A clinical trial tests a method designed to increase the pro

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 642 babies were born, and 321 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?

_____< p < ______?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.5          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019733426          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.33          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.454021118          
upper bound = p^ + z(alpha/2) * sp =    0.545978882          
              
Thus, the confidence interval is              
              
(   0.454021118   ,   0.545978882   ) [ANSWER]

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As 0.50 is still in this confidence interval, then it doesn\'t appear to be effective.