what is the sample size formula for ME=2√(P(1-P)/n) if ME= margin errors, P= percentage of people responding in a certain way, and n is the size of the sample?

Question

what is the sample size formula for ME=2√(P(1-P)/n)    if ME= margin errors, P= percentage of people responding in a certain way, and n is the size of the sample?

anonymous · Answer

I need it to solve a problem......is it like this? 
n= (p(1-p)*2^2)/(ME^2)

anonymous · Answer

Is the given formula$$ME=2 \sqrt{\frac{ P(1-P) }{ n }}$$

anonymous · Answer

Hello?

anonymous · Answer

yes

anonymous · Answer

sorry, I was answering the other problems

anonymous · Answer

And is your answer$$n=\frac{ 4P(1-P) }{ ME^2 }$$

anonymous · Answer

well, I did it like this n=  (p(1-p)*2^2)/(ME^2 )

anonymous · Answer

Right, but I'm having a hard time deciphering the brackets. Knowing that $2^2=4$, is my last post the same as your answer?

anonymous · Answer

oh, jaja you are right. Didn't noticed that XD

anonymous · Answer

Well then, you are correct! Well done!

anonymous · Answer

thank you

anonymous · Answer

You're welcome