An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $21.1$ 21.1 for a random sample of 18581858 people. Assume the population standard deviation is known to be $10.4$ 10.4. Construct the 80%80% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

The 80% confidence interval for the mean per capita income in thousands of dollars is between $20.8 and $21.4.

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:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

The lower end of the interval is the sample mean subtracted by M. So it is 21.1 - 0.3 = $20.8.

The upper end of the interval is the sample mean added to M. So it is 21.1 + 0.3 = $21.4.

The 80% confidence interval for the mean per capita income in thousands of dollars is between $20.8 and $21.4.

10 * 563.67 = 5636.7