Respuesta :
Answer:
The minimum sample size required for the estimate is of 3820.
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].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Variance is known to be 5.76
This means that [tex]\sigma = \sqrt{5.76} = 2.4[/tex]
What is the minimum sample size required for the estimate?
Maximum error of 0.1 means that we find n for which M = 0.1. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.1 = 2.575\frac{2.4}{\sqrt{n}}[/tex]
[tex]0.1\sqrt{n} = 2.575*2.4[/tex]
Multiplying both sides by 10
[tex]\sqrt{n} = 2.575*24[/tex]
[tex](\sqrt{n})^2 = (2.575*24)^2[/tex]
[tex]n = 3819.2[/tex]
Rounding up:
The minimum sample size required for the estimate is of 3820.
