A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 512 babies were born, and 256 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? ___

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Answer :

pkmnmushy
Ok so what we need to do is:

p = 312/624 = 0.500 best estimate
mean is p N = (0.500) 624 = 312
standard deviation is sqrt[Np(1-p)] = 12.5
 If we approximate the binomial by a Normal we would be using
 mu = 312, s.d = 12.5
 and the 99% C.I. is
mean +- 2.57 s.d
 we convert the mean to p by dividing by 624 so corresponding C.I. for p would be 0.500+ - (2.57)(12.5)/624 or 0.500+- 0.051
I hope this is useful for you

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