Question

Wild bears were anesthetized, and their bodies were measured and weighed. One goal of the study was to make a table (or perhaps a set of tables) for hunters, so they could estimate the Weight of a bear based on other measurements. We are particularly interested in predicting the Weight (pounds) of a bear from its Length (inches). The data file of Weights and Lengths of 143 bears is provided separately. Take the base ten logarithms of each value of Weight. We will label the transformed variable logten_Weight (I did it).

i)

1. Find the percent of the variation in the logten_Weight variable that is explained by the regression line. Does the regression model provide a good fit? Explain it (r= 0.9377764)

Answer 1

percent of the variation in the logten_Weight variable that is explained by the regression line is the coefficient of determination. It is denoted as r squared or r^2

Coefficient of determination = r^2

given that r = 0.9377764 (strong positive relationship)

this gives

Coefficient of determination = (0.9377764)^2

= 0.8794

converting to %, 0.8794*100 = 87.94%

Therefore, we can say that 87.94 percent of the variation in the logten_Weight variable is explained by the regression line.

Looking at the coefficient of determination (which is very high) , we can say that the regression model provide a good fit