Question 1092099
The population proportion estimate is 200/400 or 0.50.  The sample is the best estimate available.
98%CI interval width is z(0.99)*sqrt {p*(1-p)/n}
+/-2.33*0.5/sqrt(n), where n=400, so it is +/-2.33*0.5/20=+/-0.058. The interval is on both sides of the sample proportion mean.
(0.442, 0.558).
The basic rule here is 1/sqrt (n) is about the error so this is 1/ sqrt(400) or 1/20 or 0.05.