It appears you conducted a simple chi-square test of independence (or test of association). This test does not make any assumptions about normality or heteroscedasticity. The test does make an assumption of independence, in the sense that, for example, no two numbers in the table should include the same individual. But there is no test for this.

For the reader, the following code in R duplicates the analysis.

A = 6

B = 34 - 6

C = 10

D = 21- 10

M = matrix(c(A,B,C,D), nrow=2)

M

chisq.test(M, correct=F)

### Pearson's Chi-squared test

###

### X-squared = 5.6534, df = 1, p-value = 0.01742

I don't think its strictly true that the chi-square test of independence makes no assumptions about normality. It relies on asymptotic normality of the standardised residuals which doesn't hold when there are low expected counts. This is usually presented as continuous vs. discrete data issue. I think the independence issue could also be reframed this way in relation to the dispersion of the counts (which has to do with clustering and dispersion being different from that expended from independent observations). So in checking independence and the proportion of expected counts that are say below 5 you are indirectly checking these assumptions.

You can see the link in the test statistic as the Chi-square is just a sum of the squared z values from the standardised residuals. It would be weird if a test statistic which is derived from the standard normal (z) distribution didn't in some sense assume normality.