Posted by Rachel on .
a survey is to be conducted to determine the proportion of U.S. voters who support additional aid to African nations to fight poverty and HIV/AIDS. Of the proposed sammple sizes which is the smallest that will guarantee a margin of error of no more than 4% for a 90% confidence level?
n = [(z-value) * p * q]/E^2
...where n is the sample size you seek, z-value is obtained using a z-table for the confidence interval (90%) in the problem, p and q represent proportions, E is the maximum error, and ^2 means squared.
n = [(1.645) * .5 * .5]/.04^2
...where 1.645 represents the 90% confidence interval, .5 represents p if no value is stated, q = 1 - p = 1 -.5 = .5, .04 is 4% in decimal form and represents the maximum error of estimate for a proportion.
Calculate and choose the closest answer. There may be differences in the answer due to methods used.
I hope this will help.