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

Question

In the study 380 babies were​ born, and 342 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

Let G: Girl birth and B: Boy birth, so we write p=P(G), q=P(B)

Then a confidence interval of a proportion, in this case girls’ birth, will be given as

As long as n*P(G)≥10 and n*P(B)≥10 (and the proportion is distributed normally, which you just assume).

In this case these two conditions are met as n*P(G)=342 and n*P(B)=38.

So let's compute each element of the above formula.

First we look up Z value for 99 percentile from a Z table, which is 2.326

The estimated p (=P(G)) is 342/380=0.9 and

q (=P(B)) = (380-342)/380=0.1.

So Standard Error is a square root of (0.9*0.1/380) which is 1.54*10^-2 (or 0.0154)

Thus Z*standard error is 3.58*10^-2 (or 0.0358)

Finally we obtain a 99% confidence interval for a girl birth in this population by

0.9±3.58*10^-2=0.864, 0.936

This means that the probability (or the percentage) of a girl's birth within 99% confidence lies approximately between

0.864 and 0.936.

Since this confidence interval does not include 50% (if so you'd better guess the effectiveness of such method) and is greater than 50%, we can conclude that this method seems effective.