# use the given data to find the minimum sample size required to estimate a population proportion or percentage Margin of error: 0.005, confidence level: 99%, p and q unknown

Unless the population proportion is very skewed, you can use p=.5 and q=.5

E=.005; confidence level of 99% implies `z_(alpha/2)=2.58` and

`E=z_(alpha/2)sqrt((pq)/n)`

`.005=2.58sqrt((.5*.5)/n)`

`.0019379845=sqrt((.25)/n)`

`266256=n/.25`

n=66564

Thus to get an error of 1/2% with 99% confidence, you need a sample size of at least 66564.

** You may question the...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Unless the population proportion is very skewed, you can use p=.5 and q=.5

E=.005; confidence level of 99% implies `z_(alpha/2)=2.58` and

`E=z_(alpha/2)sqrt((pq)/n)`

`.005=2.58sqrt((.5*.5)/n)`

`.0019379845=sqrt((.25)/n)`

`266256=n/.25`

n=66564

Thus to get an error of 1/2% with 99% confidence, you need a sample size of at least 66564.

** You may question the ability to use p=q=.5 As long as the population isn't very divided, the numbers are approximately the same. Using 40% and 60% yields 266256*.24=63901. Even 70%-30% gives 266256*.21=55914, so the use of 50-50 is justified. **