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 $21.6 for a random sample of 3839 people.

Assume the population standard deviation is known to be $12.5.

Construct the 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 $21.3 and $21.9.

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 the margin of error 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 21.6 - 0.3 = $21.3.

The upper end of the interval is the sample mean added to M. So it is 21.6 + 0.3 = $21.9.

The 80% confidence interval for the mean per capita income in thousands of dollars is between $21.3 and $21.9.

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