a clinical trial tests a method to increase the probability

a clinical trial tests a method to increase the probability of concieving a girl. in the study 300 babies were born and 255 of them were girls. use the sample data to construct a 99% confidence interval estimate of percentage girls born. based on the results, does this method appear to be effective?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.85          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.020615528          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.796897919          
upper bound = p^ + z(alpha/2) * sp =    0.903102081          
              
Thus, the confidence interval is              
              
(   0.796897919   ,   0.903102081   )

As this interval does not contain 0.50 (the usual probability of a girl), thenthis method indeed appears effective. [CONCLUSION]