If n=540 and ˆ p p ^ (p-hat) =0.23, find the margin of error at a 99% confidence level Give your answer to three decimals

Answer 1

The margin of error for the confidence interval is equal to the critical z value for 99% CI (z*) * standard error of sample proportion.

z* is found by looking in the normal distribution tables for the z-value corresponding to 0.995 probability (the 99% CI corresponds to the middle 0.99, with 0.005 probability on each tail). z* = 2.575

the standard error of the sample proportion is sqrt [ phat * (1-phat*)/n]

for phat = 0.23 and n = 540, that would be sqrt [ 0.23 * 0.77/540] = 0.018

so the margin of error = 2.575 * 0.018 = 0.047