Out of 200 people sampled, 52 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

0.180144 < p < 0.339856

Step-by-step explanation:

According to the Question,

⇒ standard error of the sample is square root(0.26 * (1-0.26) / 200) = 0.031

⇒alpha(a) = 1 - 99/100 = 0.01

⇒critical probability(p*) = 1 - a/2 = 0.995

⇒assuming a normal distribution, look for the z-score associated with 0.995 cumulative probability , z-score = 2.576

⇒margin of error(ME) = 2.576 * 0.031 = 0.079856

0.180144 < p < 0.339856

i think the answer would be c