What \(r^2\) Summarizes—and What It Leaves Out
In “Low and High \(r\)-squared Values,” you learned that \(r^2\) describes the proportion of variation in a response accounted for by its linear relationship with an explanatory variable. That is useful information, but it is only a summary. It does not describe every feature of the data or explain why the variables are associated.
Three cautions are especially important. A high \(r^2\) does not guarantee that a linear model is appropriate; it does not show that the explanatory variable causes the response to change; and it does not guarantee a small prediction error for every individual value of \(x\). To evaluate those questions, look beyond \(r^2\) to the scatterplot, residual plot, study design, and the context of the prediction.
A High \(r^2\) Does Not Guarantee Linearity
A linear model describes a relationship with a straight-line pattern. A large \(r^2\) tells you that the fitted line accounts for a large proportion of the response variation, but it does not reveal whether the points follow a straight-line pattern or a systematic curve. A curved relationship can still have a high \(r^2\), especially when the response generally increases or decreases as the explanatory variable changes.
This is why a scatterplot and residual plot matter. As covered in “Curved Patterns in a Residual Plot,” a systematic curve in residuals indicates that a fitted line misses in an organized way. For a suitable linear model, residuals should instead be scattered above and below zero without a clear pattern, as described in “Reading a Residual Plot for Random Scatter.” A high \(r^2\) cannot substitute for checking that pattern.
Worked Example: A High \(r^2\) with a Curved Pattern
A fictional materials lab records a setting \(x\) and a measured response \(y\) for seven test conditions. The values are deliberately generated by the curved rule \(y=100+x^2\). A least-squares line is fitted to these observations. Does its \(r^2\) establish that a linear model is appropriate?
State. The explanatory variable is the setting \(x\), and the response is the measured value \(y\). The question is whether the coefficient of determination alone supports using a straight-line model.
Plan. Calculate the fitted line and \(r^2\), then examine the residuals. A high \(r^2\) describes variation accounted for; the residual pattern helps assess whether the line misses systematically.
Do. The data and the fitted values are:
| \(x\) | Observed \(y=100+x^2\) | Fitted \(\hat{y}=88+8x\) | Residual \(y-\hat{y}\) |
|---|---|---|---|
| 1 | 101 | 96 | 5 |
| 2 | 104 | 104 | 0 |
| 3 | 109 | 112 | -3 |
| 4 | 116 | 120 | -4 |
| 5 | 125 | 128 | -3 |
| 6 | 136 | 136 | 0 |
| 7 | 149 | 144 | 5 |
The mean of the \(x\)-values is \(4\), and the mean of the \(y\)-values is \(120\). The slope is the sum of the products of the deviations divided by the sum of squared \(x\)-deviations:
So the fitted line is \(\hat{y}=88+8x\). The sum of squared \(x\)-deviations is \(28\), and the sum of squared \(y\)-deviations is \(1876\). The sum of the products of the corresponding deviations is \(224\). Therefore:
The coefficient of determination is about \(0.9552\), or \(95.52\%\). The residuals are positive at both ends of the \(x\)-range and negative in the middle: \(5, 0, -3, -4, -3, 0, 5\). That organized pattern is curved, not random scatter around zero.
Conclude. Although the line accounts for about \(95.52\%\) of the variation in these measured responses, the residuals reveal a curved pattern. The high \(r^2\) does not establish that a linear model is appropriate. In this example, inspecting the residuals reveals what the single summary number does not.
A High \(r^2\) Does Not Establish Causation
Causation is a claim that changing one variable produces a change in another. A strong association, even one with a high \(r^2\), is not enough to establish that claim. The value of \(r^2\) describes the strength of a linear association in the observed data; it does not tell you how the data were collected or rule out other explanations.
In an observational study, researchers record variables without assigning people or cases to treatments. A third variable may help explain why the explanatory and response variables change together. In a well-designed randomized experiment, randomly assigning treatments can provide evidence for a cause-and-effect relationship, subject to the experiment’s design and execution. Neither design changes the definition of \(r^2\); the study design affects what conclusions about cause are justified.
Worked Example: A Strong Association Is Not a Cause
A fictional community center records the daily number of frozen drinks sold and the number of admissions to its outdoor pool. In an observational analysis, the linear regression has \(r^2=0.84\). A manager concludes that selling more frozen drinks causes more people to visit the pool. Is that conclusion supported?
State. The response is pool admissions, and the explanatory variable is the number of frozen drinks sold. The reported \(r^2\) is \(0.84\), or \(84\%\).
Plan. Interpret \(r^2\) as a summary of linear association, then consider the observational design and a plausible alternative explanation. Do not infer causation from the statistic alone.
Do. About \(84\%\) of the variation in pool admissions in these data is accounted for by the linear relationship with frozen-drink sales. But the records are observational: the center did not randomly assign days to different drink-sales levels. Hot weather could increase both pool attendance and demand for cold drinks. The two variables could therefore rise together even if selling drinks does not cause additional pool visits.
Conclude. The \(r^2\) supports describing a strong linear association between frozen-drink sales and pool admissions in these recorded days. It does not establish that frozen-drink sales cause admissions to increase. The plausible effect of weather illustrates why the study design and context matter.
When communicating a result, use wording that matches the evidence. “The variables are strongly associated” describes a relationship. “The explanatory variable causes the response to increase” makes a causal claim and requires support beyond a high \(r^2\). Do not turn a descriptive statistic into a conclusion the data collection method cannot justify.
A High \(r^2\) Does Not Guarantee an Accurate Prediction for Every Case
The \(r^2\) value is a summary for the data as a whole. It is not a measure of the error for one particular observation, and it does not promise that the line’s predictions are close at every value of \(x\). Individual responses can differ from the fitted value. As you learned in “Prediction Versus Observed Values,” the difference for an observed case is its residual, \(y-\hat{y}\). The residual’s size and context matter when assessing that case’s prediction.
A model may account for a large share of the overall variation while still having some sizable residuals. The residual standard deviation \(s\), discussed in “Using \(s\) to Describe Prediction Accuracy,” summarizes the typical size of prediction errors in the response’s units. It can add information about error scale, but neither \(s\) nor \(r^2\) guarantees that a particular case will have a small error. The residual plot can also reveal whether errors behave differently across the range of \(x\).
And even if a requested \(x\)-value falls within the observed range, so the prediction is interpolation, that alone does not guarantee precision. “Predicting Within the Data Range” explains why checking the range matters; a prediction within that range still needs to be considered in light of the model’s fit and the variability of individual responses.
Worked Example: A Good Overall Summary and a Large Individual Error
A fictional delivery service uses distance in kilometers to predict delivery time in minutes. The fitted line is \(\hat{y}=6+0.6x\), and the regression output reports \(r^2=0.92\). Distances in the data range from 10 to 100 kilometers. For one delivery at 90 kilometers, the observed delivery time was 78 minutes. What does the line predict, and what does this case show about \(r^2\)?
State. The response is delivery time in minutes, and the explanatory variable is distance in kilometers. The \(r^2\) value of \(0.92\) describes the data overall, while the question concerns a particular delivery.
Plan. Substitute the delivery’s distance into the fitted line to get its predicted time. Then calculate the residual as observed time minus predicted time and interpret it in context.
Do. At \(x=90\) kilometers, the predicted delivery time is:
The residual is:
Conclude. The model predicted 60 minutes for this 90-kilometer delivery, which took 78 minutes. Its residual is \(18\) minutes, so the line underpredicted this delivery time by 18 minutes. Although \(r^2=0.92\) means the linear relationship accounts for about \(92\%\) of the variation in delivery times in the data, it does not guarantee an accurate prediction for this individual delivery.
Common Mistakes and AP Exam Tips
- Treating a high \(r^2\) as proof of a straight-line pattern. A full-credit response checks the scatterplot or residual plot. If residuals show a clear curve, describe that pattern and explain that a high \(r^2\) does not make the linear model appropriate.
- Claiming that association proves causation. A high \(r^2\) describes a linear relationship in the observed data, not the effect of changing the explanatory variable. State whether the study is observational or experimental and avoid causal wording unless the design supports it.
- Calling \(r^2\) the percentage of predictions that are accurate. \(r^2\) is about variation in the response accounted for by a linear relationship. For a particular observed case, calculate and interpret its residual if the needed values are given.
- Assuming a prediction within the data range must be close. Interpolation avoids the particular concern of extrapolating beyond observed \(x\)-values, but it does not eliminate individual response variation. Consider residuals and the typical error scale.
- Overstating what a residual establishes. A large residual describes one observed response’s difference from its fitted value. It does not by itself explain why that case differed.
On an AP response, name the response and explanatory variable when interpreting \(r^2\), use “about” for a rounded value, and match the conclusion to the evidence. For a linearity question, describe the residual pattern. For a causation question, discuss the study design. For an individual prediction, report the predicted response and, when the observed value is supplied, interpret \(y-\hat{y}\) in context and units.
Check Your Understanding
Use the meaning and limitations of \(r^2\) to answer each question.
- A regression of a response on an explanatory variable has \(r^2=0.93\), but its residual plot shows a clear U-shaped pattern. What does \(r^2\) say, and what does the residual pattern add?
- A fictional observational study finds \(r^2=0.78\) between daily umbrella sales and the number of people using a covered walkway. Give one reason the association alone does not establish causation.
- A model has \(r^2=0.90\). Does that mean \(90\%\) of its individual predictions are accurate? Explain what the value does mean.
- A fitted line predicts 42 minutes for a trip that actually takes 51 minutes. Calculate the residual and interpret its sign in context.
- Why does a prediction for an \(x\)-value inside the observed range not automatically guarantee a small prediction error?