CODEWITHSUNDEEP
rstatistics
Ole Petersen
Ole Petersen

Reputation: 680

Contradiction between Pearson and Pairwise.prop.test

I have two vectors a and b with same length. The vectors contains number of times a game has been played. So for example game 1 has been played 265350 in group a while it has been played 52516 in group b.

a <- c(265350, 89148, 243182, 208991, 113090, 124698, 146574, 33649, 276435, 9320, 58630, 20139, 26178, 7837, 6405, 399)
b <- c(52516, 42840, 60571, 58355, 46975, 47262, 58197, 42074, 50090, 27198, 45491, 43048, 44512, 27266, 43519, 28766)

I want to use Pearsons Chi square test to test Independence between the two vector. In R I type

chisq.test(a,b)

and I get a p-value 0.2348 meaning that the two vectors are independent (H is true).

But when I run pairwise.prop.test(a,b) and get all the pairwise p-values and almost all of them are very low, meaning that there are pairwise dependence between the two vectors but this is in contrast to the first result. How can that be ?

Upvotes: 1

Views: 508

Answers (1)

LyzandeR
LyzandeR

Reputation: 37879

The pairwise.prop.test is not the correct test for your case.

As it mentions in the documentation:

Calculate pairwise comparisons between pairs of proportions with correction for multiple testing

And also:

x (first argument).
Vector of counts of successes or a matrix with 2 columns giving the counts of successes and failures, respectively.

And

n (second argument).
Vector of counts of trials; ignored if x is a matrix.

So, x in the number of successes out of n which is the trials, i.e. x <= (less than or equal) to each pair in n. And this is why pairwise.prop.test is used for proportions. As an example imagine tossing a coin 1000 times getting heads in 550. x would be 550 and n would be 1000. In your case you do not have something similar, you just have counts of a game in two groups.

The correct hypothesis test for testing independence is the chisq.test(a,b) that you have already used and I would trust that.

Upvotes: 1

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