Statistics class

Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were​ born, and 323 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

Assuming the normal chance of having a girl is 50%, our conclusion is that the method is effective at increasing the probability of conceiving girls. Let's see how.

To find the confidence interval, we need to find a few items and follow a few steps.

99% CI = Sample Proportion +/- Margin of Error, where Sample Proportion is 323/380 or 0.85, z-value is 2.5758, and Sample Size is 380

Sample Prop. +/- z-value * sqrt(Sample Prop. * (1 - Sample Prop.)/Sample Size)

Lets put the actual numbers into the equation:

0.85 +/- 2.5758 * sqrt((0.85 * 0.15)/380)

0.85 +/- 2.5758 * sqrt(0.0003355)

0.85 +/- 2.5758 * 0.01832

0.85 +/- 0.0472

99% CI = (0.85 - 0.0472, 0.85 + 0.0472) = (0.8028, 0.8972)

Since 50% or 0.50 does not fit within this confidence interval, we can say that the method is effective at increasing the probability of conceiving girls.

Hope this helps!

DRP