Reading Slope Inference in a Regression Table

From a Regression Table to an Inference

Regression software often compresses several important results into a small coefficient table. A row may contain an estimated coefficient, its standard error, a t statistic, and a p-value. Reading that row correctly is more than matching column names: the analyst must identify the predictor’s slope, know which null hypothesis the reported p-value tests, and explain the result in the units and context of the variables.

This tutorial focuses on inference for the slope in a simple linear regression, which has one quantitative predictor. The slope describes the change in the model’s mean response associated with a one-unit increase in the predictor. Inference uses the slope estimate’s sampling uncertainty to assess a specified claim about the population slope.

Model and conditions: For observations \((x_i,y_i)\), the simple linear regression model is \(Y_i=\beta_0+\beta_1x_i+\epsilon_i\). Here, \(\beta_0\) is the population intercept, \(\beta_1\) is the population slope, and \(\epsilon_i\) is the error for observation \(i\). For the usual exact t inference, the errors are independent, have mean zero, have a common variance \(\sigma^2\), and are normally distributed. The predictor values must vary. In a sample-based study, the observations must also be obtained in a way that supports the intended inference. The t procedure is appropriate only when these conditions are reasonable.

A coefficient table commonly has a row for the intercept and a row for the predictor. The predictor’s coefficient is the estimated slope, written \(\hat{\beta}_1\). Its standard error, \(SE(\hat{\beta}_1)\), estimates the standard deviation of the sampling distribution of that estimate. Smaller standard errors indicate less sampling variability, all else equal. The estimate and its standard error have the same units: response units per predictor unit.

Typical table heading	Meaning for the predictor row
Estimate, Coefficient, or Estimate	The fitted slope \(\hat{\beta}_1\)
Std. Error	The estimated standard error \(SE(\hat{\beta}_1)\)
t value or t statistic	The test statistic for the null slope used by the software
Pr(>\|t\|), P-value, or p-value	The probability, under that null hypothesis and model conditions, of a t statistic at least as extreme in the indicated direction(s)
Residual degrees of freedom or df	For simple linear regression, \(n-2\)

The intercept is a different parameter from the slope. It estimates the mean response when the predictor is zero, if that value is meaningful in the setting. A test or p-value in the intercept row does not answer a question about the predictor’s slope. To make a claim about the predictor, use the predictor’s row.

The Test Statistic Behind the Table

For a test of a specified population slope \(\beta_{1,0}\), the null and alternative hypotheses might be \(H_0:\beta_1=\beta_{1,0}\) and \(H_a:\beta_1\ne\beta_{1,0}\). The test statistic measures how far the observed slope estimate is from the null value, in estimated standard errors:

t=\frac{\hat{\beta}_1-\beta_{1,0}}{SE(\hat{\beta}_1)},\qquad df=n-2.

Under the null hypothesis and the model conditions, this statistic follows a t distribution with \(n-2\) degrees of freedom. The subtraction in the numerator matters: a t statistic depends on the null slope being tested, not just on the coefficient and its standard error. Most standard coefficient tables test \(H_0:\beta_1=0\). For that test, the numerator simplifies to \(\hat{\beta}_1\), and the table’s t statistic is \(\hat{\beta}_1/SE(\hat{\beta}_1)\).

For a two-sided alternative, the p-value is the probability, assuming \(H_0\) and the model conditions, of a t statistic with absolute value at least as large as the observed one. In symbols, it is \(P(|T|\ge |t|)\) for a t random variable \(T\) with the appropriate degrees of freedom. A small p-value indicates that the observed statistic would be unusual under the null; it does not give the probability that the null hypothesis is true.

Reading rule: Before using a table’s p-value, identify the null hypothesis it tests. If the question’s null is \(\beta_1=0\), the usual slope-row p-value applies. If the question specifies a different null slope, calculate a new t statistic and p-value using that null value.

Worked Examples

Worked Example: Reading a Test of a Zero Slope

A hypothetical analysis examines whether weekly practice time predicts a performance score. The response is measured in points and practice time in hours per week. There are \(n=24\) independent observations. The simple linear model’s conditions are stipulated to be reasonable for this example. Software reports:

Term	Estimate	Std. Error	t value	Pr(>\|t\|)
Intercept	51.20	4.10	12.49	<0.0001
Practice hours	1.75	0.50	3.50	0.0020

The estimate in the practice-hours row is the slope, \(\hat{\beta}_1=1.75\) points per additional hour of weekly practice. The standard error is \(0.50\) points per hour. The table’s t statistic checks arithmetically:

t=\frac{1.75-0}{0.50}=3.50.

The degrees of freedom are \(n-2=24-2=22\). The reported p-value, about \(0.0020\), is for the usual two-sided test \(H_0:\beta_1=0\) versus \(H_a:\beta_1\ne 0\). At significance level \(\alpha=0.05\), \(0.0020<0.05\), so reject \(H_0\). There is evidence that mean performance score is associated with weekly practice time in this population, under the model conditions. The estimated direction is positive: the fitted mean score increases by 1.75 points per additional hour of practice. This observational result alone does not establish that practice causes the increase.

Worked Example: Testing a Nonzero Null Slope

In a second hypothetical setting, a regression relates a treatment-duration measure to a laboratory response. The predictor is measured in hours and the response in units. The data consist of \(n=18\) independent observations, and the conditions for the simple-regression t procedure are stipulated to be reasonable. The software gives the following predictor row:

Term	Estimate	Std. Error	t value	Pr(>\|t\|)
Duration	0.80	0.20	4.00	0.0010

The displayed t statistic and p-value test \(H_0:\beta_1=0\), since \(0.80/0.20=4.00\). Suppose instead the scientific question is whether the population slope equals \(1.00\) response unit per hour. For \(H_0:\beta_1=1.00\) against \(H_a:\beta_1\ne1.00\), the zero-slope p-value is not the answer. Recalculate the statistic:

t=\frac{0.80-1.00}{0.20}=-1.00,\qquad df=18-2=16.

The two-sided p-value for \(t=-1.00\) with 16 degrees of freedom is approximately \(0.3321\). Because \(0.3321>0.05\), do not reject \(H_0\) at the 5% significance level. The data do not provide sufficient evidence that the population slope differs from 1.00 response unit per hour. This is not proof that the slope equals 1.00; it is a statement about the evidence in this sample against that specified value.

Worked Example: Reporting the Corrected Table Result

A third hypothetical analysis relates a measured exposure to a continuous outcome. There are \(n=20\) independent observations, and the simple linear model conditions are assumed reasonable. The software reports a slope estimate of \(2.40\) outcome units per exposure unit and a standard error of \(0.80\) outcome units per exposure unit. The question is whether the population slope is \(0.80\) outcome units per exposure unit.

The hypotheses are \(H_0:\beta_1=0.80\) and \(H_a:\beta_1\ne0.80\). Use the reported estimate and standard error, but substitute the specified null slope:

t=\frac{2.40-0.80}{0.80}=2.00,\qquad df=20-2=18.

For a two-sided test with \(t=2.00\) and 18 degrees of freedom, the p-value is \(0.06082\). Since \(0.06082>0.05\), do not reject \(H_0\) at \(\alpha=0.05\). The sample does not provide sufficient evidence that the population slope differs from \(0.80\) outcome units per exposure unit. The estimated slope is \(2.40\), but the test’s conclusion concerns the specified null value and the uncertainty in that estimate.

If a table reports a p-value for a zero-slope test, that displayed value would answer a different question. For this nonzero null, the result must come from the statistic calculated above and its t distribution.

Why the Degrees of Freedom and Formula Fit

In simple linear regression, the fitted line estimates two parameters from the data: an intercept and a slope. The residual degrees of freedom are therefore \(n-2\). The table’s slope standard error quantifies the estimated sampling variability of \(\hat{\beta}_1\); comparing the estimate with a null value in units of that standard error produces the t statistic. Under the stated model conditions, this standardized statistic has a t distribution with those residual degrees of freedom when the null is true.

This also explains why coefficient output often includes one t value and p-value for every row: each row corresponds to a parameter and its own test. The predictor row tests a slope claim; the intercept row tests an intercept claim. The columns do not automatically tell you whether the test answers the question you intended to ask. That depends on the null hypothesis encoded by the software or specified in the analysis.

Common Misreadings and a Reporting Pattern

A frequent mistake is to see a small p-value in a slope row and report that the slope is “significant” without stating what value it was tested against. In a standard table, that p-value usually tests zero slope. It cannot be used as the p-value for a test of a different slope value. Another mistake is to confuse the estimated slope with its standard error: the estimate describes the fitted change in the response per predictor unit, while the standard error describes uncertainty in that estimate.

A clear report identifies the predictor and response, gives the estimated slope with units, names the null and alternative hypotheses, and reports the test statistic, degrees of freedom, and p-value. It then states the decision at the chosen \(\alpha\) and gives a contextual conclusion. For example:

Reporting pattern: “The estimated slope relating [predictor] to [response] was [estimate] [response units] per [predictor unit], with standard error [SE]. For \(H_0:\beta_1=[\text{null value}]\) versus \(H_a:\beta_1\ne[\text{null value}]\), \(t=[t]\), \(df=[df]\), and \(p=[p]\). At \(\alpha=[\alpha]\), [decision]. The data [do/do not] provide sufficient evidence that the population slope differs from [null value] in this context.”

The model conditions matter to the interpretation: a table’s numerical output does not guarantee that the t-based inference is appropriate. If independence, linearity of the mean, constant error variance, or the normal-error condition needed for exact small-sample inference is seriously unreasonable, the reported t reference distribution may not describe the procedure accurately. Finally, a slope association is not automatically a causal effect; a causal conclusion requires an appropriate study design and additional justification.

Key takeaway: Find the predictor’s row, interpret its coefficient and standard error with units, identify the null slope tested, and verify the t statistic, \(df=n-2\), and p-value. A complete conclusion reports what the evidence says about that specific population-slope claim.

Check Your Understanding

Use the coefficient-table ideas to answer these questions.

A predictor row shows Estimate \(=-0.60\), Std. Error \(=0.15\), \(t=-4.00\), and \(df=12\). What null slope does the usual table p-value test?
In a simple linear regression with \(n=31\), what degrees of freedom should be used for the slope t test?
A table reports \(p=0.01\) for testing a zero slope. Can this p-value be used unchanged to test whether the population slope equals 2? Explain.
In a test of \(H_0:\beta_1=1.5\), the estimate is \(1.9\) and its standard error is \(0.20\). Calculate the t statistic and state the degrees of freedom if \(n=16\).
Why is it incorrect to say that a p-value of \(0.04\) means there is a 4% probability that the null slope is true?