Question

What sample size is needed to give a margin of error within in estimating a population mean with 95% confidence, assuming a previous sample had .

Answer 1

Answer:

$n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$(if no previous estimate use 0.5) and M is the desired margin of error.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Needed sample size:

The needed sample size is n. We have that:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{\pi(1-\pi)}{n}}$

$\sqrt{n}M = 1.96\sqrt{\pi(1-\pi)}$

$\sqrt{n} = \frac{1.96\sqrt{\pi(1-\pi)}}{M}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

$n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$(if no previous estimate use 0.5) and M is the desired margin of error.