Question

A marketing firm wants to estimate how much root beer the average teenager drinks per year. A previous study found a standard deviation of 1.12 liters. How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval

Answer 1

Answer:

At least 832 teenargers must be interviewed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval

At least n teenargers must be interviewed.

n is found when M = 0.1.

We have that $\sigma = 1.12$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.1 = 2.575*\frac{1.12}{\sqrt{n}}$

$0.1\sqrt{n} = 2.575*1.12$

$\sqrt{n} = \frac{2.575*1.12}{0.1}$

$(\sqrt{n})^{2} = (\frac{2.575*1.12}{0.1})^{2}$

$n = 831.7$

Rounding up

At least 832 teenargers must be interviewed.