posted by sammy .
a recent poll of 700 peole who work indoors found that 278 of the smoke. If the researchers want to be 98% confident of thier results to within 3.5% how large a sample is necessary?
how do i work this problem?
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(2.33)^2 * .397 * .603]/.035^2
I'll let you finish the calculation.
Note: n = sample size needed; .397 (which is approximately 278/700 in decimal form) for p and .603 (which is 1 - p) for q. E = maximum error, which is .035 (3.5%) in the problem. Z-value is found using a z-table (for 98%, the value is approximately 2.33). Symbols: * means to multiply and ^2 means squared.
I hope this will help.