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 $23.1 for a random sample of 1256 people. Assume the population standard deviation is known to be $6. Construct the 90% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

The 90% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $23.4

Explanation:

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

`\alpha = (1-0.9)/(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
`1-0.05 = 0.95`, so
`z = 1.645`

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 = 1.645*(6)/(√(1256)) = 0.3`

The lower end of the interval is the mean subtracted by M. So it is 23.1 - 0.3 = $22.8

The upper end of the interval is the mean added to M. So it is 23.1 + 0.3 = $23.4

The 90% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $23.4