In the context of least-squares regression, the slope (denoted as \( b_{v1} \) in your case) indicates the change in the predicted value of the dependent variable (usually denoted \( y \)) for a one-unit increase in the independent variable (usually denoted \( x \)).
Given that the slope \( b_{v1} = 5 \), if you have two \( x \)-values that differ by 2, the predicted difference in the \( y \)-values associated with those \( x \)-values can be calculated using the slope of the regression line.
The formula for the predicted difference in the \( y \)-values (\( \Delta y \)) when the \( x \)-values differ by a certain amount (\( \Delta x \)) is:
\[ \Delta y = b_{v1} \times \Delta x \]
Substituting the given values:
\[ \Delta y = 5 \times 2 = 10 \]
Therefore, the predicted difference in the \( y \)-values corresponding to the two \( x \)-values that differ by 2 is 10.