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:

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

Assume that a recent survey suggests that about 87% of adults have heard of the brand.

This means that $\pi = 0.87$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How many adults must he survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage?

This is n for which M = 0.05. So

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

$0.05 = 1.645\sqrt{\frac{0.87*0.13}{n}}$

$0.05\sqrt{n} = 1.645\sqrt{0.87*0.13}$

$\sqrt{n} = \frac{1.645\sqrt{0.87*0.13}}{0.05}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.87*0.13}}{0.05})^2$

$n = 122.4$

Rounding up:

He must survey 123 adults.

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2 years ago