The following may (or may not) have occurred: The moment of inertia of a hollow cylinder was derived via an integral and the following expression was found:
$I_{hollow\ cylinder}\ =\ \frac{m}{2}(r_{inner}^2+r_{outer}^2)$
This is then compared to the moment of inertia of a solid cylinder by removing the term for the inner radius:
$I_{solid\ cylinder}\ =\ \frac{m}{2}(r_{outer}^2)$
and it is noted that the moment of inertia of the hollow cylinder is higher by the $+\ r_{outer}^2\ $ - term when compared to a solid cylinder. This finding is then underpinned by an experiment of a hollow and a solid cylinder rolling down an incline, showing exactly that: The moment of inertia is higher for the hollow cylinder.
The $+\ r_{outer}^2\ $ - term is the only difference between the two formulas and without further comment, we might have quietly (and erroneously) assumed the two cylinders in question are identical except the hollow cylinder has some mass removed radially, out from the center. A look at the ansatz, something like the following:
$\int\limits_{{cyl}}r^2 \ dm\ =\ \frac{m}{V_{cyl}}\iiint\limits_{V_{cyl}}r^2\ dV\ =\ ...$
shows there is no normalization to mass. These integrals really amount to the sum of all infinitesimal mass elements multiplied by their radii squared. So how could they possibly yield a higher value for a hollow cylinder than for the same object except the latter has less mass?
The answer is, of course, that a normalization to mass was done after the derivation, when just the $+\ r_{outer}^2$ - term was removed from the formula for $I_{hollow\ cylinder}$. This results in a higher density for the hollow cylinder because the volume was reduced but the mass was left unchanged. 15
This is only implicitly stated and might quite easily get overlooked.
For the case of "corresponding" cylinders, by which I mean the hollow cylinder has the same density, in addition to the same height and outer radius, the mass is lower and this lack of mass overcompensates the $+\ r_{outer}^2$ - term and it does so for all ratios $\frac{r_{inner}}{r_{outer}}$. Here is a Desmos-plot of the ratio of the respective moments of inertia for this case (i.e. same height, same outer radius, same and constant density $\rightarrow\ $lower mass of the hollow cylinder):
$\frac{I_{hollow\ cylinder}}{I_{solid\ cylinder}}=\frac{\frac{m_h}2\ (r_{outer}^2\ +\ r_{inner}^2)}{\frac{m_s}2\ r_{outer}^2}\ =\ \frac{\frac{1}2\ (height\ \cdot\ ρ\ \cdot\ 4π\ \cdot\ (r_{outer}^2\ -\ r_{inner}^2))\ \cdot\ (r_{outer}^2\ +\ r_{inner}^2)}{\frac{1}2\ (height\ \cdot\ ρ\ \cdot\ 4π\ \cdot\ (r_{outer}^2))\ \cdot\ (r_{outer}^2)}$
Note that here, the masses are named differently. This graph shows that, obviously, the moment of inertia is always lower for a hollow cylinder if nothing but the inner radius is changed and this applies analogously to a hollow vs. a solid sphere and other similar cases.
Of course, for objects with the same mass which is, on average, located at a higher distance to the rotation axis, the moment of inertia is higher and this was correctly stated in the beginning. But I think it is quite easy to miss that the hollow cylinder in the first comparison is not just a "carved-out version" of the solid cylinder. 15