Here are plots of data for Studentized residuals against Length. Here is the same regression with all of the points at 70 removed.Dependent variable is: Weight30 total bears of which 10 are missingR-squared = 97.8% R-squared (adjusted)= 97.3%s = 2.96 with 20 - 4 = 16 degrees of freedom Sum of Mean Source Squares DF Square F-ratio Regression 7455.0 3 2485 238.26 Residual 166.89 16 10.43 \begin{array} { l r r r r } & ext { Sum of } & & { ext { Mean } } \ ext { Source } & ext { Squares } & ext { DF } & ext { Square } & ext { F-ratio } \ ext { Regression } & 7455.0 & 3 & 2485 & 238.26 \ ext { Residual } & 166.89 & 16 & 10.43 & \end{array} Source Regression Residual Sum of Squares 7455.0 166.89 DF 3 16 Mean Square 2485 10.43 F-ratio 238.26 Variable Coefficient SE(Coeff) t-ratio P-value Intercept − 169.16 3.23 − 52.37

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Here are plots of data for Studentized residuals against Length. $Here are plots of data for Studentized residuals against Length. Here is the same regression with all of the points at 70 removed. Dependent variable is: Weight 30 total bears of which 10 are missing R-squared = 97.8% R-squared (adjusted)= 97.3% s = 2.96 with 20 - 4 = 16 degrees of freedom \begin{array} { l r r r r } & \text { Sum of } & & { \text { Mean } } \\ \text { Source } & \text { Squares } & \text { DF } & \text { Square } & \text { F-ratio } \\ \text { Regression } & 7455.0 & 3 & 2485 & 238.26 \\ \text { Residual } & 166.89 & 16 & 10.43 & \end{array} \begin{array} { l r c r r } \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\ \text { Intercept } & - 169.16 & 3.23 & - 52.37 & < 0.0001 \\ \text { Chest } & 0.84 & 0.58 & 1.45 & 0.1590 \\ \text { Length } & 5.59 & 2.14 & 2.61 & 0.0148 \\ \text { Sex } & - 1.19 & 1.98 & - 0.60 & 0.5537 \end{array} Compare the regression with the previous one.In particular,which model is likely to make the best prediction of weight? Which seems to fit the data better?$ Here is the same regression with all of the points at 70 removed.
Dependent variable is: Weight
30 total bears of which 10 are missing
R-squared = 97.8% R-squared (adjusted)= 97.3%
s = 2.96 with 20 - 4 = 16 degrees of freedom

166.891610.43\begin{array} { l r r r r } & \text { Sum of } & & { \text { Mean } } \\\text { Source } & \text { Squares } & \text { DF } & \text { Square } & \text { F-ratio } \\\text { Regression } & 7455.0 & 3 & 2485 & 238.26 \\\text { Residual } & 166.89 & 16 & 10.43 &\end{array}

−1.191.98−0.600.5537\begin{array} { l r c r r } \text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \\\text { Intercept } & - 169.16 & 3.23 & - 52.37 & < 0.0001 \\\text { Chest } & 0.84 & 0.58 & 1.45 & 0.1590 \\\text { Length } & 5.59 & 2.14 & 2.61 & 0.0148 \\\text { Sex } & - 1.19 & 1.98 & - 0.60 & 0.5537\end{array}

Compare the regression with the previous one.In particular,which model is likely to make the best prediction of weight? Which seems to fit the data better?

Angel Donaire · Accepted Answer

Final Answer : Omitting the values changes the coefficients of Chest and Length significantly.Chest is no longer a large factor in weight and Length is a much larger factor.The second regression model has a higher R-squared,suggesting that it fits the data better.Without the values at 70,the second regression is probably the better model.

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