I think it is generally agreed upon that the Chi-square test (specifically, the chi-square test for a 2-by-2 contingency table) is a non-parametric test. (Though there is the assumption that each "cell" has an expected outcome of greater than 5).
z-test of two proportions
The z-test for two proportions, on the other hand, is used when you have two proportions which ascribe to the binomial distribution. You can then "approximate" the normal distribution from X~B(n, p) to X~N(np, npq) with the caveat that np and nq are both greater than 5. Some therefore say that because one is using an underlying z-distribution, or that because one has approximated the binomial to the normal distribution (and made this assumption), that the z-test for two proportions is a "parametric" test.
Yet... the Chi-square test is actually identical to the z-test for two proportions - i.e. if you calculate the p-value from either the chi-square test or the z-test of two proportions, they will yield the same p-value. Furthermore, it probably is useful to note that the z^2 statistic is identical to the chi statistic.
Assuming my stated premises are all correct (please correct me if they are not!), my questions are therefore:
Given equivalence of both tests, is it more correct to say that the z-test and the chi-square test are both parametric tests or non-parametric tests?
Follow up from Q1: if we classify them both as "parametric", is there a non-parametric counterpart? Vice versa; if we classify them both as "non-parametric", is there a "parametric" counterpart?
In special circumstances when there are only two groups, the z-test should theoretically give slightly more information. This is because the z-test also gives directionality (i.e. group 1's probability has a higher or lower probability than group 2's probability). Is this correct?
(For those who would like simulation proof of the equivalence of z-test for two proportions and the chi-square test, please copy/paste the following code into R v3.5.2... it should give mostly identical p-values between the two tests depending on rounding error).
### Contrive proportions data #p1 = probability of event occuring in group 1 (2dp) #p2 = probability of event occuring in group 2 (2dp) #note, the probabilites are constrained to >0.05 and <0.95 to fulfil assumptions p1 = round(runif(1,min=0.05,max=0.95),2) p2 = round(runif(1,min=0.05,max=0.95),2) #for simplicity's sake, let's assume n=1000 n1 = 1000 n2 = 1000 q1 = 1-p1 q2 = 1-p2 ## Conduct the z-test for two proportions using formulas #pooled probability (p-hat) p = (n1*p1 + n2*p2) / (n1 + n2) #z-test formula z = (p1 - p2) / sqrt(p*(1-p)*(1/n1 + 1/n2)) #convert z-value to p-value (assume 2 tailed) p_value_from_ztest = 2*pnorm(-abs(z)) ## Convert proportions data into matrix for chi-square test matrix = as.table(rbind(c(p1*n1, p2*n2), c(q1*n1, q2*n2))) dimnames(matrix) = list(occurrence = c("YES", "NO"), group = c("1", "2")) p_value_fromchisq = chisq.test(matrix, correct = F)$p.value ### Compare pvalue from z-test and chi-square test #round to 10sigfig to avoid the rounding errors from earlier calculations... p_ztest = signif(p_value_from_ztest, 10) p_chisq = signif(p_value_fromchisq, 10) p_ztest p_chisq identical(p_ztest, p_chisq)