Question

The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error within 0.023 ppm of mercury with 97% confidence. Assume the population standard deviation is 0.143 ppm of mercury. What sample size is needed

Answer 1

Answer:

A sample size of 183 is needed.

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:

$\alpha = \frac{1 - 0.97}{2} = 0.015$

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 $1 - 0.015 = 0.985$, so Z = 2.17.

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.

Margin of error within 0.023 ppm of mercury with 97% confidence. What sample size is needed?

We have to find n for which $M = 0.023$. We have that $\sigma = 0.143$. So

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

$0.023 = 2.17\frac{0.143}{\sqrt{n}}$

$0.023\sqrt{n} = 2.17*0.143$

$\sqrt{n} = \frac{2.17*0.143}{0.023}$

$(\sqrt{n})^2 = (\frac{2.17*0.143}{0.023})^2$

$n = 182.02$

We have to round up(182 is not quite in the desired margin of error), so a sample size of 183 is needed.