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Residuals and Transformations · Tutorial 289 of 996

Case-Deletion Checks for Regression Influence

Use case deletion to see how much a flagged observation changes a fitted regression line and the predictions that matter in context.

Intermediate 10 min read

What You'll Learn

  • Define a case-deletion check and its role in assessing regression influence.
  • Refit a simple linear regression line after removing one flagged observation.
  • Compare full-data and deleted-case slopes, intercepts, and predictions.
  • Calculate how deleting a case changes centered sums used in regression.
  • Distinguish sensitivity analysis from a justification for discarding data.
  • Interpret changes in context and identify limitations of the check.

Why Refit a Regression Line?

A regression diagnostic can flag an observation as influential: the fitted regression may change noticeably when that observation is included. A case-deletion check makes that concern concrete. Fit the line using all observations, remove the flagged case, fit the line again, and compare the results. The point is not to decide automatically that the case should be excluded. It is to find out whether the substantive description or prediction depends heavily on it.

This check follows naturally from the earlier material on Cook's Distance for Influential Observations. Cook's Distance can identify cases whose removal may matter; case deletion directly examines what the refitted line looks like. The comparison should be made in the context of the predictor and response, with particular attention to the slope and to predictions at relevant predictor values.

Definition: A case-deletion check refits a regression model after removing a specified observation, then compares the refitted results with those from the full data. In simple linear regression, the comparison commonly includes the intercept, slope, and predicted response at one or more relevant predictor values. A change in these quantities indicates sensitivity of the fitted line to that case.

Refitting Simple Linear Regression

For paired observations \((x_i,y_i)\), the least-squares regression line is \(\hat{y}=b_0+b_1x\). It chooses the intercept \(b_0\) and slope \(b_1\) to minimize the sum of squared residuals. Provided the predictor values are not all identical, its coefficients can be calculated using

$$ b_1=\frac{S_{xy}}{S_{xx}}, \qquad b_0=\bar{y}-b_1\bar{x}, $$

where \(S_{xx}=\sum (x_i-\bar{x})^2\) and \(S_{xy}=\sum (x_i-\bar{x})(y_i-\bar{y})\). The intercept is the fitted response at \(x=0\); that value may not be practically meaningful if zero is outside the relevant range of the predictor.

If observation \(i\) is removed, the remaining sample size is \(n-1\). Calculate a new predictor mean, response mean, \(S_{xx}\), and \(S_{xy}\) using only the retained cases. The resulting coefficients can be denoted \(b_{0(-i)}\) and \(b_{1(-i)}\), giving the deleted-case line \(\hat{y}_{(-i)}=b_{0(-i)}+b_{1(-i)}x\). The observations used in the refit are the same in each calculation except for the specified deletion.

Comparison quantities: The slope change is \(b_{1(-i)}-b_1\), the intercept change is \(b_{0(-i)}-b_0\), and the prediction change at a chosen predictor value \(x_0\) is \(\hat{y}_{(-i)}(x_0)-\hat{y}(x_0)\). State changes in the units of the quantity being compared: the slope has response units per predictor unit, while a predicted response has response units.

Why the Centered Sums Change

A short derivation helps explain why deleting a case requires recalculating the regression rather than simply removing its residual from a summary. Let \(X=\sum x_i\), \(Y=\sum y_i\), \(X_2=\sum x_i^2\), and \(XY=\sum x_iy_i\), all calculated over the full data. Expanding the definitions of the centered sums gives

$$ S_{xx}=X_2-\frac{X^2}{n}, \qquad S_{xy}=XY-\frac{XY}{n}. $$

In the second expression, \(XY\) on the right denotes the sum of products \(\sum x_iy_i\); the product in the numerator is the product of the separate totals \(X\) and \(Y\). Equivalently, to avoid that visual ambiguity, write the formulas with the totals named explicitly:

$$ S_{xx}=\sum x_i^2-\frac{(\sum x_i)^2}{n}, \qquad S_{xy}=\sum x_iy_i-\frac{(\sum x_i)(\sum y_i)}{n}. $$

For deletion of case \(i\), each raw total must first have that case's contribution removed, and the divisor becomes \(n-1\). For example, the new sum of products is \(\sum x_iy_i-x_iy_i\), where the first term is the full-data total and the second is the deleted case's product. The new correction term uses the new totals and \(n-1\). Thus deletion changes both the raw totals and the means used for centering. Since the slope is a ratio of the resulting centered sums, its direction or size can change substantially.

Worked Examples

Worked Example: A Flagged Case Reverses the Slope

Suppose five hypothetical participants have the following hours of practice \(x\) and performance scores \(y\), in points: \((1,2),(2,4),(3,5),(4,7),(8,2)\). The case \((8,2)\) has been flagged. Compare the full-data line with the line after deleting that case, and compare predictions at 3 hours of practice.

Fit without the flagged case. For the four retained cases, \(\sum x=10\), \(\sum y=18\), \(\sum x^2=30\), and \(\sum xy=53\). Therefore,

$$ S_{xx}=30-\frac{10^2}{4}=5,\qquad S_{xy}=53-\frac{(10)(18)}{4}=8. $$

The deleted-case slope is \(b_{1(-i)}=8/5=1.6\) score points per practice hour. The retained means are \(\bar{x}=10/4=2.5\) hours and \(\bar{y}=18/4=4.5\) points, so the intercept is \(b_{0(-i)}=4.5-(1.6)(2.5)=0.5\) points. The refitted line is \(\hat{y}_{(-i)}=0.5+1.6x\).

Fit using all five cases. The full-data totals are \(\sum x=18\), \(\sum y=20\), \(\sum x^2=94\), and \(\sum xy=69\). The sum of products includes \(8(2)=16\) from the flagged case, so \(53+16=69\). Hence,

$$ S_{xx}=94-\frac{18^2}{5}=29.2,\qquad S_{xy}=69-\frac{(18)(20)}{5}=-3. $$

The full-data slope is \(-3/29.2\approx-0.1027\) score points per practice hour. With \(\bar{x}=18/5=3.6\) hours and \(\bar{y}=20/5=4\) points, the intercept is \(4-(-0.1027)(3.6)\approx4.370\) points. The full-data line is approximately \(\hat{y}=4.370-0.1027x\).

At \(x=3\) hours, the full-data line predicts \(4.370+(-0.1027)(3)\approx4.062\) points; the deleted-case line predicts \(0.5+(1.6)(3)=5.3\) points. Deleting the flagged case changes the slope from slightly negative to positive and increases the prediction at 3 hours by about \(5.3-4.062=1.238\) points. In this hypothetical data set, both the direction of the fitted relationship and this prediction depend substantially on the flagged case.

Worked Example: Comparing a Prediction When the Slope Is Unchanged

Consider hypothetical measurements of weekly study time \(x\), in hours, and quiz score \(y\), in points: \((1,2),(2,3),(3,5),(4,4),(5,6)\). Suppose the case \((3,5)\) is flagged. Compare the lines and their predictions at 4 hours.

For all five observations, \(\sum x=15\), \(\sum y=20\), \(\sum x^2=55\), and \(\sum xy=69\). Thus \(S_{xx}=55-15^2/5=10\) and \(S_{xy}=69-(15)(20)/5=9\). The slope is \(9/10=0.9\) points per hour, and the intercept is \(20/5-(0.9)(15/5)=1.3\) points. The full-data line is \(\hat{y}=1.3+0.9x\).

After deleting \((3,5)\), the four retained cases have \(\sum x=12\), \(\sum y=15\), \(\sum x^2=46\), and \(\sum xy=54\). Therefore \(S_{xx}=46-12^2/4=10\) and \(S_{xy}=54-(12)(15)/4=9\). The deleted-case slope remains \(0.9\) points per hour. The retained means are \(\bar{x}=3\) hours and \(\bar{y}=3.75\) points, giving intercept \(3.75-(0.9)(3)=1.05\) points. The deleted-case line is \(\hat{y}_{(-i)}=1.05+0.9x\).

At 4 hours, the full-data prediction is \(1.3+(0.9)(4)=4.9\) points, and the deleted-case prediction is \(1.05+(0.9)(4)=4.65\) points. The slope is unchanged, but the predicted score decreases by \(0.25\) points after deletion. This illustrates why comparing predictions can add useful context: unchanged slope does not mean the fitted lines are identical.

Worked Example: A Substantive Conclusion Can Be Less Sensitive Than a Prediction

Suppose five hypothetical trainees have practice hours and assessment scores, in points: \((1,2),(2,3),(3,4),(4,5),(5,20)\). The final case is flagged. Compare the fitted slopes and the predicted score at 4 hours.

For all five cases, \(\sum x=15\), \(\sum y=34\), \(\sum x^2=55\), and \(\sum xy=140\). Thus \(S_{xx}=55-15^2/5=10\) and \(S_{xy}=140-(15)(34)/5=38\). The slope is \(38/10=3.8\) points per hour. Since \(\bar{x}=3\) and \(\bar{y}=6.8\), the intercept is \(6.8-(3.8)(3)=-4.6\) points. The full-data line predicts \(-4.6+(3.8)(4)=10.6\) points at 4 hours.

Without the flagged case, the remaining totals are \(\sum x=10\), \(\sum y=14\), \(\sum x^2=30\), and \(\sum xy=40\). Then \(S_{xx}=30-10^2/4=5\), and \(S_{xy}=40-(10)(14)/4=5\). The slope is \(1\) point per hour; the intercept is \(3.5-(1)(2.5)=1\) point. At 4 hours, this line predicts \(1+(1)(4)=5\) points.

The slope remains positive after deletion, so the direction of the fitted relationship is the same, but its estimated rate of increase changes from 3.8 to 1 point per hour. The prediction at 4 hours changes by \(5-10.6=-5.6\) points. Thus a broad statement that the fitted association is positive is less sensitive here than the estimated rate or the prediction at 4 hours. This comparison concerns these fitted lines; it does not establish that practice causes higher scores.

How to Interpret the Comparison

A case-deletion check is most useful when the quantities compared connect to a real question. A slope change matters when the estimated rate of response per predictor unit is central. A prediction change matters when a specific predictor value is used for planning or decision-making. If the intercept corresponds to an unrealistic value such as zero hours of practice when no one in the data is near zero, its numerical change may be less informative than changes in predictions within the observed range.

The check is descriptive: it reports how the fitted results change in this data set. It does not, by itself, determine why the case is unusual or whether it should be removed. Verify the record and the case's eligibility using the study's rules. If the observation is valid and belongs to the population being studied, an influential result may be important information about how much the analysis relies on that case. Report the comparison transparently rather than presenting only the more convenient line.

Check that the regression remains calculable after deletion: at least two observations must remain, and their predictor values must not all be identical. Also examine the data and the fit in context. As in the earlier material on residual plots, a line that changes after deletion is not the only possible model concern; a deletion comparison cannot show by itself that a linear model is appropriate. Nor does one deletion check assess every possible source of error or explain why a case affects the fit.

1
Fit the full-data model.
Record the line and identify the slope, intercept, and predictions relevant to the question.
2
Remove the specified case and refit.
Recalculate the regression using the remaining observations; do not keep the original means or centered sums.
3
Compare in context.
State how the slope, intercept, and relevant predictions change, with units, then describe which conclusions are sensitive to the deletion.

Common Mistakes

  • Using an incorrect sum of products. The deleted case contributes \(x_iy_i\), so removing it changes \(\sum x_iy_i\) by that amount. Check the raw totals against the listed data before computing the slope.
  • Changing only the numerator or only the denominator. Deletion changes the means as well as the sums of squares and products. Recalculate both \(S_{xx}\) and \(S_{xy}\) for the retained observations.
  • Calling the flagged case an error without checking. A diagnostic flag signals a case worth examining; it does not prove that its recorded values are wrong or that it should be excluded.
  • Comparing slopes without comparing relevant predictions. A slope may change little while predictions shift, or a slope may change while a prediction at a particular value remains similar. Choose comparisons that address the question.
  • Treating sensitivity as causation. A changed regression line describes dependence of the fitted association on an observation. It does not show that the predictor causes the response.
Key takeaway: A case-deletion check asks how the regression line and relevant predictions change when a flagged observation is removed and the model is refit. Report that sensitivity in context, check the arithmetic, and do not treat deletion as an automatic remedy.

Check Your Understanding

Try these questions without looking back.

  1. What quantities must be recalculated when one observation is removed from a simple linear regression?
  2. A deleted-case slope changes from 2.4 to 1.1 response units per predictor unit. What does this say about the fitted rate of change, and what does it not establish?
  3. Why can predictions change even when the full-data and deleted-case slopes are equal?
  4. In the first worked example, what are the full-data and deleted-case predictions at 3 hours, and which line predicts the higher score?
  5. Why is a case-deletion check not, by itself, a reason to discard the flagged observation?