Step 1

Given that the sample size $n=600$, the favourable size $x=192$.

Let p denotes the proportion of adult residents who are parents in the country.

Thus, the value of p is given by $p=\frac{x}{n}=\frac{192}{600}=0.32.$

From the standard normal table, the Z value at $99\mathrm{\%}$ confidence interval is $Z=2.576.$

Substitute the known values in the formula $E=Z\cdot \sqrt{\frac{p(1-p)}{n}}$ to evaluate the margin of error E.

$E=Z\cdot \sqrt{\frac{p(1-p)}{n}}$

$=2.576\cdot \sqrt{\frac{0.32(1-0.32)}{600}}$

$\approx 0.049056852$

Step 2

The confidence interval is given by $(p-E,p+E).$

Here, we have

$(p-E,p+E)=(0.32-0.049056852,0.32+0.049056852)$

$=(0.270943147,0.369056852)$

$\approx (0.271,0.369)$

Hence, the confidence interval in tri-inequality form is $0.271<p<0.369.$

The confidence interval suing point estimate 0.32 and margin of error $E\approx 0.049\text{}is\text{}p=0.32\pm 0.049.$