Respuesta :
Answer:
A sample of 217 is needed.
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{1-0.95}{2} = 0.025[/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]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/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 population standard deviation is $3,750 how large should the sample be if margin of error is $500
We have that [tex]\sigma = 3750[/tex].
We need a sample of n, and n is found when [tex]M = 500[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 1.96*\frac{3750}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 1.96*3750[/tex]
[tex]\sqrt{n} = \frac{1.96*3750}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*3750}{500})^{2}[/tex]
[tex]n = 216.1[/tex]
Rounding up
A sample of 217 is needed.
