Question

The U.S. Department of Housing and Urban Development publishes data on the fair market monthly rent for existing one-bedroom housing by metropolitan area (The Federal Register, April 30 1997). The standard deviation for the monthly rent is about $80. Assume that a sample of metropolitan areas will be selected in order to estimate the population mean of the monthly rent for existing one-bedroom housing. Use 95% confidence. a. How large should the sample be if the desired margin of error is $25?

Answer 1

Answer:

$n=(\frac{1.960(80)}{25})^2 =246.73 \approx 247$

So the answer for this case would be n=247 rounded up to the nearest integer

Step-by-step explanation:

We know that the standard deviation is :

$\sigma = 80$ represent the deviation

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 =25 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 95% of confidence interval now can be founded using the normal distribution and the critical value would be $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

$n=(\frac{1.960(80)}{25})^2 =246.73 \approx 247$

So the answer for this case would be n=247 rounded up to the nearest integer