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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.85}{2} = 0.075$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.44\frac{12.6}{\sqrt{717}} = 0.7$

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.