Introduction
A point in a scatterplot can have an ordinary response value and still deserve attention. If its predictor value lies far to the left or right of the main cloud, it represents a part of the predictor range with relatively little data. The value might be correct and important, or it might reveal a recording, unit, or data-design problem. The plot identifies a question to investigate; it does not answer that question by itself.
Here, the predictor is the variable plotted on the horizontal axis, and the response is the variable plotted on the vertical axis. The focus is on observations unusual in their predictor value. That is a marginal judgment about the horizontal coordinate, not a conclusion that the whole point is anomalous.
Read the Horizontal Spread First
Start by looking along the horizontal axis, setting the vertical positions of the points aside temporarily. Ask where most predictor values lie and whether one or a few observations sit well beyond that range. For example, if most policies in a scatterplot have insured values between $10,000 and $60,000, a policy at $150,000 may be unusual in the predictor even if its response is near the middle of the vertical spread.
The response value still matters when deciding what the observation means, but it does not determine whether the predictor is unusual. A point can have a routine response and a remote predictor. Conversely, a point can have a predictor value in the middle of the horizontal range while its response is surprising. Those are different questions and should not be conflated.
A scatterplot is a visual aid, so the apparent distance can depend on the axis limits and scale. A very wide axis may compress the main cloud; a log scale may make a long right tail easier to see. Read the axis labels and units, and compare the point with the distribution of predictor values on the scale actually used. A histogram or a boxplot of the predictor can provide a complementary view, but the scatterplot retains the pairing between each predictor and its response.
A Quantitative Screening Rule
A common screening method is the 1.5 IQR rule, applied here to the predictor values alone. It is a descriptive convention, not a probability model. It does not assume that the predictor is normally distributed, and it does not declare every value beyond a fence erroneous.
The reason for the upper fence is straightforward: it asks whether the value lies more than 1.5 middle-half widths above the third quartile. Specifically, \(x>Q_3+1.5\operatorname{IQR}\) is equivalent to \(x-Q_3>1.5(Q_3-Q_1)\). The same reasoning below \(Q_1\) gives \(Q_1-x>1.5(Q_3-Q_1)\). Thus the rule measures distance relative to the spread of the central half of the predictor values. It is a screening threshold, not a claim that observations beyond it cannot occur.
Quartile calculations can differ slightly between software conventions, especially in small samples. Use the convention specified for the analysis, or state the convention when calculating by hand. When quartiles are supplied, use those values rather than silently substituting another method.
Worked Example: A Remote Predictor With an Ordinary Response
In a hypothetical portfolio study, the predictor is annual insured payroll in thousands of dollars, and the response is a claim frequency. The predictor quartiles are \(Q_1=18\) and \(Q_3=42\). Most plotted predictor values fall between 12 and 65. One policy has predictor value 91 and response 2.7 claims per 100 policies; responses for most policies range from 1.8 to 4.2. Decide whether the predictor should be flagged.
Method. Apply the 1.5 IQR rule to the horizontal variable. The response value is not part of this calculation.
First calculate the middle-half spread: \(\operatorname{IQR}=42-18=24\). The upper fence is \(42+1.5(24)=42+36=78\), and the lower fence is \(18-36=-18\). Since \(91>78\), the predictor is above the upper fence. It is flagged for inspection.
The response, 2.7, lies within the stated typical response range of 1.8 to 4.2. That does not undo the horizontal flag. The policy could be a valid large-payroll account, or the predictor could have been recorded in dollars rather than thousands of dollars. Checking the source record and units is more informative than deciding from the response alone.
Check. The upper fence is 78 in the same units as the predictor, and 91 exceeds it by 13. The rule therefore flags this predictor value.
Scale and Context Change the Question
An extreme value on a raw scale may be less unusual on a logarithmic scale when the predictor is strongly right-skewed. This does not mean that a log transformation makes a questionable record correct. It means that “far from the main spread” depends on the scale relevant to the analysis. Inspect the measurement units and the data-generating context before choosing a scale for interpretation.
Worked Example: Comparing Raw and Logarithmic Scales
A hypothetical insurer plots contract values, measured in thousands of dollars, against a routine service-cost response. The predictor quartiles on the raw scale are \(Q_1=30\) and \(Q_3=70\). A contract has predictor value 180. Assess the flag on the raw scale, then compare it with a base-10 log-scale screen.
Raw-scale calculation. The raw-scale IQR is \(70-30=40\). The upper fence is \(70+1.5(40)=130\). Since \(180>130\), the value is flagged on the raw scale.
Log-scale calculation. The transformed quartiles are \(\log_{10}(30)\approx1.4771\) and \(\log_{10}(70)\approx1.8451\). Their IQR is \(1.8451-1.4771=0.3680\). The log-scale upper fence is \(1.8451+1.5(0.3680)=1.8451+0.5520=2.3971\). The observed value transforms to \(\log_{10}(180)\approx2.2553\), which is below 2.3971. It is not above the upper fence on this log scale.
For context, the log-scale fence corresponds to a raw value of approximately \(10^{2.3971}=249.5\) thousand dollars. Thus 180 is beyond the raw-scale fence of 130 but below the back-transformed log-scale fence of about 249.5. This difference is possible because taking logarithms changes distances.
The conclusion is not that one calculation is universally right. The raw scale asks about departures measured in dollars; the log scale asks about departures in multiplicative terms. Check whether a log scale suits the predictor’s meaning and spread. Do not transform merely to make a flagged value disappear, and do not reject a valid large contract solely because the raw-scale rule flags it.
Locate and Investigate the Observation in R
In R, a scatterplot can help identify which record lies at an unusual horizontal position. For example, use plot(x, y) to display the predictor vector x against the response vector y. If records have labels, add them with text(x, y, labels = record_id, pos = 4), or use identify(x, y) interactively to inspect points. Keep a record identifier with the plotted data: a plot can reveal a location, but the corresponding row is needed for a data audit.
A practical inspection separates three questions:
Check the original source, decimal placement, units, and whether the value was joined to the correct record. A striking value is not enough evidence to edit or delete it.
Compare it with plausible limits and with the study’s inclusion rules. A valid large account may be rare but still belong in the analysis.
Determine whether the model is expected to describe this part of the predictor range. Predictions far beyond the range represented by most observations require particular caution.
If the value is verified and relevant, retain it unless there is a justified analytical reason not to. If it is an error, correct it from reliable source information or handle it under a documented data-cleaning rule. If it cannot be verified, document the uncertainty and assess how conclusions change with and without the observation. Do not silently remove or replace a value simply because it is inconvenient.
Worked Example: Checking a Possible Data-Entry Error
A hypothetical study relates years of service to an annual benefit rate. The ten predictor values, in sorted order, are 2, 3, 4, 5, 6, 7, 8, 9, 10, and 38. The response paired with 38 is 0.08; typical responses in the study range from 0.04 to 0.12. Use the R type 7 quartile convention, and describe the flag and a sensible next step.
Calculate the quartiles. With ten sorted observations, the type 7 position for \(Q_1\) is \(1+(10-1)(0.25)=3.25\). Interpolate one-quarter of the way from the third value, 4, to the fourth value, 5: \(Q_1=4+0.25(5-4)=4.25\). The position for \(Q_3\) is \(1+(10-1)(0.75)=7.75\). Interpolate three-quarters of the way from the seventh value, 8, to the eighth value, 9: \(Q_3=8+0.75(9-8)=8.75\).
The IQR is \(8.75-4.25=4.50\). The upper fence is \(8.75+1.5(4.50)=8.75+6.75=15.50\), and the lower fence is \(4.25-6.75=-2.50\). Since \(38>15.50\), the predictor is flagged. The paired response, 0.08, is within the stated typical response range, but that does not resolve whether 38 years is accurate.
The predictor mean also illustrates why the record merits attention. The ten values sum to 92, so their mean is \(92/10=9.2\) years. The other nine values sum to 54 and have mean \(54/9=6.0\) years. The unusual value changes the summary of the predictor, but the calculation does not prove it is wrong. Check the source record and the permitted service range. If the source verifies 38, retain it as a valid observation; if the source shows a different value, correct it using that source and document the correction.
Check. The upper fence is 15.50, far below 38, while the response is ordinary by the range supplied. The correct conclusion is “investigate the predictor,” not “delete the point.”
Common Mistake: Using the Wrong Spread
A frequent error is to multiply the third quartile by 1.5 instead of multiplying the IQR. In the first example, this incorrect calculation gives \(42+1.5(42)=105\), so the value 91 would appear not to be flagged. The correct calculation is \(42+1.5(42-18)=42+36=78\), and 91 is above 78. The factor 1.5 applies to the distance between the quartiles, not to \(Q_3\) itself.
A second mistake is to let an ordinary response value overrule a horizontal-axis check. The IQR screen uses the predictor values; response values do not enter its fences. A third is to treat a fence as a deletion rule. The correct interpretation is narrower: it marks an observation for review under a stated descriptive convention.
Why Unusual Predictor Values Matter
Predictor values far from the main cloud matter because they may represent a different part of the population, a data-quality issue, or a region with little information. A fitted relationship may be sensitive to observations at the edges of the predictor range, and a prediction for a new case far outside the range represented in the data has less direct support from observed cases. An ordinary response does not remove these concerns.
Keep the interpretation appropriately limited. A horizontal extreme is not, by itself, proof that the point is unusual in its response relative to the relationship. Nor does a marginal fence determine whether the observation should be included in a model. Those decisions require the study context, verified records, and a clear statement of the prediction or estimation goal. The useful conclusion from this plot is that the predictor value deserves a deliberate check.
Check Your Understanding
For each situation, decide what the predictor screen says and interpret it in context. Treat the quartiles given in each question as the quartiles to use.
- A predictor has \(Q_1=12\) and \(Q_3=28\). A plotted observation has predictor value 50. Calculate the upper fence and state whether the value is flagged. (Answer: IQR \(=16\), upper fence \(=52\); 50 is not above the fence, so it is not flagged by this rule.)
- A predictor has \(Q_1=8\) and \(Q_3=20\). One observation has predictor value \(-12\). Calculate the lower fence and state whether it is flagged. (Answer: IQR \(=12\), lower fence \(=8-18=-10\); \(-12\) is below the fence and is flagged.)
- On a base-10 log scale, predictor quartiles are 1.2 and 1.8. An observation has raw predictor value \(10^{2.8}\). Is it above the log-scale upper fence? (Answer: The log-scale IQR is 0.6 and the upper fence is \(1.8+1.5(0.6)=2.7\); the transformed value is 2.8, so it is above the fence.)
- A model’s observed predictor values range from 1 to 9. A new case has predictor value 25 and a response value typical of the training data. What should be concluded before using the model for this case? (Answer: The predictor is outside the observed range and deserves an applicability check; a typical response does not establish that a prediction at 25 is supported.)
- A point is beyond the predictor’s upper fence, but its source record confirms the value and the study includes cases of that size. Should the point be deleted solely because it is flagged? (Answer: No. A fence flags the value for inspection; a verified, relevant observation should not be deleted solely for exceeding it.)