We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this,

Step 1
Given that the sample size ${\displaystyle n=600}$, the favourable size ${\displaystyle x=192}$.
Let p denotes the proportion of adult residents who are parents in the country.
Thus, the value of p is given by ${\displaystyle p=\frac{x}{n}=\frac{192}{600}=0.32.}$
From the standard normal table, the Z value at $99\mathrm{\%}$ confidence interval is ${\displaystyle Z=2.576.}$
Substitute the known values in the formula ${\displaystyle E=Z\cdot \sqrt{\frac{p(1-p)}{n}}}$ to evaluate the margin of error E.
${\displaystyle E=Z\cdot \sqrt{\frac{p(1-p)}{n}}}$
${\displaystyle =2.576\cdot \sqrt{\frac{0.32(1-0.32)}{600}}}$
${\displaystyle \approx 0.049056852}$
Step 2
The confidence interval is given by ${\displaystyle (p-E,p+E).}$
Here, we have
${\displaystyle (p-E,p+E)=(0.32-0.049056852,0.32+0.049056852)}$
${\displaystyle =(0.270943147,0.369056852)}$
${\displaystyle \approx (0.271,0.369)}$
Hence, the confidence interval in tri-inequality form is ${\displaystyle 0.271<p<0.369.}$
The confidence interval suing point estimate 0.32 and margin of error ${\displaystyle E\approx 0.049isp=0.32\pm 0.049.}$