Thermal resistance of anodized aluminium interface
The coating (on both sides) has a thickness of about 50um or less (according e.g. to this paper) and a thermal conductivity of 0.5 ... 1.5 W/Km (let's use 1.0 W/Km in the simple calculation below). 

The extruded aluminium housing has a thickness of 1.5mm and a thermal conductivity of up to 200 W/Km. 

The thermal grease layer has a more or less undefined thickness (also changing over time) in the range of 50 - 100um. As a rule of thumb designers often set the layer thickness to 100um and its thermal conductivity to 1 W/Km. 

This gives a thermal resistance power module "case to sink" of 

$$
\begin{align}
R_{th} &amp;=
{2 \cdot \num{50e-6} \over A}
 + {\num{1.5e-3} \over 200 \cdot A}
 + {2 \cdot \num{100e-6} \over A}
\\
&amp;= {\num{100e-6} + \num{7.5e-6} + \num{200e-6} \over A} \\
&amp;= {\num{0.3e-3} \over A}
\end{align}$$

with \$A\$ being the semiconductor module case size (footprint) in square-meters.

Without the housing (power module directly on heatsink) your actual thermal resistance is about 3 times smaller: 

$$ R_{th} = {\num{100e-6} \over A} = {\num{0.1e-3} \over A} $$