Variation Is Not a Count of Correct Predictions
In “Interpreting \(r^2\) as a Percent of Variation,” you practiced describing \(r^2\) as the percentage of variation in the response accounted for by its linear relationship with the explanatory variable. A common next mistake is to read \(r^2=0.64\) as “64% of the predictions are correct.” Those statements sound similar because both use a percentage, but they answer different questions.
The coefficient of determination summarizes how response values vary around their mean compared with how they vary around the least-squares regression line. It does not count observations, check whether predictions are exactly right, or define how close a prediction must be to count as right. In particular, “64% of predictions are correct” is not an acceptable interpretation of \(r^2=0.64\).
Recall from “Defining the Coefficient of Determination \(r^2\)” and “Interpreting \(r^2\) in Context” that \(r^2\) is about variation in the response, not the number of cases. In the usual simple linear regression setting, a reported \(r^2\) of 0.64 means about 64% of the variation in the response is accounted for by its linear relationship with the explanatory variable.
By contrast, a prediction for an individual case is \(\hat{y}\), and its error can be described by the residual \(y-\hat{y}\), as in the earlier tutorials on residuals. The residual is in the response variable’s units. It tells how far that observed response is above or below its prediction. An \(r^2\) interpretation does not tell you the residual for any particular case.
Why “Correct” Needs a Separate Rule
For a numerical response, a prediction will rarely equal the observed value exactly. If “correct” means an exact match, even small rounding differences make a prediction incorrect. If “correct” means close enough for a practical purpose, the acceptable distance must be chosen in the response’s units. For example, a prediction might count as acceptable if it is within 2 centimeters of the observed value. That tolerance is a separate rule; it does not come from \(r^2\).
Once the rule is stated, you can count how many predictions satisfy it in a particular data set. But that percentage depends on the individual prediction errors and on the chosen tolerance. Changing the tolerance can change the count without changing the fitted line or its \(r^2\). So \(r^2\) cannot, by itself, give a percent of predictions that are “right.”
The residual standard deviation \(s\), discussed in “Standard Deviation of the Residuals, \(s\)” and “Using \(s\) to Describe Prediction Accuracy,” summarizes a typical residual size. It is also not a percentage of correct predictions. It can help describe the scale of prediction errors in context, but it does not tell you what fraction of cases falls within an arbitrary tolerance.
Worked Example: What Does \(r^2=0.64\) Mean?
A fictional school project uses the number of minutes students spend reviewing a lesson to predict their quiz scores. The reported coefficient of determination is \(r^2=0.64\). A student writes, “64% of the students’ quiz scores were predicted correctly.” Explain the error and give an accurate interpretation.
State. Decide whether the student’s statement matches the meaning of \(r^2\), then interpret the reported value in context.
Plan. Identify the response, which is quiz score, and the explanatory variable, which is review time. Use the percent-of-variation interpretation. Do not describe a count of students or claim that their individual predictions were correct.
Do. Convert the coefficient to a percentage: \(0.64(100\%)=64\%\). This percentage refers to variation in quiz scores, not to the number of students, the number of predictions, or a proportion of scores matched exactly.
Conclude. About 64% of the variation in students’ quiz scores is accounted for by the linear relationship with the number of minutes they spent reviewing. The student’s statement is incorrect because \(r^2\) does not tell us what percentage of students had correct predictions. To assess a chosen level of prediction accuracy, we would need the prediction errors and a definition of “correct.”
A Small Data Set Makes the Difference Visible
The distinction is not just a wording preference. A data set can have \(r^2=0.64\) even when none of its predictions is exact, or when every prediction is within a stated tolerance. Here is a fictional example that shows both possibilities in the same data.
Worked Example: Same \(r^2\), Different Accuracy Standards
A fictional experiment records daily hours of supplemental light, \(x\), and the increase in seedling height over a week, \(y\), in centimeters. For four seedlings, the fitted line is \(\hat{y}=6+x\). The observations are shown below. Use the residuals to compare exact matches with predictions within 2 centimeters, and verify the reported \(r^2\).
| Hours of light, \(x\) | Observed increase, \(y\) (cm) | Predicted increase, \(\hat{y}\) (cm) | Residual, \(y-\hat{y}\) (cm) |
|---|---|---|---|
| 6 | 13.5 | 12 | 1.5 |
| 2 | 9.5 | 8 | 1.5 |
| 6 | 10.5 | 12 | -1.5 |
| 2 | 6.5 | 8 | -1.5 |
State. Determine whether \(r^2=0.64\) describes the fraction of predictions that are exact or within 2 centimeters.
Plan. First calculate the residuals as observed minus predicted. Then count exact matches and cases with absolute residual at most 2 centimeters. To verify \(r^2\), use \(r^2=1-\frac{SSE}{SST}\), where \(SSE\) is the sum of squared residuals and \(SST\) is the total variation in the observed responses.
Do. The observed mean is \(\bar{y}=(13.5+9.5+10.5+6.5)/4=10\) centimeters. The total variation is \(SST=(13.5-10)^2+(9.5-10)^2+(10.5-10)^2+(6.5-10)^2=12.25+0.25+0.25+12.25=25\) square centimeters. The sum of squared residuals is \(SSE=1.5^2+1.5^2+(-1.5)^2+(-1.5)^2=9\) square centimeters. Therefore, \(r^2=1-\frac{9}{25}=0.64\), or 64% when expressed as a percentage. None of the four residuals is zero, so \(0/4=0\%\) of the predictions are exact. Each absolute residual is \(1.5\) centimeters, which is at most 2 centimeters, so \(4/4=100\%\) of the predictions are within 2 centimeters.
Conclude. In this example, \(r^2=0.64\), but 0% of the predictions are exact and 100% are within 2 centimeters of the observed increase. About 64% of the variation in seedling height increase is accounted for by its linear relationship with hours of supplemental light. The exact-match and within-2-centimeter percentages answer different questions and depend on the prediction errors and the chosen accuracy rule.
The calculations also show why you should keep the units straight. The residuals are measured in centimeters, while \(SSE\) and \(SST\) use squared centimeters. Their ratio, and therefore \(r^2\), is unitless. Neither \(r^2\) nor its percentage is an error measured in centimeters.
Correct the Typical Wrong Answers
A statement can be wrong in more than one way. Some answers treat \(r^2\) as a count of successful predictions. Others call it “accuracy” without saying what accuracy means, or claim it tells how close predictions are for individual cases. A careful correction replaces the unsupported claim with a statement about response variation and, when accuracy is relevant, says what additional information is needed.
Worked Example: Repair an “Accuracy” Claim
A fictional community garden uses weekly rainfall to predict the mass of vegetables harvested, in kilograms. The model has \(r^2=0.81\). A draft report says, “The model is 81% accurate, so it predicts the harvest correctly for 81% of weeks.” Rewrite the report’s claim and explain what information would be needed to assess prediction accuracy.
State. The report treats a coefficient of determination as a rate of correct predictions. That conclusion does not follow from \(r^2\).
Plan. Identify harvest mass as the response and weekly rainfall as the explanatory variable. Interpret \(r^2\) as a percentage of variation in harvest mass. To discuss a percent of weeks meeting an accuracy standard, specify a tolerance in kilograms and check the prediction errors for the weeks.
Do. Convert \(0.81\) to a percentage: \(0.81(100\%)=81\%\). The coefficient gives no count of weeks and no individual residuals. For instance, a separate accuracy question could ask what proportion of weeks have predictions within 5 kilograms of the observed harvest; that would require counting residuals whose absolute values are at most 5 kilograms.
Conclude. About 81% of the variation in weekly vegetable harvest mass is accounted for by its linear relationship with weekly rainfall. This does not mean that the model predicts the harvest correctly in 81% of weeks. To make a claim about weeks predicted within 5 kilograms, we would need the observed and predicted harvests, calculate their residuals, and count how many satisfy that tolerance.
Notice that a good correction does not simply replace “81% accurate” with “81% explained.” It names the response and explanatory variable and says the percentage concerns variation in the response. It also avoids implying that rainfall necessarily causes the harvest differences; \(r^2\) describes the linear relationship, not a cause.
Common Mistakes and AP Exam Tips
- Equating \(r^2\) with percent correct. Do not write “64% of predictions are correct” or “64% of cases are on the line.” A full-credit interpretation says that about 64% of the variation in the response is accounted for by its linear relationship with the explanatory variable.
- Using “accuracy” without defining it. For a numerical prediction, say what error would count as acceptable and give the tolerance in response units. Then use residuals to determine which predictions meet that rule.
- Assuming a high \(r^2\) guarantees small errors for every case. \(r^2\) summarizes overall response variation; it does not report the residual for an individual observation. Use residuals to discuss particular predictions.
- Calling a prediction “correct” because it is close. “Close” is not a mathematical rule until a tolerance is stated. Specify, for example, “within 2 centimeters,” rather than treating a vague judgment as a calculated percentage.
- Mixing up units. Residuals have the response’s units, but \(r^2\) is unitless. Do not report \(r^2\) as a number of centimeters, dollars, or other measurement units.
- Reversing the variables or implying cause. Name variation in the response and the linear relationship with the explanatory variable. Avoid causal words such as “causes” unless the study design supports a causal conclusion.
On an AP response, answer the question that was asked. If asked to interpret \(r^2\), give the percent-of-variation sentence in context. If asked about the proportion of predictions within a tolerance, use the residuals and that stated tolerance. Do not use one result as a substitute for the other.
Check Your Understanding
For each item, distinguish what \(r^2\) says from what would be needed to assess individual prediction accuracy.
- A model predicts the monthly water use of a household from the number of people living there. If \(r^2=0.49\), write an accurate interpretation. Does this mean that 49% of households were predicted correctly?
- A student says, “An \(r^2\) of 0.90 means predictions are within 1 liter for 90% of observations.” Explain what additional information this statement requires.
- For a numerical response, an observation has residual \(-3\) units. Is its prediction exact? Is it within 4 units of the observed value? Explain using the residual.
- In a fictional model of delivery time from route length, \(r^2=0.25\). Write the correct kind of interpretation, and name one claim about individual predictions that cannot be concluded from this value alone.
- Why could the percentage of predictions within a chosen tolerance change if the tolerance changes, even though the model’s \(r^2\) stays the same?