Answer:
A sample size of at least 737 specimens is required.
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{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the width 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.
In this problem, we have that:
[tex]M = 3, \sigma = 31.62[/tex]
So:
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 2.575*\frac{31.62}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 81.4215[/tex]
[tex]\sqrt{n} = 27.1405[/tex]
[tex]n = 736.60[/tex]
A sample size of at least 737 specimens is required.
