PLEASE HELP - HELP IS NEEDED ASAP - FOR EVERY SECOND THIS QUESTION ISN'T ANSWERED, YOU LOSE A DAY IN YOUR LIFE - 15 POINTS. What sample size do you need to get a margin of error of 2.9%? (What do you have to remember when giving me a sample size?) If the number is larger than 1000, don't type a comma.Please don't write anything that isn't related to the question.

Answer: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. (