Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California.

Answer 1

Answer:

The 80% confidence interval for the mean per capita income in thousands of dollars is between $21.3 and $21.9.

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:

$\alpha = \frac{1-0.8}{2} = 0.1$

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 $1-0.1 = 0.9$, so $z = 1.28$

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.

$M = 1.28*\frac{12.5}{\sqrt{3839}} = 0.3$

The lower end of the interval is the sample mean subtracted by M. So it is 21.6 - 0.3 = $21.3.

The upper end of the interval is the sample mean added to M. So it is 21.6 + 0.3 = $21.9.

The 80% confidence interval for the mean per capita income in thousands of dollars is between $21.3 and $21.9.

Answer 2

Answer:

$21.6-1.28\frac{12.5}{\sqrt{3839}}=21.3$    

$21.6+1.28\frac{12.5}{\sqrt{3839}}=21.9$    

So on this case the 80% confidence interval would be given by (21.3;21.9)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=21.6$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=12.5$ represent the population standard deviation

n=3839 represent the sample size  

Solutio n to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.1,0,1)".And we see that $z_{\alpha/2}=1.28$

Now we have everything in order to replace into formula (1):

$21.6-1.28\frac{12.5}{\sqrt{3839}}=21.3$    

$21.6+1.28\frac{12.5}{\sqrt{3839}}=21.9$    

So on this case the 80% confidence interval would be given by (21.3;21.9)