Answer:
We need at least 601 incomes.
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.
How many such incomes must be found if we want to be 95% confident that the sample mean is within $500 of the true population mean?
We have to find n, for which [tex]M = 500, \sigma = 6250[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 1.96*\frac{6250}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 1.96*6250[/tex]
[tex]\sqrt{n} = \frac{1.96*6250}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*6250}{500})^{2}[/tex]
[tex]n = 600.25[/tex]
Rounding up
We need at least 601 incomes.
