Respuesta :
Answer:
D) 239
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 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.
Required sample size:
Is n
n is found when [tex]M = 6[/tex]
We have that [tex]\sigma = 36[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]6 = 2.575*\frac{36}{\sqrt{n}}[/tex]
[tex]6\sqrt{n} = 36*2.575[/tex]
Simplifying by 6
[tex]\sqrt{n} = 6*2.575[/tex]
[tex](\sqrt{n})^{2} = (6*2.575)^{2}[/tex]
[tex]n = 238.7[/tex]
We round up, and the answer is 239.
