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:

A sample size of 183 is needed.

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.97}{2} = 0.015[/tex]

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

That is z with a pvalue of [tex]1 - 0.015 = 0.985[/tex], so Z = 2.17.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] 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 [tex]M = 0.023[/tex]. We have that [tex]\sigma = 0.143[/tex]. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.023 = 2.17\frac{0.143}{\sqrt{n}}[/tex]

[tex]0.023\sqrt{n} = 2.17*0.143[/tex]

[tex]\sqrt{n} = \frac{2.17*0.143}{0.023}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.17*0.143}{0.023})^2[/tex]

[tex]n = 182.02[/tex]

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

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