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I have $3$ groups ($n_1 = 30$, $n_2 = 20$, $n_3 = 5$), and I am looking at anxiety scores, which is binary (0 = no anxiety, 1 = anxiety present). I want to perform pairwise comparisons i.e is there a significant difference in anxiety scores between group 1 v.s group 2? Group 2 v.s group 3? Group 1 v.s group 3?

I am not quite sure if t-test would be appropriate here since I don't know if the variables are iid normal, so I am thinking of using the Wilcoxon signed-rank test (wilcox.test(..., paired = FALSE) in R). What other test would be appropriate here?

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The answer would probably include repeating this post dealing with the resilience of the chi square test even when expected cell counts are below 5. So a chi square test may be the answer, and it could be approached as a goodness of fit (GOF), comparing the distribution of the frequency of "anxious" subjects to a uniform distribution across the 3 groups.

However, there will be probably a warning message when running the chi square test. I simulated the data set (the code is here, and also pasted below this paragraph for convenience. There were the same number of subjects as in the original post group a = 30, b =20 and c = 5:

subjects <- c(group_a <- rep("a", 30), group_b <- rep("b", 20), group_c <- rep("c", 5))
sam <-      c(sample(c(c(rep("N",25)), c(rep("A",5)))),
            sample(c(c(rep("N",10)), c(rep("A",10)))),
            c(sample(c("A","N"), 5, replace = T)))
(tab = table(subjects,sam))  
        sam
subjects  A  N
       a  5 25
       b 10 10
       c  3  2

The group a was set up as much less anxious: 25 N for "no anxiety".

The test output was:

    (p_observed = chisq.test(tab, correct = F))
    Pearson's Chi-squared test

data:  tab
X-squared = 7.9142, df = 2, 

$p-value = \color{orange}{0.01912}$

But there was a warning message:

Warning message:
In chisq.test(tab, correct = F) :
  Chi-squared approximation may be incorrect

So I wanted to see if we could come up with an ad hoc permutation test, and shuffle the A, N labels across all subjects, under the null hypothesis of no differences in distribution of anxious subjects across groups, and then run a chi square test on the tabulated resulting frequencies. Disregarding thus any issues with the small size of group 3, and just paying attention to the relative p value of every iteration with respect to the others, I think it is fair to say that the proportion of permutations with a p value lower than the observed in the actual data is an exact p value - a simulated simulated Fisher test, or, probably more exactly, a permutation test.

Here's the code in R, and the results:

set.seed(0)
options(warn=0)
            (tab = table(subjects,sam))  
            (p_observed = chisq.test(tab, correct = F))
chisq.test(tab, simulate.p.value = T)

pval <- c(NA, length(sam))
options(warn=-1)

            for (i in 1:1e4){
       anx <- sample(sam, length(sam), replace = F)
       tab     <- table(subjects,anx)
       pval[i] <- chisq.test(tab, correct = F)$p.value
} 
(p_value = mean(pval < p_observed$p.value))

The p_value = $\color{red}{0.0186}$ was lower than the initially calculated with chisq.test.

I was surprised that this p value was also lower than the calculation via Monte Carlo simulation within the R built-in function:

chisq.test(tab, simulate.p.value = T)

    Pearson's Chi-squared test with simulated p-value (based
    on 2000 replicates)

data:  tab
X-squared = 7.9142, df = NA, p-value = 0.02099

After getting a significant result, pairwise comparisons can be performed with Bonferroni correction (level of significance $0.05 /\text{no.hypotheses} = 0.05 / 3 = \color{blue}{0.0167}$). These can be obtained directly with the R built-in fisher.test function:

Between group a and b:

fisher.test(tab[1:2,], alternative = "two.sided")
p-value = 0.02546

Between b and c:

fisher.test(tab[2:3,], alternative = "two.sided")
p-value = 1

And a and c:

fisher.test(tab[c(1,3),], alternative = "two.sided")
p-value = 0.06654

Oddly, none of the results is significant, because of the conservative nature of the Fisher test. If we were to run a chisq.test between a and b - and the size of the samples would clearly allow it - we would get a statistically significant result:

$p-value = \color{green}{0.01174}.$

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Are you trying to compare proportions between 2 groups? If yes I think you can use Z-test where Z= (p1-p2)/sqrt(p(1-p)/(1/n1+1/n2)) and p is the proportion when 2 groups are combined.

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  • $\begingroup$ Welcome to our site! How would you propose coping with the two salient complications: namely, (1) that one of the groups has only five elements, suggesting a Z test could be a poor approximation; and (2) that three interdependent comparisons will be made, indicating the need somehow to adjust the p-values of the separate test results? $\endgroup$ – whuber Jan 29 '16 at 14:39
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    $\begingroup$ @whuber I was curious to see if the conversation got going further, but since it may not be the case... I think the answer is a GOF chi square to test whether the proportion of anxiety cases are distributed uniformly across groups. The problem with the low counts in group 3 may not be prohibitive. Pairwise comparisons with Bonferroni or other adjustments may follow. Any role for Freeman test, or any R Monte Carlo to permute tables? $\endgroup$ – Antoni Parellada Jan 29 '16 at 20:43
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    $\begingroup$ @Antoni All sound like good ideas. You could add to them other well-known approaches such as logistic regression. Whether or not the low count in a group is a problem can be decided in various ways, but what is of greatest importance in this forum is to recognize that it is a potential problem, rather than to ignore it. $\endgroup$ – whuber Jan 29 '16 at 20:46

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