Contradiction between scalar curvature and volumes on pseudo-Riemannian manifolds

I have trouble understanding how the curvature and volume work in pseudo-Riemannian geometry. Something I will say should be wrong. Let $(mathcal{M},g)$ be a pseudo Riemannian manifold of dimension $n$ and $B(epsilon)subset mathcal{M}$ be the ball of radius $epsilon$ centered on a point $pin mathcal{M}$. The wikipedia on scalar curvature gives
$$ frac{vol(B(epsilon))}{vol(B_{mathbb{R}^n}(epsilon))} = 1 – frac{S}{6(n+2)}epsilon^2+ O(epsilon^4),$$
where $S=g^{ij}R_{ij}$, $R_{ij}$ being the Ricci tensor, is the scalar curvature at $p$. I am confused by the change of metric $grightarrow -g$.
-The Ricci tensor is defined from the curvature tensor, which can be defined from the Levi-Civita connection. $grightarrow -g$ does not change the connection so it does not change the Ricci tensor.
-Hence the sign of the scalar curvature should change.
-But on the other hand the volume of balls should not change, since in a coordinate system the volume form is defined as
$$omega = sqrt{|det(g)|}dx_1 wedge … wedge dx_n.$$
This seems to be in contradiction with the above formula. Does anyone see what’s wrong in what I said?

Thank you for your help.