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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

95% confidence level

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

Sample size needed

At least n, in which n is found when $M = 0.09$

We don't know the proportion, so we use $\pi = 0.5$ , which is when we would need the largest sample size.

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

$0.09 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.09\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.09})^{2}$

$n = 118.6$

Rounding up

We need a sample size of least 119

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3 years ago