A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 512 babies were born, an
d 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? ___ ____
<span>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 </span>I hope this is useful for you
just put the numbers on the choice to know the right answer. here when you put 896 in <u> </u>*27/14/72=24 you will get the right answer which is 24. so the answer is 896.
