From a Slope to a Change
A regression slope describes how the fitted response changes as the predictor changes. The predictor change need not be one unit: if a slope is reported per hour, for example, a change of three hours corresponds to three times that slope. Keeping track of the size and direction of the predictor change makes the calculation straightforward and prevents common sign and unit errors.
This tutorial focuses on differences between two fitted responses, not on finding either fitted response separately. The key operation is to multiply the estimated slope by the change in the predictor, taking care to subtract the starting predictor value from the ending value.
The Fitted Difference Formula
For a simple linear regression, write the fitted line as \(\hat{y}=b_0+b_1x\), where \(x\) is the predictor, \(\hat{y}\) is the fitted response, \(b_0\) is the intercept, and \(b_1\) is the estimated slope. The slope gives the change in fitted response for a one-unit increase in \(x\), with the other features of this simple fitted line held fixed.
The predictor change is signed: \(x_2-x_1\) is positive for an increase and negative for a decrease. The response difference uses the same order, ending value minus starting value. Therefore, a positive fitted difference means the fitted response is higher at the ending predictor value; a negative fitted difference means it is lower.
The multiplication also carries units. If \(b_1\) is measured in response units per predictor unit and \(x_2-x_1\) is measured in predictor units, the predictor units cancel, leaving response units. Always identify those units before interpreting the result.
Why the Formula Works
The formula follows directly by subtracting the two fitted responses. Apply the same fitted line at both predictor values:
The intercept cancels because it contributes equally to both fitted responses. Thus, for a fixed predictor change, the fitted difference does not depend on the starting predictor value. This result assumes that both predictor values are being compared on the same fitted straight line. The line represents a constant slope over the interval; it does not establish that the real relationship is exactly linear or that a change in the predictor causes a change in the response. For interpretation beyond the algebra, consider whether the interval is appropriate for the fitted line and the data.
Worked Example: An Increase of Several Predictor Units
Suppose a hypothetical fitted line relates study time, \(x\) in hours, to a fitted exam score, \(\hat{y}\) in points. Its estimated slope is \(b_1=4.6\) points per hour. What is the fitted score difference for an increase in study time from 5.0 hours to 7.5 hours?
First find the signed predictor change:
Multiply this change by the estimated slope:
According to this fitted line, the fitted exam score at 7.5 hours is 11.5 points higher than the fitted score at 5.0 hours. The result is a difference between fitted scores, not a claim that every student who studies 2.5 more hours will gain 11.5 points. It also does not imply that study time caused a score difference.
Decreases and the Sign of the Difference
A decrease in the predictor must be entered as a negative change when moving from the starting value to the ending value. It is not enough to calculate the size of the decrease: the sign determines the direction of the fitted response difference. If both the slope and the predictor change are negative, their product is positive.
Worked Example: A Predictor Decrease With a Negative Slope
Consider a hypothetical fitted line relating outdoor temperature, \(x\) in degrees Celsius, to daily heating use, \(\hat{y}\) in kilowatt-hours (kWh). Suppose its estimated slope is \(b_1=-1.7\) kWh per degree Celsius. Find the fitted heating-use difference when temperature decreases from \(8^\circ\text{C}\) to \(4^\circ\text{C}\).
The signed predictor change is ending temperature minus starting temperature:
Then calculate the fitted response difference:
The fitted heating use at \(4^\circ\text{C}\) is 6.8 kWh higher than at \(8^\circ\text{C}\), according to this line. The two negative signs produce a positive difference. The interpretation is about the fitted relationship, not a guarantee about actual heating use on particular days.
A reliable way to manage signs is to write the requested comparison in words first: “fitted response at the ending predictor value minus fitted response at the starting predictor value.” Then calculate \(x_2-x_1\) in that same order. Do not reverse the predictor subtraction while keeping the response comparison unchanged.
Several Units and Unit Conversion
A slope is always tied to the units used to measure the predictor. If a slope is given per pound and the change is stated in ounces, convert the change to pounds before multiplying—or convert the slope consistently. The result should have response units, not a mixture of predictor and response units.
Worked Example: A Multi-Unit Change With a Unit Check
Suppose a hypothetical fitted line describes shipping fee, \(\hat{y}\) in dollars, using package weight, \(x\) in pounds. Its estimated slope is \(b_1=\$0.82\) per pound. What is the fitted fee difference for an increase in package weight from 6.4 pounds to 10.15 pounds?
The predictor change is:
Multiply the change by the slope:
The fitted fee is therefore \$3.075 higher at 10.15 pounds than at 6.4 pounds. If fees are reported to the nearest cent, that difference is about \$3.08. The unrounded calculation is shown first so the rounding is transparent.
The same calculation works for a change spanning any number of predictor units, including a fraction of a unit. For instance, a change of \(0.25\) units corresponds to one quarter of the slope; a change of 10 units corresponds to 10 times the slope. The multiplication is valid algebraically, although whether a particular interval is suitable for using the fitted line depends on the data and the context.
What the Calculation Does—and Does Not—Say
The fitted difference is a comparison between two values on the estimated line. It is not necessarily the difference between two observed responses. Individual observations may lie above or below the line, so their actual difference can differ from the fitted difference. Nor does the calculation, by itself, describe uncertainty in the estimated slope or fitted difference.
A fitted difference is also not automatically a causal effect. A regression line summarizes an estimated association between predictor and response. Whether changing the predictor would produce a response change requires additional justification about the study design and possible alternative explanations.
This calculation is useful because it expresses a slope in a change that matters in context. A slope reported per one predictor unit may seem small or awkward, while a difference over a realistic multi-unit interval can be easier to interpret. State the predictor interval, show the signed change, and report the fitted response difference with its units.
Common Mistakes
- Using only the size of a decrease. A decrease from 8 to 4 is a predictor change of \(4-8=-4\), not \(+4\), when the comparison is ending minus starting.
- Multiplying by the slope without checking units. A slope in dollars per pound must be paired with a weight change in pounds, or the units must first be converted consistently.
- Reversing the comparison. If the question asks for the fitted response at \(x_2\) minus that at \(x_1\), use \(x_2-x_1\). Reversing the order reverses the sign of the difference.
- Treating a fitted difference as an observed or causal difference. The product \(b_1(x_2-x_1)\) describes a contrast on the fitted line. It does not guarantee the same change for particular observations or establish causation.
- Assuming the starting value affects the result. For a fixed predictor change on the same straight line, the intercept cancels, so the fitted difference depends on the slope and the change—not on where the interval begins.
Check Your Understanding
For each question, keep the direction of the comparison and the units explicit.
- A fitted line has slope 3.2 points per hour. What is the fitted response difference for an increase of 4 hours in the predictor?
- A fitted line has slope \(-0.6\) liters per degree Celsius. Calculate and interpret the fitted response difference when the predictor decreases by 5 degrees Celsius.
- For a slope of \$1.25 per kilogram, what predictor change should be used when comparing 12 kg with 9.5 kg, taking the second value as the ending value?
- Why does the intercept not appear in the formula for a fitted response difference?
- Does a fitted difference calculated from a regression slope establish that changing the predictor causes that difference in the response? Explain.