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 of the interval is given by:

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

In this problem, we have that:

$p = 0.69$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

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

$0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}$

$0.07\sqrt{n} = 1.2996$

$\sqrt{n} = \frac{1.2996}{0.07}$

$\sqrt{n} = 18.5658$

$(\sqrt{n})^{2} = (18.5658)^{2}$

$n = 345$

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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3 years ago