Respuesta :
Answer:
The minimum sample size is 610.
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 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.
Population standard deviation is $6300, so [tex]\sigma = 6300[/tex]
Suppose we want to determine the minimum sample size required to give us a 95% confidence interval that estimates, to within $500, the average salary of a state employee.
This is n when M = 500. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 1.96*\frac{6300}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 1.96*6300[/tex]
[tex]\sqrt{n} = \frac{1.96*6300}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*6300}{500})^{2}[/tex]
[tex]n = 609.89[/tex]
Rouding up
The minimum sample size is 610.
