In 1000 tosses of a coin, 560 heads appear. Is it reasonable to assume the coin is fair?

Question

So the question is:

In 1000 tosses of a coin, 560 heads appear. Is it reasonable to assume the coin is fair?

Let $Y = \sum_{i=1}^{1000} X_i = $ number of heads. I chose $H_0: p = 0.5$ and $H_1: p > 0.5$. The $p$-value is:

$$P(Y\geq 560) = \sum_{y=560}^{1000}\binom{1000}{y}(0.5)^y(0.5)^{1000-y}$$ $$= 0.00008252494$$

Then I try to find $\alpha$ using $\frac{Y}{n} = \overline{X}$ as the estimator for $p$: $$P(Y \geq 560) = P(\overline{X} \geq 0.56)$$ $$= P(\frac{\sqrt{n}(\overline{X} - p_0)}{p(1-p)} \geq \frac{\sqrt{1000}(0.56 - 0.5)}{0.5*0.5}$$ $$= P(Z\geq 7.59)$$

That Z-score seems incredibly high and I don't think they're even included in any Z-tables.

Am I on the right track here or have I missed something?

Answer 1

From the given $p$-value we calculate a $z$-score of $$Z=\Phi^{-1}(1-0.00008252494)=3.7672\dots$$ The Z-score is not that high, but still high enough (beyond three SDs) that we can reject $H_0$ and say that the coin is biased.

Answer 2

First of all you have $P(Y\geq 560)=1-P(Y\leq 559)$. Next we use the normal approximation without the continuity correction factor.

$$1-P(Y\leq 559)\approx 1-\Phi\left(\frac{0.559-0.500}{\sqrt{\frac{ 0.5\cdot 0.5}{1000}}} \right)$$

You missed the root at $p\cdot 1-p$. But the z-score is still high, $z=3.73$. So we assume approximately that $\Phi(3.73)\approx 1$. This leads to $1-P(Y\leq 559)\approx 1-1=0<0.05$