Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 395 babies were born, and 316 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?
I am really stuck on this formula if any can help. Please and thank u!

Answer 1

\[I=\hat{p}\pm Z_{0.005} \sigma\]
Now search for the Z-score in your std normal table
p hat is simply the proportion, or 316/395
sigma is \[\sqrt{\frac{pq}{n}}\]

Answer 2

recall q=1-p
Also, I am assuming you need a 2 sided confidence interval

Answer 3

also note that in this case, sample mean is an accurate and unbiased estimator of the proportion

Answer 4

I really don't get it maybe you can lay it out for me with the numbers instead of pqn, etc

Answer 5

I appreciate your help!!

Answer 6

I got so far 395/316 = 0.8, so I imagine this is 80 percent?? so I don't know if I need to equal 1 so would that be 0.2, because when I tried to do pq/n, I got 16/395, which comes to 24.7. I don't know if i am supposed to find the square root of that or what. HELP!!

Answer 7

p is your proportion you are trying to find, so it is 0.8 for now
q = 0.2

Answer 8

Yeah I know, I times 0.2 * 0.8/395 then got the square root, which what I did was figure that it came to 24.7 and then I found the square root which was 4.979. I am not sure if this is the right way. I never got past basic math, and I am taking a Statistics class, which is my last class and then I graduate with a associates degree in Criminal Justice. If I do not pass this class, which I will, because right now I have between a C and a D, but the difference is if I get a "D" then I will have to repeat this class in my bachelor's program, and I will croak if I have to take this class again, as it has been by far the hardest class I have taken besides Algebra!!