What sample size is needed at the 95% confidence level, where the error (E) is 3 and the standard deviation is 20?

Question

What sample size is needed at the 95% confidence level, where the error (E) is 3 and the standard deviation is 20?

I can't figure out this question for the life of me and I am not sure of what formula to use. I know the answer is 171 but I would like to understand how we got there.

Answer 1

$$E =z \frac{\sigma}{\sqrt{n}}$$ We have $$E= 20$$, the z value (critcal value) for a 95% confidence is $$z = 1.96$$ ( get it from z table) and $\sigma = 20$ is given. We can get the sample size $n$, as follows: $$E^2 = z^2 \frac{\sigma^2}{n}$$ which gives $$n = z^2 \frac{\sigma^2}{E^2}$$ that is $$n = (1.96)^2 \frac{20^2}{3^2} \simeq 171$$

Answer 2

$n = (\frac{Z\cdot \sigma}{E})^2$

From the formula you can see that to reduce the error, a larger sample size is required making for a more reliable test.