Question

A survey of 1000 adults in the US conducted in March 2011 asked ‘‘Do you favor or oppose ‘sin taxes’ on soda and junk food?” The proportion in favor of taxing these foods was 32%.

If we want a margin of error of only1% (with 99% confidence), what sample size is needed?

Round your answer up to the nearest integer.

Answer 1

Solution :

Given that,

= 0.32

1 - = 1 - 0.32 = 0.68

margin of error = E = 0.01

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z _0.005 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.01)² * 0.32 * 0.68

= 14439

sample size = 14439