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Exam SRM · Inference and ANOVA · Tutorial 63 of 926

Tests of Intercepts and Specific Coefficient Values

Use regression estimates and standard errors to test a specified value for an intercept or individual coefficient, then interpret the hypothesis in the model.

Intermediate 10 min read

What You'll Learn

  • State null and alternative hypotheses for an intercept or individual regression coefficient
  • Calculate the t statistic for a specified coefficient value
  • Identify the residual degrees of freedom and conditions for exact t inference
  • Interpret an intercept test at the model’s zero-predictor setting
  • Choose a one-sided or two-sided decision rule and explain its meaning
  • Recognize common errors involving the null value, tail, and coefficient interpretation

Why Test a Specified Coefficient Value?

A regression coefficient test need not ask only whether a coefficient is zero. A decision-maker may need to assess whether an intercept equals a meaningful baseline, or whether a predictor’s effect is larger than a specified benchmark. The calculation is a t test for one coefficient: the hypothesized value determines the numerator, while the coefficient’s standard error measures uncertainty.

The hypotheses must describe a value of the population regression coefficient, not a value of the estimated coefficient. The estimate comes from the sample; the hypothesis is a claim about the model parameter. A test of an intercept also requires care about what the intercept represents: it is the expected response when every predictor is zero, which may or may not be a useful setting in the application.

Definition: In a multiple linear regression with an intercept and \(q\) predictors, let \(\beta_j\) be one specified coefficient and let \(c\) be a hypothesized value. A test of that coefficient has null hypothesis \(H_0:\beta_j=c\). A two-sided alternative is \(H_A:\beta_j\ne c\); one-sided alternatives are \(H_A:\beta_j>c\) and \(H_A:\beta_j<c\).

Model, Assumptions, and Test Statistic

Write the regression model for \(n\) observations as \(Y=X\beta+\epsilon\), where \(X\) contains a column of ones for the intercept and \(q\) predictor columns. For exact t inference, treat the design matrix \(X\) as fixed, or condition on its observed values. Assume the errors are independent and normally distributed with mean zero and common variance \(\sigma^2\). Also require that \(X\) has full column rank and that \(n>q+1\). The latter condition gives positive residual degrees of freedom, so the error variance can be estimated.

Let \(\hat{\beta}_j\) be the least-squares estimate of the coefficient being tested, and let \(SE(\hat{\beta}_j)\) be its estimated standard error. If \(s^2=SSE/(n-q-1)\), where \(SSE\) is the residual sum of squares, the standard error is obtained from the relevant diagonal element of the inverse cross-product matrix:

$$ SE(\hat{\beta}_j)=s\sqrt{\left[(X^\mathsf{T}X)^{-1}\right]_{jj}}. $$

For the null hypothesis \(H_0:\beta_j=c\), calculate:

$$ t=\frac{\hat{\beta}_j-c}{SE(\hat{\beta}_j)}, \qquad df=n-q-1. $$

Under the stated assumptions and when \(H_0\) is true, this statistic follows a t distribution with \(n-q-1\) degrees of freedom. The test is about one coefficient at a time; a claim involving several coefficients together requires a different test.

Why the Statistic Has a t Distribution

The numerator is the estimated coefficient’s distance from the value specified by the null hypothesis, measured in coefficient units. Under the model, the least-squares estimator is normally distributed conditional on \(X\), with variance \(\sigma^2[(X^\mathsf{T}X)^{-1}]_{jj}\). If \(\sigma\) were known, dividing the difference by its standard deviation would give a standard normal statistic under \(H_0\).

In practice, \(\sigma\) is unknown and is replaced by \(s\), calculated from the residuals. Under normal errors, \(SSE/\sigma^2\) has a chi-squared distribution with \(n-q-1\) degrees of freedom and is independent of the standardized coefficient estimate. Dividing a standard normal variable by the square root of an independent chi-squared variable divided by its degrees of freedom gives a t variable. This is why the test uses a t distribution rather than a standard normal distribution.

This exact result depends on the model assumptions. Full column rank alone is not enough: if \(n=q+1\), then \(df=0\), so \(s^2\) and the t statistic are not defined. With non-normal errors, the exact finite-sample t result does not generally hold, although approximate inference may sometimes be justified under additional conditions.

1
State the parameter and value.
Name the intercept or predictor coefficient and write \(H_0:\beta_j=c\), using the value specified in the problem.
2
Choose the alternative.
Use “not equal to,” “greater than,” or “less than” according to the research question. This determines whether the test is two-sided or one-sided.
3
Compute the statistic and degrees of freedom.
Subtract \(c\) from the estimate, divide by its standard error, and use \(n-q-1\) degrees of freedom.
4
Make and interpret the decision.
Compare the statistic with the appropriate critical value, or compare the p-value with the significance level. State the conclusion in the context of the coefficient and model.

Choosing the Test Direction and Interpreting an Intercept

For a two-sided test at significance level \(\alpha\), reject \(H_0\) when \(|t|\) exceeds the t critical value with upper-tail probability \(\alpha/2\). For \(H_A:\beta_j>c\), reject for a sufficiently large positive statistic; for \(H_A:\beta_j<c\), reject for a sufficiently large negative statistic. The direction must come from the question and be set before examining the estimate. A one-sided alternative is not a way to make an observed result more significant.

An intercept test is mathematically the same test with \(j\) referring to the intercept. Its interpretation depends on the predictor values that define zero. If predictors are uncentered, zero may be outside the observed data range or have no practical meaning. If continuous predictors have been centered, the intercept instead describes the expected response at their centering values (and at the reference category for any coded categorical predictors). A test can be computed either way, but its scientific interpretation follows the model’s coding.

Worked Example: Testing a Hypothetical Claim-Cost Intercept

A hypothetical insurer models claim cost, measured in thousands of dollars, using two predictors: vehicle age and an indicator for a particular repair category. For this fitted model, \(n=20\), \(q=2\), \(SSE=153\), the estimated intercept is \(14.0\), and the intercept diagonal element of \((X^\mathsf{T}X)^{-1}\) is \(0.64\). Test \(H_0:\beta_0=8.0\) against a two-sided alternative at \(\alpha=0.05\). Treat the stated regression assumptions as satisfied.

Method and degrees of freedom. This is a t test for one specified coefficient. The residual degrees of freedom are \(20-2-1=17\), which are positive. The estimate of error variance is \(s^2=153/17=9\), so \(s=3\).

Standard error. Using the supplied intercept diagonal element, \(SE(\hat{\beta}_0)=3\sqrt{0.64}=3(0.8)=2.4\).

Test statistic. The estimate is \(14.0\), while the null value is \(8.0\). Therefore, \(t=(14.0-8.0)/2.4=6.0/2.4=2.50\), with 17 degrees of freedom.

Decision and interpretation. The two-sided 5% critical value is approximately \(2.110\). Since \(|2.50|>2.110\), reject \(H_0\). There is evidence that the model intercept differs from 8.0 thousand dollars. In context, the intercept describes expected cost at vehicle age zero and repair-category indicator zero; the result does not by itself establish that this baseline is practically relevant. The common mistake avoided here is treating a significant intercept test as evidence about all fitted claim costs.

Worked Example: Testing a Predictor Against a Nonzero Benchmark

In a hypothetical linear model for annual repair cost, measured in thousands of dollars, one predictor is vehicle age in years. The model has \(n=30\) observations and \(q=3\) predictors. For the age coefficient, the estimate is \(0.44\), \(SSE=104\), and the relevant diagonal element of \((X^\mathsf{T}X)^{-1}\) is \(0.0036\). Test whether the age coefficient exceeds \(0.20\) thousand dollars per year at the 5% significance level. Assume the exact-inference conditions hold.

Hypotheses and degrees of freedom. The specified one-sided hypotheses are \(H_0:\beta_{\text{age}}=0.20\) and \(H_A:\beta_{\text{age}}>0.20\). The residual degrees of freedom are \(30-3-1=26\).

Standard error. First, \(s^2=104/26=4\), so \(s=2\). Then \(SE(\hat{\beta}_{\text{age}})=2\sqrt{0.0036}=2(0.06)=0.12\).

Test statistic and decision. The statistic is \(t=(0.44-0.20)/0.12=0.24/0.12=2.00\). For an upper-tailed test with 26 degrees of freedom, the 5% critical value is approximately \(1.706\). Because \(2.00>1.706\), reject \(H_0\). The data provide evidence that the modeled increase in expected repair cost per additional year of age exceeds \(0.20\) thousand dollars, holding the other predictors fixed. The common mistake avoided is using a two-sided critical value for a prespecified one-sided alternative.

Worked Example: Testing a Centered-Model Intercept

A hypothetical model predicts annual benefit amounts in thousands of dollars from a centered health score and a binary indicator for a coverage option. At the centered score and the reference coverage option, the fitted intercept is \(52.0\). Suppose \(n=16\), \(q=2\), \(SSE=117\), and the intercept diagonal element of \((X^\mathsf{T}X)^{-1}\) is \(0.25\). Test \(H_0:\beta_0=50.0\) against \(H_A:\beta_0\ne 50.0\) at \(\alpha=0.05\).

Degrees of freedom and error standard deviation. The residual degrees of freedom are \(16-2-1=13\). Thus \(s^2=117/13=9\), and \(s=3\).

Standard error and statistic. The intercept standard error is \(3\sqrt{0.25}=3(0.5)=1.5\). The test statistic is \(t=(52.0-50.0)/1.5=2.0/1.5=1.333\) to three decimals.

Decision and interpretation. The two-sided 5% critical value for 13 degrees of freedom is approximately \(2.160\). Since \(|1.333|<2.160\), do not reject \(H_0\). These data do not provide sufficient evidence that the expected benefit at a centered health score and the reference coverage option differs from 50.0 thousand dollars. The common mistake avoided is interpreting this intercept as the expected benefit for a person with health score zero; the score was centered, so the baseline is the centering value.

Common Errors to Avoid

  • Subtracting the estimate from the null in only one chosen direction. The numerator is \(\hat{\beta}_j-c\). Reversing the subtraction reverses the sign; it can change the conclusion for a one-sided test.
  • Testing against zero when the stated value is nonzero. Use the exact benchmark \(c\) in the numerator and in the hypotheses.
  • Choosing a one-sided test after seeing the estimate. The alternative must reflect the question before the result is examined.
  • Using the wrong degrees of freedom. With \(q\) predictors plus an intercept, use \(n-q-1\), and verify that \(n>q+1\).
  • Overstating what an intercept test means. It concerns the expected response at the model’s zero settings, not necessarily an average response across the observed population.
  • Equating failure to reject with proof of equality. A non-significant result means the sample does not provide sufficient evidence against the null at the chosen level; it does not prove the null value is correct.
Key takeaway: For a single intercept or predictor coefficient, test \(H_0:\beta_j=c\) with \(t=(\hat{\beta}_j-c)/SE(\hat{\beta}_j)\) and \(n-q-1\) degrees of freedom. State the alternative, check the assumptions and positive residual degrees of freedom, and interpret the coefficient at the predictor settings defined by the model. This treatment is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each question, use the stated regression assumptions and a 5% significance level unless otherwise specified. Compare with a t critical value and give a contextual decision.

  1. A model has \(n=14\) and one predictor. Its estimated intercept is \(6.8\), with standard error \(1.7\). Test \(H_0:\beta_0=4.25\) against a two-sided alternative. (Answer: \(df=12\), \(t=1.50\); do not reject because the two-sided critical value is about \(2.179\).)
  2. A model has \(n=22\) and three predictors. A coefficient estimate is \(-0.30\), with standard error \(0.10\). Test whether its population coefficient differs from \(-0.50\). (Answer: \(df=18\), \(t=2.00\); do not reject at 5% two-sided because the critical value is about \(2.101\).)
  3. A model has \(n=28\) and two predictors. A coefficient estimate is \(0.75\), with standard error \(0.20\). Test \(H_0:\beta_j=0.35\) against \(H_A:\beta_j>0.35\). (Answer: \(df=25\), \(t=2.00\); reject at 5% one-sided because the critical value is about \(1.708\).)
  4. A model has \(n=12\) and two predictors. Its intercept estimate is \(3.2\), with standard error \(0.8\). Test whether the intercept differs from \(2.0\). (Answer: \(df=9\), \(t=1.50\); do not reject at 5% two-sided because the critical value is about \(2.262\).)
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