#### Answer

$0.47\leq p\leq0.57$

#### Work Step by Step

We have $n=600, \hat p=312/600=0.52, \hat q=0.48$
With c=0.99, $Z_{\alpha/2}=2.58$
the margin of error is $E=2.58\times\sqrt \frac{0.52\times0.48}{600}=0.05$
the confidence interval is $(0.52-0.05, 0.52+0.05)$
The true proportion of married Americans with 99% confidence is $0.47\leq p\leq0.57$