Predicting a Slope Change from a Point’s Position
In “Leverage Without Large Residuals,” you saw that a point can be far from the center of the \(x\)-values yet close to the linear pattern. This tutorial focuses on a related question: if a point is added to or removed from a data set, will the slope increase or decrease?
The answer depends on two directions: whether the point is to the left or right of the center of the other \(x\)-values, and whether it is below or above the line fitted to those other observations. A point’s residual tells you the second direction: a positive residual means the point is above the line, and a negative residual means it is below.
The Sign Rule for Adding a Point
Suppose the existing data have \(n\) observations, mean explanatory-variable value \(\bar{x}\), slope \(b\), and sum of squared deviations in \(x\), \(S_{xx}\). Consider adding a point \((x_0,y_0)\). Its horizontal displacement from the existing center is \(x_0-\bar{x}\). Its residual relative to the existing fitted line is \(e=y_0-\hat{y}_0\), where \(\hat{y}_0\) is the line’s predicted response at \(x_0\).
The change in slope has the same sign as the product \((x_0-\bar{x})e\). The full update formula makes the reason explicit: the denominator is positive, so only the numerator determines whether the slope changes up or down.
You do not need to calculate this formula every time. To predict the direction, assign a positive or negative sign to each factor and multiply. For example, a point on the right has \(x_0-\bar{x}>0\); if it is above the line, then \(e>0\), so their product is positive and the slope increases. If that same right-side point is below the line, the product is negative and the slope decreases.
| Point’s position relative to existing data | Effect when added |
|---|---|
| Right of \(\bar{x}\), above the line | Slope increases |
| Right of \(\bar{x}\), below the line | Slope decreases |
| Left of \(\bar{x}\), above the line | Slope decreases |
| Left of \(\bar{x}\), below the line | Slope increases |
“Increases” and “decreases” refer to the numerical value of the slope, not automatically to whether the trend becomes steeper. For a positive slope, increasing the slope makes it steeper. For a negative slope, increasing the slope can make it less negative and therefore flatter. Keep the algebraic direction separate from the visual description.
Predicting the Effect of Removing a Point
For removal, first be clear about which line you are using. If you have the regression line for the data without the point, use that line as the starting fit and treat the point as one being added. The effect of removing it is the opposite of the effect of adding it to the reduced data.
If instead you are given the line fitted to the full data set, the point’s residual must be measured from that full-data line. In that case, removing the point changes the slope in the opposite direction from the sign of \((x_0-\bar{x}_{\mathrm{full}})e_{\mathrm{full}}\). This is why it matters to identify the data set used for the line and the mean \(x\)-value before deciding what removal will do.
Worked Examples: Predicting Slope Direction
Worked Example: A Right-Side Point Above a Positive Trend
Original AP-style question. In a fictional skills practice, \(x\) is the number of practice sessions and \(y\) is the number of tasks completed correctly. Four participants have \((1,2)\), \((2,4)\), \((3,5)\), and \((4,7)\). Predict what happens to the slope when a participant with \((8,15)\) is added.
State. We will compare the added participant’s \(x\)-value with the existing mean \(x\), then compare the participant’s response with the existing regression line.
Plan. Calculate the existing slope and fitted line, then find the new point’s residual. The sign of the product of its horizontal displacement and residual predicts the direction of the slope change.
Do. The existing means are \(\bar{x}=2.5\) and \(\bar{y}=4.5\). The sums of squared \(x\)-deviations and cross-product deviations are \(S_{xx}=5\) and \(S_{xy}=8\). Thus the existing slope is \(b=8/5=1.6\), and the intercept is \(4.5-(1.6)(2.5)=0.5\). The existing fitted line is \(\hat{y}=0.5+1.6x\). At \(x=8\), it predicts \(0.5+(1.6)(8)=13.3\), so the added point’s residual is \(15-13.3=1.7\). The point is to the right of the existing mean because \(8-2.5=5.5>0\), and it is above the line because its residual is positive. Their product is \(5.5(1.7)=9.35>0\), so the slope increases.
We can check the direction by calculating the new slope. The updated sums are
Here \(10.5=15-4.5\), the point’s vertical deviation from the old mean. Therefore,
Conclude. Adding the participant raises the slope from \(1.6\) to about \(1.8562\) tasks per practice session. In this fictional data set, the point is right of the existing \(x\)-center and above the existing line, so it pulls the fitted line toward a larger numerical slope.
Worked Example: A Right-Side Point Above a Negative Trend
Original AP-style question. A fictional equipment check records \(x\), years of use, and \(y\), battery capacity in percentage points. Four devices have \((1,90)\), \((2,88)\), \((3,87)\), and \((4,85)\). Predict what happens to the slope when a device with \((8,82)\) is added.
State and plan. We will determine the added point’s side relative to the existing mean \(x\), its residual from the existing line, and the sign of their product. Since the original slope is negative, we will describe the result as an increase or decrease in the numerical slope, rather than assuming that increase means a steeper trend.
Do. The existing means are \(\bar{x}=2.5\) and \(\bar{y}=87.5\). The sums are \(S_{xx}=5\) and \(S_{xy}=-8\), giving slope \(b=-8/5=-1.6\) percentage points per year. The intercept is \(87.5-(-1.6)(2.5)=91.5\), so the line is \(\hat{y}=91.5-1.6x\). At \(x=8\), the predicted capacity is \(91.5-(1.6)(8)=78.7\). The new point’s residual is \(82-78.7=3.3\) percentage points. It is to the right of the existing mean, and it is above the line; therefore \((8-2.5)(3.3)=18.15>0\), so adding it increases the numerical slope.
To check, the new sums are
The updated slope is
Conclude. The slope increases numerically from \(-1.6\) to about \(-1.1027\) percentage points per year. It becomes less negative, so the fitted trend is flatter, not steeper. The point is above the original line and to its right; that combination raises the numerical value of the slope even though the original association is negative.
Worked Example: Removing a Left-Side Point Below the Full-Data Line
Original AP-style question. A fictional training log relates practice hours \(x\) to tasks completed correctly \(y\). Five observations are \((2,12)\), \((3,14)\), \((4,15)\), \((5,17)\), and \((0,3)\). Predict how the slope changes if the observation \((0,3)\) is removed.
State. The question asks about removing a point from the full data set, so we will use the full-data regression line and mean \(x\)-value. The removal rule uses the opposite direction from the sign of the point’s horizontal displacement multiplied by its full-data residual.
Plan. Calculate the full-data slope and residual for \((0,3)\), determine the sign of the product, and then verify the direction by comparing the full-data slope with the slope after removal.
Do. For the first four observations, \(\bar{x}=3.5\), \(\bar{y}=14.5\), \(S_{xx}=5\), and \(S_{xy}=8\). Their fitted line is \(\hat{y}=8.9+1.6x\). Adding \((0,3)\) gives full-data means \(\bar{x}_{\mathrm{full}}=14/5=2.8\) and \(\bar{y}_{\mathrm{full}}=61/5=12.2\). The updated sums are
Thus the full-data slope is \(40.2/14.8\approx2.7162\), and its intercept is \(12.2-(2.7162)(2.8)\approx4.5946\). At \(x=0\), the full-data line predicts about \(4.5946\), so the point’s residual is \(3-4.5946\approx-1.5946\). Its horizontal displacement is \(0-2.8=-2.8\). Both quantities are negative, so their product is positive. The point has pushed the full-data slope upward; removing it should lower the slope.
Conclude and check. Removing the point leaves the first four observations, whose slope is \(8/5=1.6\). The slope therefore falls from about \(2.7162\) to \(1.6\), as predicted. The removed observation was left of the full-data mean and below the full-data line; both signs together gave a positive product, meaning the point had raised the slope when included.
Common Mistakes and AP Exam Tip
- Using “above” or “below” without naming the reference line. State whether the point is above or below the line fitted to the data without it, or the full-data line, depending on the question.
- Forgetting to compare \(x_0\) with the mean \(x\). “Right side” means \(x_0>\bar{x}\), and “left side” means \(x_0<\bar{x}\). It does not simply mean that a point looks far to the right or left on a sketch.
- Reversing the removal direction. A point that raised the slope when added to the reduced data will lower it when removed from the full data. Make sure you identify which data set supplies the starting line.
- Calling every slope increase “steeper.” Slope increases numerically when it moves upward on the number line. For example, changing from \(-1.6\) to \(-1.1\) is an increase, but the negative trend becomes flatter.
- Claiming a point must have a large effect. The sign rule predicts direction, not size. A point close to the center of the \(x\)-values may change the slope only slightly; the actual change depends on the data.
A full-credit response identifies the line used for the comparison, states whether the point is left or right of the relevant mean \(x\), and gives its residual sign. It then connects those signs to an increase or decrease in the numerical slope and describes the change in context. Do not make a causal claim: as discussed in “Causal Versus Associational Wording,” a regression describes an association unless the study design supports a causal conclusion.
Check Your Understanding
For each situation, identify the direction of the numerical slope change and explain the signs you used.
- An added point is right of the existing mean \(x\) and below the existing regression line. Does the slope increase or decrease?
- An added point is left of the existing mean \(x\) and above the existing regression line. Does the slope increase or decrease?
- A negative slope changes from \(-2.4\) to \(-1.7\). Did it increase or decrease numerically, and did the line become steeper or flatter?
- A point in a full data set is left of the full-data mean \(x\) and above the full-data line. What direction should the slope move if the point is removed?
- Why must a response say which fitted line was used to decide whether a point is above or below it?