Question

What it the minimum number of times you would need to flip a coin, if you
wanted a 90% chance that the ratio of heads to total flips were between .499 and .501.

Answer 1

Solution:

Given confidence interval for population proportion is (0.499 , 0.501)

First need to find sample proportion and margin of error.

So,

Sample proportion:

= (Lower limit+Upper limit) / 2

= (0.499+0.501)/2 = 0.5

Margin of error:

E = (Upper limit-Lower limit)/2

E = (0.501-0.499) / 2 = 0.001

As confidence level is 0.90,

Level of significance = α= 0.10

Therefore, Z_c = Z_α/2 = 1.64 ...Using excel formula, =ABS(NORMSINV(0.1/2))

Therefore, sample size is,

n = 672400

Hence, required number of flip of coin is 672400

Done