#### Answer

A

#### Work Step by Step

We first write the equation in the form $y=a \sin [b(x-d)]$. Therefore, $y=-2\sin (4x-3)$ becomes $y=-2\sin [4(x-\frac{3}{4})]$.
Comparing the two equations, $a=-2,b=4$ and $d=\frac{3}{4}$.
The amplitude is $|a|=|-2|=2.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{4}=\frac{\pi}{2}$.
The phase shift is $|d|=|\frac{3}{4}|=\frac{3}{4}.$
Therefore, the description in option A is the correct answer.