The printed values appear to favor model 2 because $-12.529<336.381$. They are not directly comparable: model 2 reports the Gaussian likelihood of $Z_i=\log T_i$, whereas model 3 reports a density for $T_i$. The <change-of-variables formula for a probability density> gives
$$
\ell_T=\ell_Z-\sum_{i=1}^{35}\log T_i,
$$
so the transformed model's AIC on the original response scale is
$$
\operatorname{AIC}_{2,T}
=\operatorname{AIC}_{2,Z}+2\sum_i\log T_i.
$$
Because the standardized predictors have zero sample means and the ordinary-least-squares residuals sum to zero,
$$
\sum_i\log T_i=35\widehat\alpha_0=35(4.99902).
$$
Therefore
$$
\operatorname{AIC}_{2,T}
=-12.52943+70(4.99902)=337.402,
$$
which is slightly worse than model 3's $336.381$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 218 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-218/3/d/solution