Question

Natalia estimates that she wins a board game 72% of the time. How large of a random sample is required to obtain a margin of error of at most 0.06 with 95% confidence?

Answer 1

Explanation:

Margin of error = Z * √p(1-p)/n

z is the z score at 95% confidence = 1.96

p = 0.72

1-p = 0.28

Substitute and get n

Margin of error = Z * √p(1-p)/n

0.06 = 1.96 * √0.72(0.28)/n

0.06/1.96 = √0.72(0.28)/n

0.0306² = 0.2016/n

n = 0.0306²/0.2016

n =0.2016/0.00093636

n = 215.3

Hence the random sample is about 215