Why Test a Regression Coefficient?
A fitted regression coefficient describes an estimated relationship in a model. A test asks whether the data provide sufficient evidence that the corresponding population coefficient differs from a specified value. That value need not be zero. For example, an analyst might test whether the change in a predicted response per unit increase in a predictor differs from a pre-established benchmark.
A two-sided test treats departures above and below the specified value as evidence against the null hypothesis. The sign of the estimated difference tells you which direction the estimate lies, but the decision rule considers departures in either direction.
Model, Assumptions, and Test Statistic
Use the usual linear regression model \(Y=X\beta+\varepsilon\), where \(X\) is the design matrix and \(\beta\) is the vector of population coefficients. For the exact t-test procedure, assume the design matrix has full column rank, the errors are independent and normally distributed with mean zero and common variance \(\sigma^2\), and the model form is appropriate. In particular, the conditional mean is linear in the coefficients and the error variance is constant. With \(p\) coefficients estimated, including the intercept if present, the residual degrees of freedom are \(n-p\), which must be positive.
The standard error used in the test is estimated from the regression fit. It reflects the estimated error variance and the design of the predictors. The test statistic centers the coefficient estimate at the value asserted by the null hypothesis, then measures that difference in standard-error units:
When \(H_0\) is true and the stated assumptions hold, this statistic has a t distribution with \(n-p\) degrees of freedom. The numerator is not generally \(\hat{\beta}_j\): it is the distance between the estimate and the specified null value. Under the special null hypothesis \(\beta_j=0\), those happen to be the same.
The division of \(\alpha\) between two tails is essential. Under the null, a sufficiently large positive statistic is evidence of a value above \(\beta_{j,0}\); a sufficiently large negative statistic is evidence of a value below it. The critical-value rule includes both possibilities.
Why the t Statistic Has This Form
The centering and scaling in the statistic follow from the sampling distribution of the least-squares estimator. With a fixed full-rank design and the model assumptions above, the estimator is normally distributed around the true coefficient. Its variance depends on the unknown error variance \(\sigma^2\). If that variance were known, standardizing the difference from the null value would use \(\sigma\). In regression it is estimated from the residuals, which introduces the t distribution.
More explicitly, let \(s^2\) be the residual mean square, the estimate of \(\sigma^2\), and let \(c_{jj}\) be the \(j\)th diagonal element of \((X^\mathsf{T}X)^{-1}\). Then \(\operatorname{SE}(\hat{\beta}_j)=s\sqrt{c_{jj}}\). Under \(H_0\), the standardized numerator using the true standard deviation is standard normal. Independently, the residual sum of squares divided by \(\sigma^2\) has a chi-squared distribution with \(n-p\) degrees of freedom. Dividing the standard normal quantity by the square root of that independent chi-squared quantity divided by its degrees of freedom gives a t statistic. Replacing the unknown \(\sigma\) with its residual estimate therefore yields the stated t distribution.
This exact result depends on the model assumptions, especially normal errors for a small-sample exact t distribution. If those assumptions do not hold, the same calculation may not have the stated exact distribution. The test should not be treated as automatically valid simply because software reports a t statistic.
Applying the Critical-Value Procedure
A complete test identifies the coefficient and its units, states both hypotheses, uses the correct residual degrees of freedom, computes the statistic against the null value, and compares its absolute value with the two-sided critical value. The final conclusion should name the response and predictors in context. “Reject” is a statistical decision, not a statement that a relationship is causal.
Worked Example: Testing a Service-Time Slope Against a Benchmark
A hypothetical insurer models claim review time in hours using policy age and claim complexity. The fitted coefficient for policy age is \(1.20\) hours per year, with standard error \(0.30\) hours per year. The model uses \(n=18\) claims and estimates \(p=3\) coefficients. At \(\alpha=0.05\), test whether the policy-age coefficient differs from \(0.50\) hours per year.
Method and hypotheses. This is a two-sided t test of one coefficient against a specified value: \(H_0:\beta_{\text{age}}=0.50\) versus \(H_A:\beta_{\text{age}}\ne0.50\).
Degrees of freedom. The fitted coefficients include the intercept, policy age, and claim complexity, so \(p=3\). Thus \(n-p=18-3=15\) residual degrees of freedom.
Test statistic. Subtract the null value from the estimate, then divide by the standard error: \((1.20-0.50)/0.30=0.70/0.30=2.3333\).
Critical-value decision. For a two-sided test at \(\alpha=0.05\) with 15 degrees of freedom, the critical value is approximately \(2.131\). Since \(\lvert2.3333\rvert=2.3333>2.131\), reject \(H_0\).
Conclusion and check. Under the regression assumptions, the data provide evidence at the 5% level that the policy-age coefficient differs from \(0.50\) hours per year. The estimate is above the specified value. The statistic is positive, as expected from \(1.20>0.50\), which checks the direction of the calculation. The common mistake this test avoids is treating the null value as zero when it is actually \(0.50\).
Worked Example: A Coefficient Estimate Near the Null Value
A hypothetical model predicts monthly repair cost in dollars from equipment age, usage, and service category. The estimated usage coefficient is \(-0.18\) dollars per usage unit, with standard error \(0.08\). There are \(n=24\) observations and \(p=4\) fitted coefficients. At \(\alpha=0.05\), test whether the coefficient differs from \(-0.10\) dollars per usage unit.
Method and hypotheses. Set \(H_0:\beta_{\text{usage}}=-0.10\) and \(H_A:\beta_{\text{usage}}\ne-0.10\). Because the null value is negative, retain its sign when subtracting.
Degrees of freedom. The residual degrees of freedom are \(24-4=20\).
Test statistic. The difference is \(-0.18-(-0.10)=-0.08\). Dividing by \(0.08\) gives \(t=-0.08/0.08=-1.00\).
Critical-value decision. The two-sided 5% critical value with 20 degrees of freedom is approximately \(2.086\). Since \(\lvert-1.00\rvert=1.00<2.086\), do not reject \(H_0\).
Conclusion and check. The data do not provide sufficient evidence at the 5% level that the usage coefficient differs from \(-0.10\) dollars per usage unit. This does not establish that the coefficient equals \(-0.10\). The negative statistic is consistent with the estimate being below the null value. A frequent arithmetic error is to calculate \(-0.18-0.10\); the correct subtraction is \(-0.18-(-0.10)\).
Worked Example: Avoiding an Incorrect Zero-Value Test
A hypothetical regression predicts the number of days needed to resolve an account issue from its complexity score. The estimated complexity coefficient is \(0.30\) days per score point, with standard error \(0.08\). With \(n=17\) observations and \(p=3\) fitted coefficients, test at \(\alpha=0.10\) whether the coefficient differs from \(0.20\) days per point.
Method and hypotheses. Use \(H_0:\beta_{\text{complexity}}=0.20\) and \(H_A:\beta_{\text{complexity}}\ne0.20\). The test concerns the stated benchmark, not whether the coefficient is zero.
Degrees of freedom. There are \(17-3=14\) residual degrees of freedom.
Test statistic. The null-centered difference is \(0.30-0.20=0.10\). Therefore, \(t=0.10/0.08=1.25\).
Critical-value decision. For \(\alpha=0.10\) and 14 degrees of freedom, the two-sided critical value is approximately \(1.761\). Since \(\lvert1.25\rvert<1.761\), do not reject \(H_0\).
Conclusion and check. The data do not provide sufficient evidence at the 10% level that the complexity coefficient differs from \(0.20\) days per point. The common mistake is to use \(0\) in the numerator: \(0.30/0.08=3.75\) would test \(H_0:\beta_{\text{complexity}}=0\), a different hypothesis, and would give the wrong decision for the question asked. Centering at \(0.20\) gives the appropriate statistic, \(1.25\).
Reading the Decision Correctly
The test answers a specific question: whether the observed coefficient estimate is sufficiently far from the specified null value, relative to its standard error, to cross the chosen two-sided threshold. Keep the hypotheses, coefficient, and units together throughout the calculation. A numerical coefficient without its predictor and response units can be easy to misinterpret.
- Reject \(H_0\): the data provide evidence that the coefficient differs from the specified value, under the model assumptions and at the chosen significance level.
- Do not reject \(H_0\): the data do not provide sufficient evidence of a difference at that level. This is not proof that the null value is exactly correct.
- Use the absolute statistic: the two-sided decision checks both positive and negative departures.
- Use \(n-p\): count all fitted coefficients, including the intercept, when finding residual degrees of freedom.
The critical value depends on both \(\alpha\) and the residual degrees of freedom. A smaller \(\alpha\) places more probability in the tails and requires a more extreme statistic for rejection. The decision is therefore tied to the chosen significance level; it is not a description of how large or practically important the coefficient difference is.
Check Your Understanding
For each problem, identify the correct null-centered statistic, use the stated significance level, and make the two-sided critical-value decision.
- A model has \(n=14\) observations and \(p=2\) fitted coefficients. A slope estimate is \(2.40\), its standard error is \(0.15\), and the specified null value is \(2.00\). Test at \(\alpha=0.05\). (Answer: \(df=12\), \(t=2.6667\), critical value approximately \(2.179\); reject \(H_0\).)
- A model has \(n=22\) and \(p=3\). An estimated coefficient is \(0.07\), its standard error is \(0.04\), and the null value is \(0.10\). At \(\alpha=0.05\), what is the decision? (Answer: \(df=19\), \(t=-0.75\), critical value approximately \(2.093\); do not reject \(H_0\).)
- With \(n=12\) and \(p=3\), an estimate is \(-1.50\), its standard error is \(0.20\), and the null value is \(-1.00\). Test at \(\alpha=0.01\). (Answer: \(df=9\), \(t=-2.50\), critical value approximately \(3.250\); do not reject \(H_0\).)
- A coefficient estimate is \(0.64\), its standard error is \(0.055\), and its specified null value is \(0.50\). The regression has \(n=30\) observations and \(p=5\) fitted coefficients. Test at \(\alpha=0.05\). (Answer: \(df=25\), \(t=0.14/0.055\approx2.5455\), critical value approximately \(2.060\); reject \(H_0\).)
- A model has \(n=20\) observations and includes four predictors and an intercept. A coefficient estimate is \(0.90\), its standard error is \(0.10\), and the null value is \(1.10\). Test at \(\alpha=0.05\). (Answer: \(p=5\), \(df=15\), \(t=-2.00\), critical value approximately \(2.131\); do not reject \(H_0\).)