates the slope of the log-likelihood at parameter values satisfying the null hypothesis. It scales that slope using information about its variability. It is also called a Rao test or, in common formulations, a Lagrange multiplier test.

What it means

The test asks whether the data provide a sufficiently strong local indication that the model should move away from the null restriction. For example, a null hypothesis might fix a success probability or set selected regression coefficients to zero.

Here, “score” means a derivative of the log-likelihood with respect to a parameter. It is not a prediction accuracy score, a credit score or a human rating. Its statistical meaning comes from the specified probability model.

How it works

Fit the model subject to the null restriction. Evaluate the score and an appropriate information matrix at that restricted fit. In a single-parameter problem without nuisance parameters, a common statistic is U squared divided by I, where U is the score and I is expected Fisher information under the null.

With several parameters, the calculation uses a quadratic form and accounts for nuisance parameters. Under suitable regularity conditions, its null distribution is approximately chi-square, with degrees of freedom corresponding to the independent restrictions tested.

The familiar reference distribution is not automatic for boundary parameters, singular models or inadequate samples. The null model and the calibration both require checking.

Three practical examples

1. Success probability: test a prespecified reference probability for independent binary outcomes using the binomial score calculation.

2. Regression extension: assess whether adding specified predictors is supported, using derivatives evaluated at the model fitted under their null restrictions.

3. Model diagnostics: use a specifically derived score test for a proposed departure from a fitted model. The diagnostic requires its own assumptions and cannot be replaced by an arbitrary slope calculation.

Think of it this way

Imagine standing at the location required by the null hypothesis on a likelihood landscape. The score describes the local slope. The test asks whether that slope is unusually large relative to its expected variation, rather than simply asking whether the slope is nonzero.

Worked example

Suppose a fictional evaluation records fifty-two successes in eighty independent trials. Each trial is modeled as having the same success probability p. Before observing results, the team specifies a two-sided test of p = 0.5 against p differing from 0.5.

Ignoring a constant that does not depend on p, the log-likelihood is 52 log(p) + 28 log(1 - p). Its derivative is U(p) = 52/p - 28/(1 - p).

At the null value, U(0.5) = 104 - 56 = 48. The expected Fisher information for eighty binomial trials is I(p) = 80/[p(1 - p)]. At p = 0.5, this equals 320.

The score statistic is therefore 48 squared divided by 320, or 7.2. Equivalently, it is (52 - 40) squared divided by [80 × 0.5 × 0.5] = 144/20 = 7.2.

Using an approximate chi-square reference distribution with one degree of freedom gives an upper-tail probability of about 0.00729. This is the two-sided score-test probability for the specified binomial hypothesis. At a prespecified five-percent significance level, the procedure rejects the null.

The equivalent signed normal statistic is (52 - 40)/sqrt(20), approximately 2.6833. Squaring it gives 7.2. Its positive sign indicates that the observed success proportion is above the null value; the chi-square statistic alone does not retain that direction.

The observed proportion is 52/80 = 0.65, fifteen percentage points above 0.5. That estimate describes the observed difference, while the test assesses compatibility with the null under its model. The probability 0.00729 is not the probability that the null hypothesis is true.

For comparison, substituting the observed proportion into the variance produces the ordinary binomial Wald statistic: 80(0.65 - 0.5)^2/[0.65 × 0.35], approximately 7.9121. A likelihood-ratio calculation gives approximately 7.3121. These are different statistics, even though their large-sample behavior is related.

No continuity correction is used in this worked score calculation. An exact binomial procedure or a corrected approximation need not produce the same probability. Repeated or correlated trials would also require a model or analysis that represents their dependence; eighty recorded outcomes do not automatically mean eighty independent observations.

Illustrative case study

Fictional Birch Evaluation receives an automated report calling the result a “99.27% probability of improvement.” Its reviewer removes that interpretation because the reported tail probability is conditional on the null model, not a posterior probability of improvement.

The report instead identifies the success count, estimated proportion, reference hypothesis, score statistic and approximate probability. Birch also checks whether repeated evaluations share tasks or users. If the independence assumption fails, the numerical calculation above is not sufficient for its actual study. This invented case demonstrates interpretation, not a real performance claim.

Common misconceptions and mistakes

“The score is evaluated at the unrestricted best fit.” A conventional score test uses the restricted null fit; the unrestricted interior maximum has zero score.

“Score, Wald and likelihood-ratio tests always agree.” Finite-sample statistics and decisions can differ.

“A small probability proves an important effect.” Statistical evidence and practical importance are different questions.

“Chi-square calibration works for every parameter.” Boundaries and other nonregular settings need special treatment.

Business use

Specify the null restriction and intended alternative before interpreting a result. Ask which model was fitted, how nuisance parameters were handled and which information estimate and reference distribution were used.

For automated analysis, retain the estimated effect and uncertainty alongside the test. Check sampling, dependence and model assumptions rather than treating a software label as validation. A computational advantage of score testing is that the unrestricted model need not be fitted, but simpler computation does not remove the need for a defensible statistical question.

Check your understanding

Why is the information evaluated at 0.5?

Answer: That is the probability fixed by the null hypothesis.

Why does squaring the normal statistic lose direction?

Answer: Positive and negative values have the same square.

Why is 0.00729 not a probability of the hypothesis?

Answer: It is a tail probability calculated under the assumed null model.

Related terms

Wald test
Null hypothesis
Score function
Fisher information
maximum likelihood
nuisance parameter
likelihood-ratio test
chi-square distribution

Learn next

Repeat the binomial calculation with forty-eight successes in eighty trials. Keep the null probability at 0.5 and compare the resulting statistic with the worked result.

Sources

Charles J. Geyer, Likelihood Inference, section 6.5, binomial score and information: https://www.stat.umn.edu/geyer/f21/5421/notes/likelihood.html

Marco Taboga, Score test, restricted estimation and asymptotic conditions: https://www.statlect.com/fundamentals-of-statistics/score-test

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