Question

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.

Answer 1

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

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.05`

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
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error 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 = 1.96*(10)/(√(976)) = 0.6`

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.