Answer:
The upper end point in a 99% confidence interval for the average income is 15,364.
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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/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.
[tex]M = 2.575*\frac{1000}{\sqrt{50}}= 364.15[/tex]
The upper end of the interval is the mean added to M. So it is 15000 + 364.15 = 15,364.15 dollars.
So the upper end point in a 99% confidence interval for the average income is 15,364.
