Two Summaries of a Linear Association
In “Effect of an Outlier on r-squared,” you saw that an observation can change \(r^2\), the coefficient of determination. This tutorial compares \(r^2\) with the correlation \(r\). They are closely connected: in simple linear regression, \(r^2\) is the square of \(r\). But they emphasize different features of an association, so it matters which one you use and how you interpret it.
Correlation \(r\) has a sign and a magnitude. Its sign describes direction: positive \(r\) indicates a positive linear association, and negative \(r\) indicates a negative linear association. The magnitude, \(|r|\), describes how closely the data follow a straight-line pattern. The closer \(|r|\) is to 1, the stronger the linear association; a value closer to 0 indicates a weaker linear association.
The coefficient of determination \(r^2\) has no sign. Squaring removes the direction information, but \(r^2\) describes the fraction of variation in the response that is accounted for by its linear relationship with the predictor. As you learned in “Interpreting r-squared in Context,” report this as a percentage of response variation in context, not as a percentage of predictions that are correct.
Because squaring preserves the order of nonnegative values, comparing \(r^2\) values gives the same ranking of linear-association strength as comparing \(|r|\) values. For example, if one data set has \(|r|=0.8\) and another has \(|r|=0.5\), their \(r^2\) values are 0.64 and 0.25. The first association is stronger by either comparison. But \(r^2\) does not tell you whether either association is positive or negative.
Neither summary tells the whole story of a data set. As discussed in “What r-squared Does Not Tell You,” a numerical summary does not show the shape of the scatterplot or establish that a linear model is appropriate. Use the plot and context as well as the number when describing an association.
When Each Summary Is More Useful
The best summary depends on the question. If you need to describe whether the response tends to increase or decrease as the predictor increases, \(r\) is more useful because its sign gives the direction. If the question asks how much of the response variation is accounted for by a linear relationship, \(r^2\) is more useful because it directly represents that fraction.
For a question about strength alone, either can help, provided you compare \(r\) by its magnitude, not by its signed value. For instance, \(-0.9\) is a stronger linear association than \(0.7\), because \(0.9\) is greater than \(0.7\) in magnitude. Their \(r^2\) values, 0.81 and 0.49, also show that the first has the stronger linear association. Comparing the signed values \(-0.9\) and \(0.7\) as ordinary numbers would give a misleading answer about strength.
The summaries use different kinds of interpretation. Correlation is unitless and describes direction and strength of a linear pattern. The \(r^2\) interpretation names the response variable and describes variation in that response. Neither describes a cause-and-effect relationship just because the association is strong.
If it asks for direction, you need a signed summary such as \(r\). If it asks about the fraction of response variation accounted for, use \(r^2\).
Compare \(|r|\) values or compare their \(r^2\) values. Do not treat a more negative \(r\) as weaker just because it is numerically smaller.
For \(r\), name the direction and strength of the linear association. For \(r^2\), name the response and the percentage of its variation accounted for by the linear relationship.
Worked Examples: Choosing Between r and r-squared
Worked Example: Calculate Both Summaries
A fictional school counselor records the number of practice sessions \(x\) and a student’s score \(y\) on a short skills assessment for five students. The data are \((1,2),(2,3),(3,5),(4,4),(5,6)\). Calculate \(r\) and \(r^2\), then explain what each tells us.
State. The predictor is number of practice sessions, and the response is assessment score. We want to describe both the direction and strength of the linear association, and the fraction of variation in scores accounted for by the linear relationship with practice sessions.
Plan. Calculate the correlation from the centered sums \(S_{xx}\), \(S_{yy}\), and \(S_{xy}\), then square \(r\). This calculation summarizes the linear association in these five observations; it does not establish that practice caused the scores.
Do. The means are \(\bar{x}=3\) and \(\bar{y}=4\). The deviations from the means are \(-2,-1,0,1,2\) for \(x\), and \(-2,-1,1,0,2\) for \(y\). Therefore,
Similarly, \(S_{yy}=(-2)^2+(-1)^2+1^2+0^2+2^2=10\), and \(S_{xy}=(-2)(-2)+(-1)(-1)+(0)(1)+(1)(0)+(2)(2)=9\).
The positive sign indicates a positive linear association. Squaring gives \(r^2=(0.9000)^2=0.8100\). As a check, the denominator in the correlation calculation is \(\sqrt{100}=10\), so \(9/10=0.9\), and \(0.9\times0.9=0.81\).
Conclude. For these five students, there is a strong positive linear association between practice sessions and assessment score: students with more sessions tended to have higher scores. About 81% of the variation in their assessment scores is accounted for by the linear relationship with practice sessions. These summaries describe association in the observed data; they do not show that more practice alone caused higher scores.
Worked Example: Compare Strength When Directions Differ
Consider two fictional data sets. In a set of athlete observations, weekly training time and sprint time have correlation \(r=-0.80\). In a separate set of consumer observations, the number of product features and time needed to learn a device have correlation \(r=0.60\). Compare the strength and direction of the linear associations using both \(r\) and \(r^2\).
State. The first association is negative and the second is positive, as indicated by the signs of their correlations. We will compare their strength using the magnitudes of \(r\), then calculate \(r^2\) for each.
Plan. For correlation, compare \(|-0.80|\) with \(|0.60|\), not \(-0.80\) with \(0.60\) as signed numbers. For the coefficients of determination, square each correlation and interpret each percentage for its own response variable.
Do. The magnitudes are \(0.80\) and \(0.60\), so the training-time and sprint-time association is stronger. The coefficients of determination are
The calculations agree on the strength comparison: 0.6400 is greater than 0.3600. The negative correlation in the first setting indicates that greater weekly training time tends to be associated with lower sprint times. The positive correlation in the second setting indicates that more product features tend to be associated with longer learning times.
Conclude. The first data set has a stronger linear association because \(|r|=0.80\) is greater than \(|r|=0.60\), and its \(r^2\) is also larger. About 64% of the variation in sprint times is accounted for by the linear relationship with weekly training time. About 36% of the variation in learning times is accounted for by the linear relationship with number of product features. The \(r^2\) values compare strength but do not convey direction; the signs of \(r\) do.
Worked Example: Same r-squared, Different Directions
Two fictional neighborhood projects examine the relationship between outdoor temperature and daily energy use. In one setting, the response is heating energy use, with reported \(r^2=0.49\) and a negative fitted slope. In the other, the response is cooling energy use, with \(r^2=0.49\) and a positive fitted slope. Find \(r\) for each association and explain why both summaries are useful.
State. Both associations have the same reported \(r^2\), so their linear strengths are the same. Their directions may differ, and the fitted slope signs provide the direction needed to determine the sign of each correlation.
Plan. As covered in “Finding r From r-squared and the Slope Sign,” take the square root of \(r^2\) to obtain the magnitude of \(r\), then use the slope sign to select the sign of \(r\). Finally, interpret \(r\) and \(r^2\) in their respective contexts.
Do. The magnitude of each correlation is
The fitted slope for heating energy use is negative, so its correlation is \(r=-0.70\). The fitted slope for cooling energy use is positive, so its correlation is \(r=0.70\). Squaring either value checks the reported coefficient of determination: \((-0.70)^2=0.4900\) and \((0.70)^2=0.4900\).
Conclude. Both settings have the same strength of linear association because the correlations have equal magnitudes, \(0.70\), and both \(r^2\) values are 0.49. The negative \(r\) indicates that heating energy use tends to decrease as outdoor temperature increases; the positive \(r\) indicates that cooling energy use tends to increase as outdoor temperature increases. In each setting, about 49% of the variation in the named energy-use response is accounted for by its linear relationship with outdoor temperature. Reporting only \(r^2\) would hide the contrast in direction.
Common Mistakes and AP Exam Tips
- Using \(r^2\) to state direction. Squaring removes the sign. Say whether the association is positive or negative using \(r\) or the direction of the fitted line, not \(r^2\).
- Comparing signed correlations to judge strength. Compare \(|r|\). A correlation of \(-0.9\) represents a stronger linear association than \(0.7\), not a weaker one.
- Calling \(r^2\) a percentage of accurate predictions. State the percentage of variation in the response accounted for by its linear relationship with the predictor.
- Assuming equal \(r^2\) means identical relationships. Equal values indicate equal linear-association strength, but the directions may be opposite and the contexts may differ.
- Making a cause-and-effect claim. A large \(|r|\) or \(r^2\) describes an association. It does not, by itself, show that changes in the predictor cause changes in the response.
A full-credit response matches the summary to the question and uses accurate context. For \(r\), name the direction and describe the strength of the linear association. For \(r^2\), identify the response variable and state what fraction or percentage of its variation is accounted for by the linear relationship with the predictor. If comparing strength using \(r\), compare magnitudes.
Check Your Understanding
Use the distinction between correlation and coefficient of determination to answer these questions.
- A data set has \(r=-0.75\). What are the direction and strength of its linear association? What is \(r^2\)?
- Which association is stronger: one with \(r=-0.85\) or one with \(r=0.70\)? Explain using the appropriate comparison.
- Two settings both have \(r^2=0.36\), but one has positive \(r\) and the other negative \(r\). What do the values have in common, and what differs?
- Write a contextual interpretation of \(r^2=0.64\) for a study relating outdoor temperature to household water use.
- A question asks whether a response tends to increase or decrease as a predictor increases. Which summary is more directly useful, \(r\) or \(r^2\), and why?