= Solution
The null model's 99 residual <degrees of freedom> imply $n=100$. Each predictor has a one-degree-of-freedom linear contribution in the parametric table, plus the reported nonlinear contribution. Consequently the termwise <effective degrees of freedom> are
$$
\boxed{\mathrm{edf}_{\mathrm{experience}}\approx1+8.8=9.8,\qquad
\mathrm{edf}_{\mathrm{education}}\approx1+0.4=1.4.}
$$
Include the intercept once, giving \b[about $12.2$ total effective <degrees of freedom>]. The more precise residual <degrees of freedom> in the output give
$$
\boxed{\mathrm{edf}_{\mathrm{total}}=100-87.7644=12.2356.}
$$
There is no conflict: $8.8$ and $0.4$ are rounded, so they cannot recover the precise total. Individual unrounded termwise values are not provided.
The package's residual mean-square estimate of error <variance> is
$$
\boxed{\widehat\sigma^2=\frac{906437.7}{87.7644}\approx10328.08,}
$$
matching the rounded $10328$ in the table. The nonlinear experience contribution is highly significant, while the smaller nonlinear education contribution has approximate $p=0.03923$.
This calculation uses the effective residual <degrees of freedom> reported by this fit. For a general non-idempotent <linear smoother> $S$, the noise contribution to expected RSS is $\sigma^2\operatorname{tr}[(I-S)^T(I-S)]$, and the expectation can additionally contain squared smoothing bias. Thus dividing by $n-\operatorname{tr}(S)$ is not a universal exact-unbiasedness theorem. The output's separation of linear and nonlinear term <degrees of freedom> is documented at https://cran.r-project.org/web/packages/gam/gam.pdf.
Back to article page
