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Consider the guessing milk tea example

$$\begin{array}{c|c|c|} & \text{Guess Milk} & \text{Guess Tea} \\ \hline \text{Milk} & 3 & 1 \\ \hline \text{Tea} & 1 & 3 \\ \hline \end{array}$$

I want to test that $H_0: \theta = 1$ (independent) vs. $H_a: \theta \neq 1$ (associated)

The formula for $P(n_{11} = t) = \frac{{n_{1+} \choose t} {n_{2+} \choose n_{+1}-t}}{{n \choose n_{+1}}}$

It's fixed on $n_{11}$

The range is given by $m_- = max(0, n_{11} - n_{22}) = max(0,0) = 0$ and $m_+ = min(n_{11}+n_{12}, n_{11}+n_{21}) = min(4,4) = 4$.

$$0 \leq n_{11} \leq 4$$


This is what my notes did after that

$P(n_{11} = 0) = 0.0143, P(n_{11} = 1) = 0.2285, P(n_{11} = 2) = 0.5143, P(n_{11} = 3) = 0.2285, P(n_{11} = 4) = 0.0143$

Thus, the two sided p-value is $$P(n_{11} = 0) + P(n_{11} = 1) + P(n_{11} = 3) + P(n_{11} = 4) = 0.4857$$.

Why was $P(n_{11} = 2)$ excluded ?

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  • $\begingroup$ It is excluded since it is less extreme than the given data. $\endgroup$
    – NCh
    Feb 21, 2020 at 3:24
  • $\begingroup$ What does that mean? $\endgroup$
    – bobby
    Feb 21, 2020 at 3:34
  • $\begingroup$ en.wikipedia.org/wiki/Fisher%27s_exact_test : the variant $n_{11}=2=n_{12}$ is favorable to independence. You have $n_{11}=3, n_{12}=1$. The variant $n_{11}=4, n_{12}=0$ is more extreme. To the opposite direction, $n_{11}=0, n_{12}=4$ is more extreme and also $n_{11}=1, n_{12}=3$ is as extreme as the given one. $\endgroup$
    – NCh
    Feb 21, 2020 at 4:34

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