Testing The Regression Intercept

Why Test a Regression Intercept?

A regression intercept describes the model’s mean response when the predictor equals zero. A test of the intercept can therefore address a specific baseline question: for example, whether a hypothetical response is zero at zero exposure, or whether a model’s baseline level equals a specified benchmark. The test uses the estimated intercept’s own standard error, not the standard error of the sample mean response.

The interpretation depends on the predictor’s scale and origin. If zero is meaningful and within the relevant range, the intercept may have a direct practical interpretation. If zero is outside the observed range, the test still has a mathematical definition under the model, but its practical interpretation may rely on extrapolation.

Model, Assumptions, and Test Statistic

Model: For fixed predictor values \(x_1,\ldots,x_n\), suppose \(Y_i=\beta_0+\beta_1x_i+\varepsilon_i\), where the errors are independent and \( \varepsilon_i\sim N(0,\sigma^2)\). Thus the errors have mean zero, a common variance, and a normal distribution. The usual exact t test requires these assumptions and at least two distinct predictor values, so \(S_{xx}>0\).

Write \(b_0\) and \(b_1\) for the least-squares estimates of the intercept and slope. Define \(\bar{x}=n^{-1}\sum_{i=1}^n x_i\) and \(S_{xx}=\sum_{i=1}^n(x_i-\bar{x})^2\). Let \(\operatorname{SSE}=\sum_{i=1}^n(Y_i-b_0-b_1x_i)^2\), and estimate the error variance by \(s^2=\operatorname{SSE}/(n-2)\). The estimated standard error of the intercept is

\operatorname{SE}(b_0) =s\sqrt{\frac{1}{n}+\frac{\bar{x}^{\,2}}{S_{xx}}}.

To test a specified value \(\beta_{00}\), use \(H_0:\beta_0=\beta_{00}\). A two-sided alternative is \(H_A:\beta_0\ne\beta_{00}\); one-sided alternatives are \(H_A:\beta_0>\beta_{00}\) or \(H_A:\beta_0<\beta_{00}\). The test statistic and its null reference distribution are

t_{\mathrm{obs}} =\frac{b_0-\beta_{00}}{\operatorname{SE}(b_0)} \quad\text{and, under }H_0,\quad T\sim t_{n-2}.

The degrees of freedom are \(n-2\) because the residual variance is estimated after fitting two regression coefficients. The \(t\) reference distribution is exact under the stated normal-error model. Without normal errors, this exact finite-sample result need not hold; a large-sample approximation requires separate justification.

Why the Intercept Standard Error Has This Form

The least-squares intercept can be written as \(b_0=\bar{Y}-\bar{x}b_1\). Under the model assumptions, \(\operatorname{Var}(\bar{Y})=\sigma^2/n\) and \(\operatorname{Var}(b_1)=\sigma^2/S_{xx}\). The covariance between these estimates is zero: the error contribution to \(b_1\) is proportional to \(\sum (x_i-\bar{x})\varepsilon_i\), and the centered predictor values sum to zero.

Consequently, the variance of the intercept estimate is the variance of \(\bar{Y}-\bar{x}b_1\), with no covariance term:

\begin{aligned} \operatorname{Var}(b_0) &=\operatorname{Var}(\bar{Y})+\bar{x}^{\,2}\operatorname{Var}(b_1)\\ &=\sigma^2\left(\frac{1}{n}+\frac{\bar{x}^{\,2}}{S_{xx}}\right). \end{aligned}

Replacing \(\sigma\) by its estimate \(s\) gives the standard error formula. Under the normal-error model, \(b_0\) is normally distributed, and the residual variance estimate supplies the independent \(t\)-distributed scaling. This is why the statistic, when the null value is correct, has a \(t\) distribution with \(n-2\) degrees of freedom.

A Reliable Testing Procedure

Define the null and alternative.
State the specified intercept value and whether the question calls for a two-sided or one-sided test.

Calculate the intercept standard error.
Use the sample size, predictor mean, \(S_{xx}\), and residual standard error in the intercept formula.

Calculate the test statistic.
Subtract the hypothesized value from \(b_0\), then divide by \(\operatorname{SE}(b_0)\).

Use the correct tail and degrees of freedom.
Compare with the appropriate \(t_{n-2}\) critical value, or use the matching tail probability. State the decision in context.

Worked Tests of an Intercept

Worked Example: A Two-Sided Baseline Test

In a hypothetical study, \(Y\) is a response measured in dollars and \(x\) is an exposure index. Assume the fixed-predictor normal-error model above. For \(n=10\), suppose \(\bar{x}=2\), \(S_{xx}=30\), \(b_0=8\), and \(s=2\). Test \(H_0:\beta_0=5\) against \(H_A:\beta_0\ne5\) at the 5% significance level.

Method. Calculate the estimated standard error, form the two-sided intercept t statistic, and compare its absolute value with the \(t\) critical value using \(10-2=8\) degrees of freedom.

First calculate the standard error:

\begin{aligned} \operatorname{SE}(b_0) &=2\sqrt{\frac{1}{10}+\frac{2^2}{30}}\\ &=2\sqrt{\frac{3}{30}+\frac{4}{30}}\\ &=2\sqrt{\frac{7}{30}} \approx 0.9661. \end{aligned}

The observed statistic is

t_{\mathrm{obs}}=\frac{8-5}{0.9660918\ldots} \approx 3.105.

For a two-sided test at 5% with 8 degrees of freedom, the critical value is \(t_{0.975,8}\approx2.306\). Since \(|3.105|>2.306\), reject \(H_0\) at the 5% level. The data provide evidence that the population mean response at \(x=0\), namely \(\beta_0\), differs from 5 dollars. The conclusion is about the model’s mean at zero exposure, not about the mean response over all observed exposure values.

Common mistake avoided. The standard error is not \(s/\sqrt{n}=2/\sqrt{10}\approx0.6325\). That is the standard error of a sample mean in a different setting. Substituting it here would give \(3/0.6325\approx4.743\), not the correct intercept statistic, because the intercept’s uncertainty also depends on the predictor’s location and spread.

Worked Example: A One-Sided Test

A hypothetical regression relates a monthly service measure \(Y\) to a standardized workload index \(x\). Assume the fixed-predictor normal-error model. The fitted regression has \(n=10\), \(\bar{x}=1\), \(S_{xx}=15\), \(b_0=4.5\), and \(s=1.5\). Test whether the population intercept exceeds 4, using a 5% significance level.

Method. The hypotheses are \(H_0:\beta_0=4\) and \(H_A:\beta_0>4\), so this is an upper-tail test. The degrees of freedom are \(10-2=8\).

Calculate the standard error:

\begin{aligned} \operatorname{SE}(b_0) &=1.5\sqrt{\frac{1}{10}+\frac{1}{15}}\\ &=1.5\sqrt{\frac{1}{6}}\\ &\approx 0.6124. \end{aligned}

Then

t_{\mathrm{obs}}=\frac{4.5-4}{0.6123724\ldots} \approx0.8165.

The upper-tail critical value is \(t_{0.95,8}\approx1.860\). Because \(0.8165<1.860\), do not reject \(H_0\). At the 5% level, the evidence is insufficient to conclude that the population response at workload index zero exceeds 4. A positive estimate alone does not determine the test decision; its size must be assessed relative to its standard error and the chosen alternative.

Common mistake avoided. The one-sided alternative requires an upper-tail comparison. Using a two-sided critical value would answer a different testing question.

Worked Example: The Predictor Origin Changes the Meaning

In a hypothetical model, \(Y\) is a response measured in thousands of dollars and \(x\) is an exposure index. Assume the stated fixed-predictor normal-error model. Suppose \(n=11\), \(\bar{x}=3\), \(S_{xx}=22\), \(b_0=14\), \(b_1=1.5\), and \(s=2\). Test \(H_0:\beta_0=12\) against a two-sided alternative at the 5% level.

Method. Test the original intercept, which is the population mean response at \(x=0\). There are \(11-2=9\) degrees of freedom.

The standard error simplifies exactly:

\begin{aligned} \operatorname{SE}(b_0) &=2\sqrt{\frac{1}{11}+\frac{3^2}{22}}\\ &=2\sqrt{\frac{2}{22}+\frac{9}{22}}\\ &=2\sqrt{\frac{1}{2}}=\sqrt{2} \approx1.4142. \end{aligned}

Thus

t_{\mathrm{obs}}=\frac{14-12}{\sqrt{2}} =\sqrt{2} \approx1.4142.

The two-sided 5% critical value is \(t_{0.975,9}\approx2.262\). Since \(1.4142<2.262\), do not reject \(H_0\). The test does not establish that the intercept differs from 12 thousand dollars.

Here the fitted mean at \(x=3\) is \(b_0+3b_1=14+3(1.5)=18.5\) thousand dollars. If the predictor is recentered as \(x^*=x-3\), the new intercept estimate is 18.5, and it represents the mean at the original \(x=3\), not at the original \(x=0\). A test of that recentered intercept would test a different population quantity. Replacing the original intercept with 18.5 while keeping the null value 12 would therefore be a mistake.

Common mistake avoided. The intercept’s meaning depends on where the predictor is zero. Centering may make an intercept more practically interpretable, but it changes the parameter being tested.

Interpretation and Common Exam Traps

A test of \(H_0:\beta_0=\beta_{00}\) concerns the model’s population mean response at predictor value zero. It does not directly test whether the fitted line passes through a particular observed point, whether the slope is zero, or whether the overall regression is significant. Those are distinct questions.

Using the wrong standard error. Use \(s\sqrt{1/n+\bar{x}^{\,2}/S_{xx}}\), not \(s/\sqrt{n}\). The additional term matters whenever \(\bar{x}\ne0\).
Using the wrong degrees of freedom. The residual degrees of freedom are \(n-2\) for a simple linear regression with an intercept and a slope.
Ignoring the direction of the alternative. A one-sided claim uses one tail; a two-sided claim allocates significance to both tails.
Giving the intercept an unsupported practical meaning. If zero is not a meaningful or observed predictor value, describe the intercept as the model’s extrapolated mean at zero rather than as an observed baseline.
Confusing a failure to reject with proof. Not rejecting \(H_0\) means the data do not provide sufficient evidence against it at the chosen level; it does not prove the null value is exactly correct.

The intercept test is aligned with the SOA VEE Mathematical Statistics syllabus because it applies coefficient standard errors and the \(t\) reference distribution to a clearly specified regression parameter. Accurate interpretation requires both the calculation and attention to the predictor’s zero point.

Check Your Understanding

For each problem, assume a fixed-predictor simple linear regression with independent normal errors of common variance unless stated otherwise. Use the appropriate \(t\) reference distribution.

A hypothetical regression has \(n=8\), \(\bar{x}=2\), \(S_{xx}=16\), \(s=2\), and \(b_0=6\). Test \(H_0:\beta_0=5\) against a two-sided alternative at 5%. State the standard error, statistic, degrees of freedom, and decision. (Answer: \(\operatorname{SE}(b_0)\approx1.2247\), \(t\approx0.8165\), 6 degrees of freedom; do not reject, since \(0.8165< t_{0.975,6}\approx2.447\).)
A hypothetical model has \(n=12\), \(\bar{x}=1\), \(S_{xx}=22\), \(s=2\), and \(b_0=9\). Test \(H_0:\beta_0=8\) against \(H_A:\beta_0>8\) at 5%. (Answer: \(\operatorname{SE}(b_0)\approx0.7177\), \(t\approx1.393\), 10 degrees of freedom; do not reject, since \(1.393<t_{0.95,10}\approx1.812\).)
A hypothetical regression has \(n=16\), \(\bar{x}=2\), \(S_{xx}=60\), \(s=2\), \(b_0=4\), and \(b_1=3\). Test \(H_0:\beta_0=0\) against a two-sided alternative at 1%. What population quantity does the null describe? (Answer: \(\operatorname{SE}(b_0)\approx0.7188\), \(t\approx5.565\), 14 degrees of freedom; reject because \(5.565>t_{0.995,14}\approx2.977\). The null states that the mean response at \(x=0\) is zero.)
A hypothetical regression has \(n=9\), \(\bar{x}=0\), \(S_{xx}=18\), \(s=3\), and \(b_0=2\). Test \(H_0:\beta_0=0\) against \(H_A:\beta_0>0\) at 5%. (Answer: \(\operatorname{SE}(b_0)=1\), \(t=2.000\), 7 degrees of freedom; reject because \(2.000>t_{0.95,7}\approx1.895\).)
A hypothetical fit has \(b_0=5\), \(b_1=2\), and predictor mean \(\bar{x}=4\). What is the fitted mean at the predictor average, and does that equal the fitted intercept? (Answer: the fitted mean at \(x=4\) is \(5+2(4)=13\); it is not the intercept, which is 5 and is the fitted mean at \(x=0\).)

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