Mathematics Asked by dshunevich on December 19, 2020
A coin was tossed $N$ times. The number of heads is 10% more than the number of tails. At what $N$ can we say that the coin is "dishonest" (heads and tails fall out with different probability)?
Sample size for a dichotomous variable with $p$ proportion of success:
$$N = p(1-p)bigg({Z over E}bigg)^2$$
where,
$Z$ is the $z$-value corresponding to your confidence interval (eg. $z$-value $1.96$ for $95%$ confidence interval),
$E$ is the margin of error (eg: $0.05$ for $5%$ MoE)
For the given example, let $p$ be the proportion of tails.
$$therefore 1 - p = 1.1 times p$$ $$implies 1 = 2.1p implies p = {10 over 21}$$
Substituting, we get
$$N = bigg({10 over 21}bigg)bigg(1 - {10 over 21}bigg)bigg({1.96 over 0.05}bigg)^2 = 383.28 approx 384$$
So, 384 trials will tell you with 5% margin of error in 95% Confidence Interval. This will change if you change your CI and margin of error parameters.
Correct answer by vvg on December 19, 2020
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