Respuesta :
Answer:
24
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.
Minimum sample size:
n for which [tex]M = 10, \sigma = 19[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 2.575*\frac{19}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 19*2.575[/tex]
Simplifying by 10
[tex]\sqrt{n} = 1.9*2.575[/tex]
[tex](\sqrt{n})^{2} = (1.9*2.575)^{2}[/tex]
[tex]n = 23.94[/tex]
Rounding up, the answer is 24.
