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 $Z = 1.96$ .

How large a sample should she take to achieve this?

She needs a sample of size at least n.

n is found when M = 0.04.

We do not know the true population proportion, so we use $\pi = 0.5$ , which is the case for which we are going to need the largest sample size.

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

$0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.05\sqrt{n} = 1.96*0.5$

Dividing both sides by 0.05

$\sqrt{n} = 19.6$

$(\sqrt{n})^{2} = 19.6^{2}$

$n = 384.2$

Rounding up

She needs a sample of at least 385.

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3 years ago