How to Compare more than two Averages of Proportions or Means of Proportions? My experiment deals with $60$ subjects, one third of them belong to type A subjects, $i=1,\dotsc,20$, another third to type B, $i=21,\dotsc,40$ and finally type C, $i=41,\dotsc, 60$.
Each of these $60$ subjects from the study have a different number of observable unities, $m_i$, $i=1,\dotsc, 60$, and each of these unities can be classified as successful o non-successful ones, $n_i$ is the number of successful ones out of $m_i$. I guess this means that my experiment is somehow related to a binomial experiment.
Therefore I am working with the relative number $p_i=n_i/m_i$. I consider this is my independent variable, variable of interest or response of my experiment.
I can then average the proportions of type A subjects by averaging the $20$ $p_i$, $i=1,\dotsc, 20$, obtaining $\overline{p}_A$, and in the same way I can obtain $\overline{p}_B$ and $\overline{p}_C$, which I call means of proportions or averages of proportions.
I want to check this null hypothesis $H_0: \overline{p}_A=\overline{p}_B=\overline{p}_C$ and if the null hypothesis is rejected I would like to perform some kind of post-hoc tests.
I am not able to find out the correct statistical tool that needs to be used, and the hypothesis that need to be checked. I would appreciate if anyone could put some light in my way.
 A: Are you essentially seeing if subjects in group A, group B, and group C flip heads on a coin more often? If so, how many times does each subject flip the coin? (Coin flipping is whatever task can be successful or unsuccessful.) This sounds lIke a job for a logistic regression with categorical predictors, as proportions often get tested by testing the raw success/failure events, not proportions like 0.7.
This is basically ANOVA for a binary outcome. (Since I posted this as a comment, I learned that my favorite way of testing this goes by the name G-test.)
A: To test an omnibus null hypothesis like $\text{H}_{0}\text{: }p_A = p_B = p_C$, against the alternative hypothesis $\text{H}_{\text{A}}\text{: } p_A \ne p_B \text{, or }p_A \ne p_C \text{, or }p_B \ne p_C\text{, or }p_A \ne p_B \ne p_C$, you can use a $\chi^2$ test for contingency tables.
Your test statistic for the $2\times 3$ contingency table, is:
$$\chi^2 = \sum_{\text{i=row, j=col}}^{\text{rows, cols}}\frac{(O_{ij}- E_{ij})^2}{E_{ij}}$$
where $O_ij$ is the observed counts (say, counts of participants whose outcome is $1$ in row $1$, and whose outcome is $0$ in row $2$, across each column $A, B\text{, and }C$, and where $E_{ij}$ is the expected counts in those same rows and columns, assuming $\text{H}_{0}$ is true. If $\text{H}_{0}$ is true, then these expectations are obtained by assuming that each of the three groups has the same proportion positive ($1$) and negative ($0$), and these proportions are estimated by using the marginal totals for each row (e.g., total counts with $1$ across all three groups, divided by total observations across all three groups gives probability of $1$ under the null).
The $p$ value for this test is $p = Pr(X^2_{\text{df=2}} > \chi^2)$, where $\text{df} = (\text{rows} - 1)\times(\text{cols} - 1) = 2$. For example, in R pchisq(x,df,lower.tail=FALSE) where x is your test statistic, and df your degrees of freedom.
Reject $\text{H}_0$ if $p\le \alpha$.
If you rejected $\text{H}_0$, you conclude you found evidence that at least one group was sampled from a population with a true proportion different than at least one other group at the $\alpha$ level of significance. If you failed to reject $\text{H}_0$, you conclude you failed to find evidence that at least one group was sampled from a population with a true proportion different than at least one other group at the $\alpha$ level of significance. You can continue on to post hoc $2\times2$ $\chi^2$ contingency table tests (with adjustments for multiple comparisons) if you reject.
The $G$ statistic in Dave's answer is the same procedure as this test, but instead calculates the test statistic instead as:
$$G = 2\sum_{\text{i=row, j=col}}^{\text{rows, cols}}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$$
$G$ is also distributed approximately $\chi^2$, so the remaining steps are identical for the likelihood ratio test.
A: You can use the multinomial test in this case. Think of your experiment as having 4 outcomes. You already specified successful unities for outcomes $A$, $B$ and $C$ , so add a category $F$ for when there is failure for all other outcomes. Now you want to test $H_0: \overline{p}_A= p_1$ and $\overline{p}_B = p_2$ and $\overline{p}_C = p_3$ and $\overline{p}_F = p_4$ so you can do that using the Multinomial goodness of fit test. However you would have to specify what are $p_1,p_2,p_3$ and $p_4$ are.
