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?

Answer 1

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

Answer 2

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

Answer 3

Hello?

Answer 4

yes

Answer 5

sorry, I was answering the other problems

Answer 6

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

Answer 7

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

Answer 8

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?

Answer 9

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

Answer 10

Well then, you are correct! Well done!

Answer 11

thank you

Answer 12

You're welcome