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Statistics LibreTexts

General linear test

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