Question

A Florida neighborhood is comprised of a total of 250 houses of which 12% are in foreclosure. A random sample of 91 homes from this neighborhood was selected. The standard error of the proportion is ________.

Answer 1

Answer:

the standard error of the proportion is 0.0272

Step-by-step explanation:

We have that if the sample size is greater than 5% of the entire population, a finite population correction factor (fpc) is multiplied with the standard error :

fpc = $\sqrt{\frac{N -n}{N -1} }$

We know that N = 250 n = 91, replacing:

fpc = $\sqrt{\frac{250 - 91}{250 -1} }$

fpc = 0.799

Now, the formula would then be:

SE  = $\sqrt{\frac{p * (1 -p)}{n} }$*fpc

Now replacing, knowing that p = 0.12

SE= $\sqrt{\frac{0.12 * (1 - 0.12)}{91} }$*0.799

SE = 0.0272

So the standard error of the proportion is 0.0272