An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $23.4$ 23.4 for a random sample of 976976 people. Assume the population standard deviation is known to be $10$ 10. Construct the 95%95% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

The 95% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $24.

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 23.4 - 0.6 = $22.8.

The upper end of the interval is the sample mean added to M. So it is 23.4 + 0.6 = $24.

The 95% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $24.

john earns $14.00 for two hours of work.