Best Multiple Regression Equation For the regression equation given in Exercise 1, the P-value is 0.000 and the adjusted \({R^2}\)value is 0.925. If we were to include an additional predictor variable of neck size (in.), the P-value becomes 0.000 and the adjusted\({R^2}\)becomes 0.933. Given that the adjusted \({R^2}\)value of 0.933 is larger than 0.925, is it better to use the regression equation with the three predictor variables of length, chest size, and neck size? Explain.

A regression equation is computed to predict the weight of a bear (inlb) using the predictor variables “weight”, “length,” and “chest size.”

The p-value and the adjusted\({R^2}\)value are 0.000 and 0.925, respectively.

Further, another predictor variable “neck size” is added to the equation, and the p-value and the adjusted \({R^2}\) are 0.000 and 0.933, respectively.

To identify the best regression equation, consider the equation with the highest value of adjusted\({R^2}\).

Here, the regression equation with two independent variables, “length” and “chest size”, has the adjusted \({R^2}\) value of 0.925. Moreover, the p-value for this regression is equal to 0.000, indicating that the regression is significant.

Further, another predictor variable “neck size” is added to the previous equation.

Now, the regression equation with three independent variables of “length”, “chest size” and “neck size” has a new adjusted \({R^2}\) value at 0.933.The new p-value is 0.000, which implies that the new regression is still significant.

Here,

\(\begin{array}{c}{\rm{New}}\;{\rm{Adjusted}}\;{R^2} > {\rm{Adjusted}}\;{R^2}\\0.933 > 0.925\end{array}\)

Since the new regression equation with the variable “neck size” has a larger value of adjusted\({R^2}\)and the regression is significant, the new regression equation is the best. It explains more variation in the response variable as compared to the equation with only two predictors.