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Prediction Intervals for New Outcomes · Tutorial 739 of 991

Sensitivity of Prediction Intervals to Influential Training Cases

Compare prediction intervals from fits with and without a carefully investigated training case, and learn what changes in the model can—and cannot—justify omitting it.

Intermediate 10 min read

What You'll Learn

  • Define sensitivity of a prediction interval to a training observation
  • Recompute an interval after a justified case omission
  • Track changes in the fitted value, residual standard error, leverage, and critical value
  • Explain why interval width can increase or decrease after omission
  • Distinguish an influential valid observation from a data error or an observation outside the target population
  • Report sensitivity results without treating them as proof that a case should be deleted

Why Examine Sensitivity?

A prediction interval depends on the regression model fitted to the training data. A single training observation can sometimes have a noticeable effect on the fitted regression coefficients, the estimate of residual variation, or the uncertainty at a particular predictor setting. Consequently, the interval for a new outcome can change when that observation is included or omitted.

That change is useful information, but it is not a deletion rule. An observation may be influential because it has unusual predictor values, an unusual outcome relative to the fitted model, or both. If it is a valid observation from the population the model is meant to describe, its influence may be substantively important rather than a nuisance to remove. The aim here is to compare fits transparently and investigate why they differ.

Definition: The sensitivity of a prediction interval to a training case is the change in the interval when a regression model is refitted after that case is omitted, with the prediction target and interval level held fixed. A case is influential for this prediction if its inclusion materially changes the fitted value, interval endpoints, or another relevant feature of the prediction.

What Changes When a Case Is Omitted?

For a regression model with \(p\) estimated coefficients, a usual two-sided prediction interval for a new outcome at predictor vector \(\mathbf{x}_0\) has the form

$$ \hat{y}_0 \ \pm\ t_{1-\alpha/2,\,df}\,s\sqrt{1+h_0}. $$

Here \(\hat{y}_0\) is the fitted value at the prediction target, \(s\) is the residual standard error, \(df=n-p\), and \(h_0\) measures the target's leverage in the fitted design. In multiple regression, \(h_0=\mathbf{x}_0^{\mathsf T}(X^{\mathsf T}X)^{-1}\mathbf{x}_0\), using a design matrix \(X\) that includes the intercept when the model has one. The term \(1\) accounts for the variation of a new individual outcome; the leverage term accounts for uncertainty in the fitted mean. This tutorial compares the usual interval from each refitted model, assuming its regression conditions are adequate.

After omitting one case, the number of observations falls by one, and the coefficient estimates generally change. The fitted value \(\hat{y}_0\) may move. The residual standard error \(s\), target leverage \(h_0\), degrees of freedom, and hence the \(t\) critical value may also change. Thus there is no general rule that omission must make an interval narrower or wider.

Key comparison: Hold the prediction target and nominal interval level fixed. Then compare the fitted value (the interval's center), the residual standard error, target leverage, degrees of freedom and critical value, and finally the interval endpoints and width.

Why the Width Can Move Either Way

The half-width of the interval is \(M=t_{1-\alpha/2,df}s\sqrt{1+h_0}\). Each factor can move differently after refitting. For example, omitting a case can reduce \(s\), but it can also increase \(h_0\) at the target or slightly increase the critical value by reducing \(df\). Even when the interval narrows, its center can shift; a narrower interval does not necessarily mean that the prediction has become more accurate or that the omitted case was erroneous.

A useful way to compare widths is the ratio of the half-widths:

$$ \frac{M_{\text{without}}}{M_{\text{with}}} = \frac{t_{\text{without}}}{t_{\text{with}}} \frac{s_{\text{without}}}{s_{\text{with}}} \sqrt{\frac{1+h_{0,\text{without}}}{1+h_{0,\text{with}}}}. $$

This follows by dividing the expression for \(M\) after omission by the same expression before omission. It makes clear that a change in width is a combined result, not a direct measure of just one observation's effect on residual variation.

Worked Example: Comparing Intervals for a Study-Time Prediction

Suppose a hypothetical training data set contains 24 students, with exam score in points as the outcome and study time in hours as the predictor. One valid student record has unusually high study time. The target is a new student's score at a fixed study time of 5 hours. The fitted-model summaries below are rounded software output; both intervals are 95% intervals.

With all 24 records, suppose \(\hat{y}_0=18.4\) points, \(s=4.2\) points, and \(h_0=0.09\). There are two estimated coefficients, so \(df=24-2=22\), and \(t_{0.975,22}\approx2.074\). The half-width is

$$ M=2.074(4.2)\sqrt{1+0.09} =2.074(4.2)\sqrt{1.09} \approx 9.09\text{ points}. $$

The full-data interval is therefore \(18.4\pm9.09=(9.31,\ 27.49)\) points, with endpoints rounded to two decimals.

After the unusual-study-time record is omitted and the model refitted, suppose the software reports \(\hat{y}_0=19.1\), \(s=3.8\), and \(h_0=0.11\). Now \(n=23\), \(df=21\), and \(t_{0.975,21}\approx2.080\). Thus

$$ M=2.080(3.8)\sqrt{1+0.11} =2.080(3.8)\sqrt{1.11} \approx 8.33\text{ points}, $$

giving the interval \(19.1\pm8.33=(10.77,\ 27.43)\) points. Omitting the case moved the fitted value up by 0.7 points and reduced the half-width from about 9.09 to 8.33 points. The interval width fell from about 18.19 to 16.65 points. Although target leverage increased from 0.09 to 0.11 and the degrees of freedom decreased, the lower residual standard error more than offset those effects on width in this comparison.

The interpretation remains about the specified target: under the model and its conditions, the interval from the full fit predicts a new student's score at 5 hours of study, while the second interval gives the corresponding prediction using the refitted model. This numerical sensitivity alone does not establish that the valid record should be excluded.

An Influential Case Can Widen the Interval When Included

Influence can also work in the other direction. A valid case may increase the fitted model's residual variation, or shift the prediction target relative to the fitted relationship. Omitting it may make the interval narrower, but that does not show that the narrower interval is preferable: it may instead describe a different population or conceal real variability.

Worked Example: A Valid Extreme-Concentration Plant

Imagine a hypothetical data set of 30 plants used to relate fertilizer concentration to final height in centimeters. One plant received a concentration near the high end of the scientifically relevant range and has a valid height measurement. For a new plant at a fixed concentration in that range, suppose the full-model output gives \(\hat{y}_0=42.0\) cm, \(s=5.0\) cm, and \(h_0=0.04\). With \(df=30-2=28\) and \(t_{0.975,28}\approx2.048\), the half-width is

$$ M=2.048(5.0)\sqrt{1.04}\approx10.44\text{ cm}. $$

The full-data interval is \(42.0\pm10.44=(31.56,\ 52.44)\) cm.

Suppose a subject-matter investigation confirms that the extreme-concentration plant was measured correctly and belongs to the population of interest. To understand its effect, the analyst also computes a clearly labelled sensitivity fit without it. Suppose that fit gives \(\hat{y}_0=41.6\) cm, \(s=5.4\) cm, and \(h_0=0.045\). Now \(df=29-2=27\), and \(t_{0.975,27}\approx2.052\). The half-width is

$$ M=2.052(5.4)\sqrt{1.045}\approx11.33\text{ cm}. $$

The sensitivity interval is \(41.6\pm11.33=(30.27,\ 52.93)\) cm. In this hypothetical comparison, omitting the case raises the residual standard error from 5.0 to 5.4 cm. The modest changes in leverage and critical value do not counteract that increase, so the interval widens. The fitted value also shifts down by 0.4 cm.

Because the plant's record is valid and its concentration is within the intended scope, the full-data fit remains the primary analysis if that scope includes such plants. The omission fit is a sensitivity analysis, not a replacement justified by the fact that its interval differs.

Track the Center and Width Separately

An interval has two distinct features: its center and its width. A case can move the center while barely changing the width, or change the width while leaving the center nearly fixed. Reporting only one feature can therefore hide an important aspect of sensitivity.

Worked Example: A Shifted Prediction With a Narrower Interval

Consider a hypothetical training set of 18 machines, with operating temperature as predictor and output in units per hour as outcome. One high-temperature machine is a valid observation. At a fixed target temperature, suppose the full-data model gives \(\hat{y}_0=72.0\) units per hour, \(s=6.0\) units per hour, and \(h_0=0.30\). With \(df=16\) and \(t_{0.975,16}\approx2.120\), the half-width is

$$ M=2.120(6.0)\sqrt{1.30}\approx14.50\text{ units per hour}, $$

so the interval is \(72.0\pm14.50=(57.50,\ 86.50)\) units per hour.

Suppose omission and refitting produce \(\hat{y}_0=70.5\), \(s=5.5\), and \(h_0=0.36\). With \(df=15\) and \(t_{0.975,15}\approx2.131\),

$$ M=2.131(5.5)\sqrt{1.36}\approx13.67\text{ units per hour}. $$

The new interval is \(70.5\pm13.67=(56.83,\ 84.17)\) units per hour. Its width is smaller, even though target leverage rose from 0.30 to 0.36, because the residual standard error fell from 6.0 to 5.5. At the same time, the predicted output fell by 1.5 units per hour. The full and reduced intervals overlap substantially, but their centers and endpoints are not identical. Both changes should be reported if the case is substantively important.

Investigate Before Deciding Whether to Omit

A sensitivity comparison is most informative when it follows an explicit investigation. First verify the observation against its source and check units, coding, and data entry. Then ask whether the observation belongs to the population and predictor range the model is intended to address. Examine whether the assumed mean relationship and error behavior remain reasonable with the case included. If a value is a documented error, correction or exclusion may be justified and should be explained. If it is valid, do not remove it merely because it changes the prediction interval.

A valid observation may reveal that the fitted relationship is sensitive in an important region, that the data provide limited support for a prediction, or that a single model does not summarize all cases well. The appropriate response may be to report both fits as a sensitivity analysis, gather more data, or reconsider whether the model is suitable for the stated prediction. Any alternative analysis must still be justified by the question and assumptions.

1
Fit the primary model.
State the target predictor values, interval level, and model conditions; calculate and report the prediction interval with the case included.
2
Investigate the case.
Check the measurement and determine whether the observation belongs to the population relevant to the prediction. Influence is a reason to investigate, not by itself a reason to delete.
3
Refit for sensitivity.
If informative, omit the case temporarily, refit the model, and compute the interval for the same target at the same nominal level.
4
Explain the differences.
Compare the fitted value, residual standard error, leverage, degrees of freedom, critical value, endpoints, and width. State which fit is primary and why.

Common Mistake

A common mistake is to say, “The case is influential, so it should be deleted; the smaller interval after deletion is better.” This reasoning confuses a change in the fit with evidence that the observation is invalid. Influence describes how much a case affects a fitted result; it does not establish that the data are wrong, that the case is outside the population of interest, or that assumptions improve when it is removed.

The correction is to investigate and document the case, then interpret the sensitivity comparison in context. If the observation is valid and relevant, the full-data result generally answers the stated question more directly. If it is an error or does not belong to the target population, an omission may be defensible, but the reason should be stated and the resulting interval interpreted for the population represented by the refitted model.

Key takeaway: Refit-and-compare analysis reveals how a training case affects a prediction interval. Track the center and the components of the width, but do not treat influence alone as permission to delete a valid observation.

Check Your Understanding

Use the formulas and comparisons in this tutorial to answer the following.

  1. When a training case is omitted, name four quantities that may change in a prediction interval.
  2. Why can a reduced residual standard error coincide with a wider prediction interval?
  3. In the study-time example, how did the fitted value and interval width change after omission?
  4. A valid observation greatly changes a prediction at a predictor value near the edge of the observed data. Is that change alone a reason to delete the observation? Explain.
  5. Why should the prediction target and nominal interval level remain fixed when comparing the two fits?