Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 320 babies were​ born, and 272 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

Answer 1

Sample proportion = 272 / 320 = 0.85

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.85 - 2.5758 * sqrt [ 0.85 ( 1 - 0.85) / 320 ] < p < 0.85 + 2.5758 * sqrt [ 0.85 ( 1 - 0.85) / 320 ]

0.799 < p < 0.901

99% CI is ( 79.9% , 90.1% )

Since 0.50 does not contain in confidence interval, the method appear to be effective.