Question

An economist would like to estimate a 99% confidence interval for the average real estate taxes collected by a small town in California. In a prior analysis, the standard deviation of real estate taxes was reported as $1,260.
What is the minimum sample size required by the economist if he wants to restrict the margin of error to $560?

Answer 1

Answer:

$n=(\frac{2.58(1260)}{560})^2 =33.698 \approx 34$

So the answer for this case would be n=34 rounded up to the nearest integer

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".

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

$\mu$ population mean (variable of interest)

$\sigma=1260$ represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =560 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(1260)}{560})^2 =33.698 \approx 34$

So the answer for this case would be n=34 rounded up to the nearest integer