What sample size do you need to get a margin of error of 2.9%? If the number is larger than 1000, do NOT type a comma.

Question

I really need some help

Answer 1

What does the comma have to do with anything? LOL

 

OK. You want a margin of error of 2.9%. But you do not have a confidence interval, which you need to find a sample size. I'm going to assume a confidence interval of 90%, because that's a popular one. If I chose the wrong one, hopefully you can repeat whatever I did.

 

For a 90% confidence interval, there is a Z-multiplier of 1.645. In your lessons, at least for now, you don't really have to know what a Z-multiplier represents. But keep in mind that (a) you need it, and (b) the day will eventually come when you need to understand why you need it.

 

The formula is

 

N = ((z-multiplier * standard deviation) / margin of error)^2

 

In this case that's

 

(1.645 * standard deviation / 0.029)^2

 

If I chose a standard deviation of 1, N would be

 

(1.645 / 0.029)^2 = about 3218. (I left out the comma. LOL)

 

And you don't have a standard deviation. So sorry - can't finish the problem. In future, be sure to give ALL relevant info if you want an answer. :-(