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.
