# General linear test

[ "article:topic", "authorname:pauld" ]

We are interested in testing for the dependence on the predictor variable from a different viewpoint. We call the model

$$Y_i = \beta_0+\beta_1X_i+\epsilon_i$$

the full model. We want to test $$H_0 : \beta_1 = 0$$ against $$H_1 : \beta_1 \neq 0$$. Under $$H_0 : \beta_1 = 0$$, we have the reduced model:

$$Y_i = \beta_0 + \epsilon_i$$

Under the full model, $$SSE_{full} = \sum_i(Y_i - \hat{Y_i})^2 = SSE$$. Under the reduced model $$SSE_{red} = \sum_i(Y_i - \hat{Y_i})^2 = SSTO$$

### General structure of test statistic

Observe that d.f.$$(SSE_{full}) = n-2$$, d.f.$$(SSE_{red}) = n-1$$ and $$SSE_{red} - SSE_{full} = SSR$$.

$$F^\ast =\frac{\frac{SSE_{red}-SSE_{full}}{d.f.(SSE_{red})-d.f.(SSE_{full})}}{\frac{SSE_{full}}{d.f.(SSE_{full})}} = \frac{\frac{SSR}{d.f.(SSR)}}{\frac{SSE}{d.f.(SSE)}}=\frac{MSR}{MSE}$$

Under normal error model, and under $$H_0 : \beta_1 = 0, F^\ast$$ has the $$F$$ distribution with (paired) degress of freedom $$(d.f.(SSE_{red}) - d.f.(SSE_{full}), d.f.(SSE_{full}))$$.

### Descriptive measure of association between $$X$$ and $$Y$$

Define the coefficient of determination:

$$R^2 = \frac{SSR}{SSTO}= 1-\frac{SSE}{SSTO}$$

Observe that $$0\leq R^2\leq 1$$, and the correlation coefficient, Corr$$(X, Y)$$ between $$X$$ and $$Y$$ is the (signed) square root of $$R^2$$. That is (Corr$$(X, Y))^2 = R^2$$. Larger value of $$R^2$$ generally indicates higher degree of linear association between $$X$$ and $$Y$$. Another (and considered better) measure of association is the adjusted coefficient of determination :

$$R^2_{ad} =1- \frac{MSE}{MSTO}$$

$$R^2$$ is the proportion of variability in $$Y$$ explained by its regression on $$X$$. Also, $$R^2$$ is unit free, i.e. does not depend on the units of measurements of the variables $$X$$ and $$Y$$.

For the housing price data, $$SSR = 352.91, SSTO = 556.08, n = 19$$, and hence $$SSE = 203.17$$,  $$d.f.(SSE) = 17$$,  $$d.f.(SSTO) = 18$$. So, $$R^2 = \frac{352.91}{556.08} = 0.635$$ and $$R^2_{ad} = 1-\frac{11.95}{30.8} = 0.613$$.

Cathy Wang