Question

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. 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? ​b) Assume that a recent survey suggests that about 87​% of adults have heard of the brand.

Answer 1

Answer:

He must survey 123 adults.

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:

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

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.