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Least-Squares Regression Lines · Tutorial 266 of 996

Comparing Predicted Responses Across Two Values

Use the fitted slope and the difference between two explanatory values to compare the responses predicted by a regression line.

Beginner 9 min read

What You'll Learn

  • Calculate two fitted responses using a regression equation.
  • Find their difference directly from the fitted slope and the change in the explanatory value.
  • Interpret the sign, size, and units of a predicted-response difference.
  • Explain why this difference is not an individual prediction or a guaranteed individual change.
  • Distinguish the slope-based comparison from comparing observed responses.

Comparing Two Predictions

A fitted regression line can provide a predicted response at a chosen explanatory value. Sometimes the question is not simply “What response does the line predict at this value?” but “How different are the responses predicted at these two values?” The slope gives a direct answer: multiply the change in the explanatory value by the fitted slope.

This comparison concerns two predictions from the same fitted line. It describes how the line’s predicted response changes as the explanatory value changes. It does not say that a particular individual’s response will change by that amount, nor does it compare two observed responses.

The Fitted Difference Formula

Write the fitted regression line as \(\hat{y}=b_0+b_1x\). Here, \(x\) is a value of the explanatory variable, \(\hat{y}\) is the response predicted by the line at that value, \(b_0\) is the fitted intercept, and \(b_1\) is the fitted slope. The slope gives the line’s predicted response change for a one-unit increase in \(x\), measured in response units per explanatory-variable unit.

Definition: For two explanatory values \(x_1\) and \(x_2\), the fitted difference in predicted responses is \(\hat{y}(x_2)-\hat{y}(x_1)\). For a fitted line \(\hat{y}=b_0+b_1x\), this difference equals \(b_1(x_2-x_1)\). Its units are response units.

The order matters. The expression \(\hat{y}(x_2)-\hat{y}(x_1)\) compares the prediction at \(x_2\) with the prediction at \(x_1\). If the slope is positive and \(x_2>x_1\), the difference is positive. If the slope is negative and \(x_2>x_1\), it is negative. Reversing the order of the two explanatory values reverses the sign.

You may calculate both fitted responses and subtract, or use the slope formula directly. The direct method is often quicker; calculating both predictions can make the comparison easier to check and explain.

Why the Intercept Cancels

The direct formula follows from substituting each explanatory value into the same fitted line and subtracting. This is an algebraic property of the line, not a separate statistical assumption.

$$ \begin{aligned} \hat{y}(x_2)-\hat{y}(x_1) &=(b_0+b_1x_2)-(b_0+b_1x_1)\\ &=b_0-b_0+b_1x_2-b_1x_1\\ &=b_1(x_2-x_1). \end{aligned} $$

The intercept cancels because the same fitted line is used for both predictions. Thus, to find their difference, you do not need to know the intercept. The slope and the difference between the two explanatory values are enough. If the two explanatory values are equal, their predicted responses are equal and the difference is zero.

This result applies to any two values substituted into the same fitted linear equation. It does not itself establish that the line is an appropriate description of the relationship in a particular application. The calculations below take the fitted line as given; their interpretations are specifically about the predictions made by that line.

Worked Example: Study Time and Predicted Quiz Score

Suppose a fitted line predicts quiz score, in points, from study time, in hours:

$$ \hat{y}=42+1.8x. $$

Compare the predicted scores for a student who studies 12 hours and one who studies 17 hours. Treat these as two inputs to the same fitted line.

Calculate the explanatory-value difference. Taking \(x_1=12\) and \(x_2=17\), the change in study time is \(17-12=5\) hours.

Use the slope. The fitted slope is \(1.8\) quiz points per study hour, so the difference in predicted scores is:

$$ b_1(x_2-x_1)=1.8(17-12)=1.8(5)=9.0\text{ points}. $$

Check by finding both predictions. At 12 hours, \(\hat{y}(12)=42+1.8(12)=42+21.6=63.6\) points. At 17 hours, \(\hat{y}(17)=42+1.8(17)=42+30.6=72.6\) points. Subtracting gives \(72.6-63.6=9.0\) points, as before.

Interpretation. According to this fitted line, the predicted quiz score at 17 hours of study is 9.0 points higher than the predicted score at 12 hours. This is a difference between two fitted predictions. It does not say that a particular student will gain 9.0 points by studying five additional hours.

Direction, Units, and Meaning

A slope-based comparison should be reported with its direction and units. The sign tells whether the prediction at the second value is higher or lower than the prediction at the first. The magnitude gives the size of the difference in response units. Keep the explanatory-variable units in the multiplication as well: the slope’s units cancel the explanatory units, leaving response units.

FeatureWhat to report
OrderState which explanatory value is being compared with which.
SignSay whether the second fitted prediction is higher or lower.
MagnitudeGive the numerical difference in response units.
ScopeIdentify the result as a difference between predictions from the fitted line.

For example, if a fitted slope is \(-0.6\) kilograms per week and \(x\) is measured in weeks, a positive increase of 4 weeks gives a fitted difference of \((-0.6)(4)=-2.4\) kilograms. The negative sign means the predicted response at the later value is 2.4 kilograms lower. The line does not guarantee that every individual’s response decreases by that amount.

Worked Example: A Negative Slope

Suppose a fitted line predicts the number of minutes required to complete a task from the number of practice sessions:

$$ \hat{y}=38-0.7x. $$

Compare the predicted completion times at 10 and 16 practice sessions. The response is measured in minutes, and the explanatory value is a number of sessions.

The second explanatory value is 6 sessions greater than the first: \(16-10=6\). The fitted slope is \(-0.7\) minutes per session. Therefore:

$$ \hat{y}(16)-\hat{y}(10)=(-0.7)(16-10)=(-0.7)(6)=-4.2\text{ minutes}. $$

Check using the predictions: \(\hat{y}(10)=38-0.7(10)=38-7=31\) minutes, and \(\hat{y}(16)=38-0.7(16)=38-11.2=26.8\) minutes. Their difference is \(26.8-31=-4.2\) minutes.

According to this line, the predicted completion time at 16 practice sessions is 4.2 minutes lower than at 10 sessions. The minus sign is meaningful: it gives the direction for the stated subtraction order. This is a fitted comparison, not a claim that every person who practices six more times will finish 4.2 minutes faster.

Changing the Explanatory-Value Units

The slope’s numerical value depends on the units used for the explanatory variable, but the predicted-response difference does not change when the same explanatory values are correctly converted. The slope and the size of the explanatory-value change adjust in opposite ways.

Worked Example: Converting Hours to Minutes

A fitted line predicts a response in points from elapsed time in minutes, with slope \(0.4\) points per minute. Compare the fitted predictions at 30 minutes and 45 minutes.

The time difference is \(45-30=15\) minutes, so the predicted-response difference is:

$$ 0.4\text{ points per minute}\times15\text{ minutes} =6.0\text{ points}. $$

Now express time in hours. The two times are \(0.5\) hours and \(0.75\) hours, a difference of \(0.25\) hours. Since one hour contains 60 minutes, the slope becomes \(0.4(60)=24\) points per hour. Using the converted values gives:

$$ 24\text{ points per hour}\times0.25\text{ hours} =6.0\text{ points}. $$

Both calculations describe the same comparison: the fitted prediction at 45 minutes is 6.0 points higher than the prediction at 30 minutes. The units cancel to leave points in either calculation.

Difference Between a Prediction and a Predicted Difference

A prediction answers a question about one explanatory value: “What response does the line predict at \(x=17\)?” A predicted difference answers a question about two values: “How far apart are the line’s predictions at \(x=17\) and \(x=12\)?” These are different quantities, even though both come from the same equation.

The predicted difference also differs from a difference between observed responses. The fitted line supplies \(\hat{y}(x_1)\) and \(\hat{y}(x_2)\), not the actual observed outcomes for two individuals. Actual outcomes can differ from the fitted predictions. The comparison formula does not remove that individual variation or state what either person’s observed response will be.

Nor does a positive or negative slope, by itself, show that changing \(x\) causes a change in \(y\). The calculation describes the fitted relationship between the explanatory and response variables. A causal claim requires support beyond the arithmetic of a regression line.

Common Mistake: Reporting the Difference as an Individual Change

A common error is to calculate a predicted difference correctly and then say, “An individual who increases \(x\) by five units will change their response by this amount.” That conclusion treats the fitted comparison as a guaranteed within-person outcome. The line only gives two predicted responses; it does not guarantee that either observed response equals its prediction.

A better interpretation names both values and keeps the result attached to the line: “The fitted line predicts a response that is 9.0 points higher at 17 hours than at 12 hours.” Also check the subtraction order and the units before writing the sentence. If the calculated difference is negative, state that the second prediction is lower, rather than dropping the sign and leaving the direction unclear.

Key takeaway: For a fitted line \(\hat{y}=b_0+b_1x\), the difference between predictions at \(x_2\) and \(x_1\) is \(b_1(x_2-x_1)\). Interpret it as a signed difference between fitted responses, in response units—not as a guaranteed change for an individual.

Check Your Understanding

Use the fitted-line interpretation carefully when answering each question.

  1. A fitted line has slope \(2.5\) kilograms per month. What is the fitted difference between the predictions at 4 months and 10 months? State the direction and units.
  2. A line predicts response in seconds and has slope \(-1.2\) seconds per unit of \(x\). Find and interpret the difference \(\hat{y}(7)-\hat{y}(2)\).
  3. Explain why the intercept is not needed to calculate the difference between two predictions from the same fitted line.
  4. A fitted difference is \(-3.5\) points for \(\hat{y}(x_2)-\hat{y}(x_1)\). What does the negative sign say about the two predictions?
  5. Why would it be incorrect to conclude from a predicted difference that each individual’s observed response will change by exactly that amount?