Answer:
The minimum number of samples required is 487.
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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
Standard deviation 4.5 mg/L.
This means that [tex]\sigma = 4.5[/tex]
What is the minimum number of samples required to estimate today's level to within 0.4 mg/L with 95% confidence?
This is n for which M = 0.4. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 1.96\frac{4.5}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.96*4.5[/tex]
[tex]\sqrt{n} = \frac{1.96*4.5}{0.4}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*4.5}{0.4})^2[/tex]
[tex]n = 486.2[/tex]
Rounding up
The minimum number of samples required is 487.
