# How can adjusted r squared be negative?

${R}^{2}$, the coefficient of multiple determination, is defined as $\frac{S {S}_{R E G}}{S {S}_{T O T A L}}$ or, equivalently, $1 - \frac{S S E}{S S T O}$. ${R}^{2}$ measures the proportionate reduction in variation of Y, associated with the set of X predictors. ${R}^{2}$ will be inflated as more X variables are added. The adjusted ${R}^{2}$ was therefore derived, as ${R}_{a \mathrm{dj}}^{2} = 1 - \left\{\left[\frac{n - 1}{n - p}\right] \left[\frac{S S E}{S S T O}\right]\right\}$.
${R}_{a \mathrm{dj}}^{2} = 1 - \left(1.18\right) \frac{S S E}{S S T O}$. If the ratio $\frac{S S E}{S S T O}$ is close enough to 1, then you can see how the ${R}_{a \mathrm{dj}}^{2}$. can be negative. [In which case it can be interpreted as zero.]