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 $21.6 for a random sample of 3839 people.

Assume the population standard deviation is known to be $12.5.

Construct the 80% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

`21.6-1.28(12.5)/(√(3839))=21.3`

`21.6+1.28(12.5)/(√(3839))=21.9`

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

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)(\sigma)/(√(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(12.5)/(√(3839))=21.3`

`21.6+1.28(12.5)/(√(3839))=21.9`

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

Answer 2

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

Explanation:

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

`\alpha = (1-0.8)/(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*(\sigma)/(√(n))`

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

`M = 1.28*(12.5)/(√(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.