#### Answer

C

#### Work Step by Step

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