Respuesta :
Answer:
A sample of 3851 is required.
Step-by-step explanation:
We have that to find our 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 Z-table as such z has a p-value of .
That is z with a pvalue of , so Z = 2.327.
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.
Variance is 5.76 kWh
This means that [tex]\sigma = \sqrt{5.76} = 2.4[/tex]
They would like the estimate to have a maximum error of 0.09 kWh. How large of a sample is required to estimate the mean usage of electricity?
This is n for which M = 0.09. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.09 = 2.327\frac{2.4}{\sqrt{n}}[/tex]
[tex]0.09\sqrt{n} = 2.327*2.4[/tex]
[tex]\sqrt{n} = \frac{2.327*2.4}{0.09}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327*2.4}{0.09})^2[/tex]
[tex]n = 3850.6[/tex]
Rounding up:
A sample of 3851 is required.
