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When given the following information:

When spun on edge 250 times, a Belgian one-euro coin came up heads 140 times and tails 110. `It looks very suspicious to me,' said Barry Blight, a statistics lecturer at the London School of Economics. If the coin were unbiased, the chance of getting a result as extreme as that would be less than 7%.'

  • N = 250
  • z = 110
  • p-value ~ 0.0664

and the model:

p(theta | m0) =   1   if theta = 0.5
                  0   otherwise

How do I calculate p(z, N | m0) ?

I actually do need to calculate this and nothing else from this given information

Moore's Lawyer
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1 Answers1

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The exact $p$-value can be calculated as $$2 \Pr[X \ge 140 \mid p = 1/2] = \frac{1}{2^{249}} \sum_{x=140}^{250} \binom{250}{x} \approx 0.0664212.$$ This is simply based on the binomial distribution CDF: $X \sim \operatorname{Binomial}(n,p)$ is the random number of heads obtained in $n$ spins, and the hypothesis to be tested is $$H_0 : p = 1/2 \quad \text{vs.} \quad H_a : p \ne 1/2.$$ There's no need to use a Bayesian framework or perform any additional computation.

We can approximate the above probability by using a continuity-corrected normal distribution, e.g. $$\Pr[X \ge 140 \mid p = 1/2] \approx \Pr\left[Z \ge \frac{139.5 - 125}{\sqrt{250(1/2)(1 - 1/2)}}\right] \approx \Pr[Z \ge 1.83412] \approx 0.033318,$$ twice of which is $0.066636$.

In closing, it is worth noting that we should not expect the outcome of spinning the coin to be unbiased. Tossing it in the air might be fairer. Why? Because coins are made by stamping a metal blank. This means the edge of the coin is not necessarily perfectly perpendicular to the faces. Moreover, the weight distribution of the coin may not be balanced across its geometric axis of spin due to the embossing of the sides. Spinning the coin on its edge, therefore, is more likely to result in a biased outcome than flipping it.

heropup
  • 146,967
  • That wasn't my question, as I had already calculated the p-value and stated it, I'll make a little more clearer. Thanks anyways. – Moore's Lawyer Jun 11 '20 at 10:18
  • @TempoRary Your model doesn't make sense, which is why I answered the way I did. You write $$p(\theta \mid m_0) = \begin{cases}1, & \theta = 0.5 \ 0, & \text{otherwise}\end{cases}$$ which is unclear, since you don't say what $m_0$ is, and your notation $p(\theta \mid m_0)$ is meaningless if $\theta = 0.5$: You are writing $p(0.5 \mid m_0) = 1$. – heropup Jun 11 '20 at 17:50