Betti numbers change after homogenization

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I'm trying to understand the following sentence: "Betti numbers usually change after homogenization". I understood it as follows, and I hope you can tell me if there is another interpretation or if I am right.

Take a homogeneous ideal $I$ in a positively graded polynomial ring $R=k[x_1,\dots,x_n]$, where $\deg(x_i)>0$ for all $i$. Then drop the degrees of each variable to $1$, and consider the standard graded ring $S=k[x_1,\dots,x_n]$ where $\deg(x_i)=1$ for all $i$. Now $I$ is not homogeneous in $S$, in general, so add a new variable $w$ of degree $1$ and consider in the standard graded ring $T=k[x_1,\dots,x_n,w]$ the homogenization $$I^\mathrm{hom}=(f^\mathrm{hom}\mid f\in I).$$ My interpretation is: The Betti numbers of $I$ are different from the Betti numbers of $I^\mathrm{hom}$.

I made the following example. I took the ring $R=k[x,y,z]$ with $$\deg(x)=1,\quad\deg(y)=2,\quad\deg(z)=3$$ and the homogeneous ideal $I=(x^3-z,x^2-y)$. Then a computer algebra system tells me the Betti table of $R/I$ is $$\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right).$$ In the standard graded ring $T=k[x,y,z,w]$ I consider the ideal $I^\mathrm{hom}$ and I find that the Betti table of $T/I^\mathrm{hom}$ is $$\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 3 & 2 \end{array}\right).$$ Is this the meaning of the sentence I read? Thank you very much in advance!