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

`n=((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

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)(\sigma)/(√(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=((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=((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