Questions: 31 point Predictor Coef SE Coef T P Constant 0.07 8.30 1.7 0.1 graduation 1.7 0.0011 7.8 0.000 S=11.50 R-Sq =49 % R-Sq(adj)=49.1 %

Solution Steps

To solve this question, we need to interpret the coefficients and statistical values provided in the table. The "Coef" column represents the coefficients of the predictors, the "SE Coef" column represents the standard error of the coefficients, the "T" column represents the t-values, and the "P" column represents the p-values. Additionally, the values $S=11.50$, R-Sq $=49 \%$, and R-Sq (adj) $=49.1 \%$ are related to the overall model fit.

We can see that the predictor "graduation" has a coefficient of 1.7, a very low standard error of 0.0011, a high t-value of 7.8, and a very low p-value of 0.000. This indicates that the predictor "graduation" is statistically significant in predicting the response variable.

The constant term has a coefficient of 0.07, a standard error of 8.30, a t-value of 1.7, and a p-value of 0.1. This suggests that the constant term may not be statistically significant in the model.

The values $S=11.50$, R-Sq $=49 \%$, and R-Sq (adj) $=49.1 \%$ provide information about the overall model fit, with $R^2$ indicating the proportion of variance explained by the model.

The coefficient for the constant term is \(0.07\) with a standard error of \(8.3\). The coefficient for the predictor "graduation" is \(1.7\) with a standard error of \(0.0011\).

The t-value for the constant term is \(1.7\) with a p-value of \(0.1\). The t-value for the predictor "graduation" is \(7.8\) with a p-value of \(0.0\).

The model has a residual standard error of \(S = 11.5\), an \(R^2\) value of \(49\%\), and an adjusted \(R^2\) value of \(49.1\%\).

Final Answer