Slope of a Regression Line vs Coefficient of Determination

In $\hat{y} = a + bx$ the slope $b$ is the change in the predicted response for a one-unit increase in the explanatory variable. The word predicted is load-bearing: $b$ describes how the line moves, not what happens to any individual. The slope also carries units, always units of $y$ per unit of $x$, so a bare number is an incomplete answer.

Six students study 1, 2, 3, 4, 5 and 6 hours and score 60, 72, 65, 78, 71 and 86. The least-squares line is $\hat{y} = 58 + 4x$, and $b = 4$ points per hour. Said in context: for each additional hour of study, the model predicts a score about 4 points higher. It comes from $b = r \cdot \frac{s_y}{s_x}$, here $0.8107 \times \frac{9.2304}{1.8708}$, which is why $b$ and the correlation $r$ always share a sign but rarely share a size.

The interpretation that loses credit is "each extra hour of study raises a student's score by 4 points." It makes two claims the line cannot support. Raises is causal, and these are observational data where a student who studies more may differ in other ways. And it promises 4 points to an individual, while the six real students sit as much as 7 points off the line. Predicted change, not guaranteed change.

The slope is not symmetric in the two variables, which surprises people. Regress hours on score instead of score on hours and the slope is 0.1643 hours per point, not the reciprocal $1/4 = 0.25$. The two slopes multiply to $4 \times 0.1643 = 0.657$, which is $r^2$. The correlation, by contrast, is 0.8107 whichever variable you call explanatory. Rescaling changes the slope too: measure study time in minutes and $b$ becomes $4/60$ points per minute while $r$ does not budge.

$r^2$ (read r squared) is the proportion of the variation in the response variable $y$ that is explained by the linear relationship with the explanatory variable $x$. Equivalently, it measures how much of the squared prediction error is removed by using the least-squares line instead of predicting $\bar{y}$ for everyone:

Six students study 1, 2, 3, 4, 5 and 6 hours and score 60, 72, 65, 78, 71 and 86. The least-squares line is $\hat{y} = 58 + 4x$. The scores have mean 72, and their squared deviations from 72 sum to 426. The line leaves residuals of -2, 6, -5, 4, -7 and 4, whose squares sum to 146. So $r^2 = 1 - 146 / 426 = 0.657$: about 66 percent of the variation in scores is explained by the linear relationship with study time, and the other 34 percent is not.

The correlation for these data is $r = 0.8107$, and squaring it returns the same 0.657, which is where the name comes from. Squaring also throws away the sign, so a data set with $r = -0.8107$ would report the identical $r^2 = 0.657$. The statistic reports strength of linear fit and never direction, so quote $r$ alongside it whenever direction matters.

What 66 percent does not mean: that 66 percent of the points lie on the line, that study time causes 66 percent of the score, or that predictions from the line are 66 percent accurate. It is a statement about variation and about nothing else.

An $r^2$ close to 1 also does not confirm that a line was the right model. A curved relationship can still produce a high $r^2$ while the residual plot shows an obvious arc, and it is that plot, not the size of $r^2$, that decides whether a linear model belongs on the data. $r^2$ is also unitless, which makes it easy to compare across settings but useless for saying how wrong a single prediction is likely to be. The standard deviation of the residuals answers that second question, in the units of $y$, and the two statistics are usually reported together.