If the candidate only wants a 0.2% margin of error at a 95% confidence level, what size of sample is needed?

n = 22

Step-by-step explanation:

Using

n = p(1-p)(Zc/E)²

From table E = 0.2 while p= 0.37

So substituting

We have n= 22.4

Thats a sample size of approximately 22

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.