Question

Brewed decaffeinated coffee contains some caffeine. We want to estimate the amount of caffeine in 8 ounce cups of decaf coffee at local fast food restaurants. Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg. If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result

Answer 1

Answer:

We would have to take a sample of 62 to achieve this result.

Step-by-step explanation:

Confidence level of 95%.

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

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.

Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.

This means that $\sigma = 2$

If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?

We would need a sample of n.

n is found when $M = 0.5$. So

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

$0.5 = 1.96\frac{2}{\sqrt{n}}$

$0.5\sqrt{n} = 2*1.96$

Dividing both sides by 0.5

$\sqrt{n} = 4*1.96$

$(\sqrt{n})^2 = (4*1.96)^2$

$n = 61.5$

Rounding up

We would have to take a sample of 62 to achieve this result.