Respuesta :
Answer:
A sample size of at least 2809 would be required in order to estimate the mean per capita income at the 98% level of confidence with an error of at most $0.36
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{1-0.98}{2} = 0.01[/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]1-0.01 = 0.99[/tex], so [tex]z = 2.327[/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.
Sample size with margin of error of at most $0.36
Sample size of at least n, in which n is found when [tex]M = 0.36, \sigma = 8.2[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.36 = 2.327*\frac{8.2}{\sqrt{n}}[/tex]
[tex]0.36\sqrt{n} = 2.327*8.2[/tex]
[tex]\sqrt{n} = \frac{2.327*8.2}{0.36}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.327*8.2}{0.36})^{2}[/tex]
[tex]n = 2809[/tex]
A sample size of at least 2809 would be required in order to estimate the mean per capita income at the 98% level of confidence with an error of at most $0.36
