Sixty heads in a hundred

Idea. Standardise the number of heads. Sixty heads is two standard deviations above the mean, and the two-sided p-value is the chance of being at least two standard deviations away on either side.

Step 1. For a fair coin, the number of heads $H$ in 100 tosses has mean $100\cdot \frac{1}{2}=50$ and variance $100\cdot \frac{1}{4}=25$, so standard deviation 5.

Step 2. The z-score is $$z=\frac{60-50}{5}=2.$$

Step 3. The two-sided p-value is $$2(1-\Phi (2))\approx 2\times 0.02275=\mathrm{0.0455.}$$ So at the 5% level the hypothesis of a fair coin is rejected, just barely.

Not so fast. The conclusion is fragile. With the continuity correction, $z=\frac{59.5-50}{5}=1.9$ and the p-value is about $0.057$; the exact binomial p-value is about $0.057$ too. Both are above 5%. When the answer sits this close to the threshold, the approximation matters.