Answer
$0.41\leq p\leq 0.53$
Work Step by Step
Given $n=400, \hat p=189/400=0.4725, \hat q=0.5275, c=0.98$
we have $Z_{\alpha/2}=2.33$ and $E=2.33\times\sqrt \frac{0.4725\times0.5275}{400}=0.058$
The confidence interval is $(0.4725-0.058, 0.4725+0.058)$
The true proportion of adults participating in some kind of formal
education program with 98% confidence is $0.41\leq p\leq 0.53$
