Question

An economist is interested in studying the incomes of consumers in a particular country. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in a mean income of $15,000. What is the upper end point in a 99% confidence interval for the average income (as a whole number without a dollar sign)?

Answer 1

The upper end point in a 99% confidence interval for the average income is 15,364.

Step-by-step explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575*(1000)/(√(50))= 364.15`

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.