EXAMPLE 3 Using a regression equation

In Example 1, we used the “up-and-over” method in Figure 15.1 to predict the humerus length for a fossil whose femur length is 50 cm. The equation of the least-squares line is

humerus length = −3.66 + (1.197 × femur length)

The slope of this line is b = 1.197. This means that for these fossils, humerus length goes up by 1.197 cm when femur length goes up 1 cm. The slope of a regression line is usually important for understanding the data. The slope is the rate of change, the amount of change in the predicted y when x increases by 1.

344

The intercept of the least-squares line is a = −3.66. This is the value of the predicted y when x = 0. Although we need the intercept to draw the line, it is statistically meaningful only when x can actually take values close to zero. Here, femur length 0 is impossible (recall that the femur is a bone in the leg), so the intercept has no statistical meaning.

To use the equation for prediction, substitute the value of x and calculate y. The predicted humerus length for a fossil with a femur 50 cm long is

humerus length = −3.66 + (1.197)(50)

= 56.2 cm

To draw the line on the scatterplot, predict y for two different values of x. This gives two points. Plot them and draw the line through them.