On this website the appropriate statistical test for comparing two independent proportions is described as a $Z$-test (a normal distribution is used to obtain a $p$-value).
However, in R prop.test
, a Chi-square distribution is invoked to obtain the $p$-value comparing two independent proportions.
An R example is:
prop.test(x = c(474, 376), n = c(750, 750))
I was wondering how a Chi-square based approach comparing two independent proportions works (e.g., is the test a chi-square test or a different test that produces a statistic that follows a Chi-square distribution) and how it compares with the Z-test approach (e.g., in terms of accuracy and power) discussed on that website?
Best Answer
Caveat: This is my non-statistician view of the question.
The following website discusses the two approaches. The
prop.test
approach is the statistically fundamental one. The z-test approach is, in my eyes, an approximation of this, and relies on certain assumptions about the data.www.r-bloggers.com/comparison-of-two-proportions-parametric-z-test-and-non-parametric-chi-squared-methods/
Since R has good support for tests of proportions (confidence intervals,
binom.test
,multinomial.test
), I don't see why one would want to use the z-test approach. But that may just be my bias and particular education.Considering using confidence intervals in R, the following two pages have examples of confidence intervals for proportions,
binom.test
, andmultinomial.test
in R. (Caveat: I am the author of these two webpages.)rcompanion.org/handbook/H_03.html
rcompanion.org/handbook/H_02.html