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

A sample size of 183 is needed.

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 2.17.

Now, find the margin of error M as such

In which 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 . We have that . So

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

21

jk it's 20

step-by-step explanation: