Respuesta :
Answer:
b. 224
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 = 2.58[/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.
In this problem:
We want to find n, when [tex]M = 0.5, \sigma = 2.9[/tex] So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 2.58*\frac{2.9}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 7.482[/tex]
[tex]\sqrt{n} = 14.964[/tex]
[tex](\sqrt{n})^{2} = (14.964)^{2}[/tex]
[tex]n = 223.9[/tex]
The nearest integer number to 223.9 is 224.
So the correct answer is:
b. 224
