1. The problem asks to express $\left(\frac{y}{2}\right)^{-4}$ in the form $a y^n$ where $a$ and $n$ are integers. 2. Recall the rule for negative exponents: $x^{-m} = \frac{1}{x^m}$. 3. Apply the negative exponent rule: $$\left(\frac{y}{2}\right)^{-4} = \frac{1}{\left(\frac{y}{2}\right)^4}$$ 4. Rewrite the denominator: $$\left(\frac{y}{2}\right)^4 = \frac{y^4}{2^4} = \frac{y^4}{16}$$ 5. Substitute back: $$\frac{1}{\frac{y^4}{16}} = \frac{16}{y^4}$$ 6. Express as $a y^n$: $$16 y^{-4}$$ 7. Therefore, $a = 16$ and $n = -4$. Final answer: $16 y^{-4}$