Respuesta :
Answer:
We need a sample size of at least 425.
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} = 0.01[/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]1-0.01 = 0.99[/tex], so [tex]z = 2.575[/tex]
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.
If the manager would like to be 99% confident that their estimate of the mean is within 0.05 ounces of the true mean, how large of a sample is needed?
We need a sample size of at least n.
n is found when [tex]M = 0.05, \sigma = 0.4[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.05 = 2.575*\frac{0.4}{\sqrt{n}}[/tex]
[tex]0.05\sqrt{n} = 2.575*0.4[/tex]
[tex]\sqrt{n} = \frac{2.575*0.4}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.4}{0.05})^{2}[/tex]
[tex]n = 424.36[/tex]
Rounding up
We need a sample size of at least 425.
