Answer:
The 90% confidence interval for the mean usage of electricity is between 17.4 kwH and 17.6 kwH
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{1-0.9}{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 1-0.05 = 0.95, so z = 1.645
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.
So
[tex]M = 1.645*\frac{2.4}{\sqrt{796}} = 0.14[/tex]
The lower end of the interval is the mean subtracted by M. So 17.5 - 0.14 = 17.4 kwH
The upper end of the interval is M added to the mean. So 17.5 + 0.14 = 17.6 kwH
The 90% confidence interval for the mean usage of electricity is between 17.4 kwH and 17.6 kwH
