Question 1132085: A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 365 babies were born, and 292 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 by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! half-interval for 99%CI is z*SE, where z=2.576 (0.995) and SE is sqrt (p*(1-p)/n), where p is the point estimate, of 292/365 or 0.8
half-interval is 2.576*sqrt[.8*.2/365]=2.576*0.0209=0.0538
The CI is (0.7462, 0.8538). Because 0.5 is not in the interval (the rough estimate of what would be likely), then we can be 99% confident the true value is greater than 0.5. It would appear to be very effective.
|
|
|
