anonymous
  • anonymous
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? ___

____

Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
p = 312/624 = 0.500 best estimate mean is p N = (0.500) 624 = 312 obvious 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
anonymous
  • anonymous
what about plugging in the numbers 256/512...I cant figure out how you did the standard deviation
anonymous
  • anonymous
Formula for the standard deviation of a binomial distribution with N data and probability p is sqrt(Npq), where q = 1-p

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anonymous
  • anonymous
ok so what about the 99% part.....here is what I have so far....I got a different set of numbers because i got the first one wrong....so: p = 256/512 = .500 n= (,500)512) = 256 \[\sqrt{(256)(.500)(.5)}\] mean = 256 sd = .5 at this point I am stuck...whats the rest?
anonymous
  • anonymous
For a Normal distribution the 99.5%tile is mean + 2.57 s.d and the 0.5%tile is mean - 2.57 s.d., so the 99% confidence interval (0.5% to 99.5%) is the mean plus or minus 2.57 times the s.d.
anonymous
  • anonymous
OK....I come up with .500 < p < .0502 but I dont think that is right :-/
anonymous
  • anonymous
(mean - 2.57 s.d)/624 < p < (mean + 2.57 s.d)/624 I got 0.495 < p < 0.505
anonymous
  • anonymous
oh wait....try using these numbers....the 624 is not the right one here is the problem again... 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?
anonymous
  • anonymous
so that makes it .494 < p < .505...did you get that? I am doing something wrong I think
anonymous
  • anonymous
".494" vs. ".495" we seem to agree. Check for rounding, yours or mine!
anonymous
  • anonymous
The 512 and 256 values give the same p= 0.500 and almost the same s.d. when you work it out as we did for the 624 and 312 case. Just recalculate mean and s.d. and then divide by 512, not 624.
anonymous
  • anonymous
ok thats what I did...so lets see if that works...well......it says its not right :( I rounded it correctly...
anonymous
  • anonymous
ok so i think where i am going wrong is in the sqrt part.....I am calculating n=256 p= .500 q= .5 ...do you see where I am confused?
anonymous
  • anonymous
if I do the sqrt equation like this: \[\sqrt{(256)(.5)(.5)}\] then I come up with 8...but where do I plug that in?
anonymous
  • anonymous
Wrong square root, need sqrt(512 x 0.5 x 0.5) = s.d. =11.31 p limits are (mean +- 2.57 s.d)/512 as done above. Good night.
anonymous
  • anonymous
Ok great!! Thank YOU!!

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