Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study

564

babies were​ born, and

282

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 = 282 / 564 = 0.50

99% confidence interval for p is

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

0.50 - 2.5758 * sqrt( 0.50 * 0.50 / 564) < p < 0.50 + 2.5758 * sqrt( 0.50 * 0.50 / 564)

0.446 < p < 0.554

99% CI is ( 0.446 , 0.554 )

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