= Solution
Let $h=(b-a)/n$ and let $M_4=\max_{x\in[a,b]}|f^{(4)}(x)|$. A <composite Simpson's rule> panel covers two strips, so it has width $2h$ and absolute error at most
$$
\frac{(2h)^5}{2880}M_4=\frac{h^5}{90}M_4.
$$
There are $n/2$ panels. By the <triangle inequality>, the absolute error of their sum is at most the sum of their absolute errors. Hence
$$
\boxed{|I-S_n|\leq\frac n2\frac{h^5}{90}M_4
=\frac{(b-a)^5}{180n^4}M_4.}
$$
This assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth <derivative> is bounded on the interval. It is the <error bound for composite Simpson's rule>, expressed in strip count rather than panel count.<NIST Digital Library of Mathematical Functions/3.5 Quadrature, subsection (ii) Simpson’s Rule>[NIST DLMF §3.5(ii), equations 3.5.6–3.5.8: elementary and composite Simpson error terms]{ref}
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