Question

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 for a random sample of 717 people. Assume the population standard deviation is known to be $12.6. Construct the 85% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

The 85% confidence interval for the mean per capita income in thousands of dollars is between $20.4 and $21.8.

Explanation:

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

`\alpha = (1-0.85)/(2) = 0.075`

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
`1-0.075 = 0.925`, so
`z = 1.44`

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 = 1.44(12.6)/(√(717)) = 0.7`

The lower end of the interval is the sample mean subtracted by M. So it is 21.1 - 0.7 = $20.4.

The upper end of the interval is the sample mean added to M. So it is 21.1 + 0.7 = $21.8.

The 85% confidence interval for the mean per capita income in thousands of dollars is between $20.4 and $21.8.