Respuesta :
Answer:
A sample of 784 is required to estimate the mean usage of water.
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 Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
The standard deviation is 2 gallons
This means that [tex]\sigma = 2[/tex]
They would like the estimate to have a maximum error of 0.14 gallons. How large of a sample is required to estimate the mean usage of water?
This is n for which M = 0.14. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.14 = 1.96\frac{2}{\sqrt{n}}[/tex]
[tex]0.14\sqrt{n} = 1.96*2[/tex]
[tex]\sqrt{n} = \frac{1.96*2}{0.14}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*2}{0.14})^2[/tex]
[tex]n = 784[/tex]
A sample of 784 is required to estimate the mean usage of water.
