A survey of shoppers is planned to determine what percentage use credit cards. Prior surveys suggest that 59% use credit cards.
The required margin of error is 0.02 at 95% confidence. What is the minimum sample size required to assure that your sample proportion "p" will be within the required margin of error, around the population proportion "p".
What I got for the answer was:
p = x/n = 59/100 = .59
q = 1-p = .41
E (margin of error) = .02
a = .05
n = [ (z .025)squared X (.59) (.41) ] / E squared
= 1.96 squared X (.59) (.41) / .02 squared
= 2323.2076 (minimum sample size required.
However, prior comments I have seen on this problem use 1.645 for the z value, when a = .05, even though z a/2 = z .025 = 1.96 (from table A-2).
Secondly, other comments I have seen suggest using Upper Limit and Lower Limit to produce the estimation of the population mean from sample mean. I don't understand the need to do that, when the problem is asking only for the minimum sample size required. Do I also need to use Upper Limit - Lower Limit to produce the Margin of Error, when the margin of error is already given as 0.02 (I don't understand the need to do this when the information is already given).
Lastly, I have seen suggestions that the margin of error, is L = 2z(p)(1-p) / n X .5 and later squaring this equation and rearranging the equation.
Again, I don't see the need to do this, when L (margin of error is already given as 0.02)
So can someone tell me if my original answer was correct, and if not, why? And why the need for the upper limits/lower limits, and rearranging the equation.
The minimum sample size is found. The expert finds the required margin of error for confidence.