Test for two proportions I have a question with three outcomes (yes, no, do not know) and observed probabilities p_yes, p_no and p_dontknow. How do I test that for the true probabilities pi_yes > pi_no holds? I think a two sample binomal test is not appropriate, since the sum of all probabilities is one.
Or could I just throw the dont know answers away and test if pi_yes* > 0.5 with a one sample binomial test?
 A: Since you have a mulitnomial outcome, the groups are not independent so I don't think a binomial test ignoring the third category would be best.
The data are multinomial with 3 categories.  This means that the estimates of the multinomial proportions $\boldsymbol{p}$ are $E(y_j) = y_j/n = p_j$ $$\boldsymbol{\Sigma}(\boldsymbol{p})=\left[\begin{array}{ccc}
p_{1}\left(1-p_{1}\right) & -p_{1} p_{2} & -p_{1} p_{3} \\
-p_{1} p_{2} & p_{2}\left(1-p_{2}\right) & -p_{2} p_{3} \\
-p_{1} p_{3} & -p_{2} p_{3} & p_{3}\left(1-p_{3}\right)
\end{array}\right]$$
Because the proportions are constrained to sum to 1, $\boldsymbol{\Sigma}$ is degenerate and so the last column is superfluous.
I think that means we can leverage the fact that the sample mean is asymptotically normal to use the following test statistic
$$z = \dfrac{\sqrt{n}(p_1 - p_2)}{\sqrt{p_1(1-p_1) + p_2(1-p_2) -2p_1p_2}} $$
Here, I've used the fact that $\operatorname{Var}(p_1-p_2) = \operatorname{Var}(p_1) + \operatorname{Var}(p_2) - 2\operatorname{Cov}(p_1,p_2)$ all of which are available from the submatrix of $\boldsymbol{\Sigma}(\boldsymbol{p})$.
Some simulation shows that the false positve rate is just slightly above the nominal (but this is maybe just due to the sample size being relatively small).

z = replicate(100000,{
  #generate data
  x = rmultinom(100, 1, c(0.4, 0.4, 0.2))
  #estimate proportions
  p = apply(x, 1, mean)
  #get covariance matrix
  s = cov(t(x))[1:2, 1:2]
  # test statistic
  z = sqrt(100)*(p[1] - p[2])/ sqrt(s[1,1] + s[2,2] - 2*s[1,2])
  
  pnorm(abs(z), lower.tail = F)*2
  
})


mean(z<0.05)
>>>0.0543 # Depending on your random seed

But I would wait for someone else to double check me on this.
