One can interpret r = 0.60 as follows:

(a) A60% reduction in error occurs in using x to predict y.

(b) A 9% reduction in error occurs in using x to predict y compared to using ¯y to predict y.

(c) 9% of the time ˆy = y.

(d) y changes 0.60 units for every one-unit increase in x.

(e) When x predicts y, the average residual is 0.6.

(f) x changes exactly 0.60 standard deviations when y changes one standard deviation.