Question: A clinical trial test a method designed to increase the probability of conceiving a Gil. In the study 310 abuse were born, and 279 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.  Edit

Answer: Z value for 99 percentile from a Z table, which is 2.326
p is 279/310=0.9 and
q = (310-279)/310=0.1.
Standard Error = sqrt (0.9*0.1/310) =0.017
So, 99% confidence interval = 0.9±2.326*0.017
=[0.08604, 0.09395]  Edit