THERE ARE ESTIMATES OF THE MARGIN OF ERROR FOR CONFIDENCE LEVELS OTHER THAN 95%. FOR A 90% CONFIDENCE LEVEL AND A SAMPLE SIZE N, THE MARGIN OF ERROR IS APPROXIMATELY 82/N%.SUPPOSE WE WISH TO HAVE A MARGIN OF ERROR OF 1% WITH A 90% CONFIDENCE LEVEL. APPROXIMATELY HOW MANY PEOPLE SHOULD WE INTERVIEW?

Answer 1

ME=Z√(0.25/n)
For 90% confidence level, and ME=1%
Z=1.644854 [ from tables for two tale at 95%]
=>
0.01=1.644854√(0.25/n)
=>
n=(1.644854/0.01)^2*0.5
=6763.86 ~ 6764 persons