I'm curious about the validity of removing/ignoring ties when calculating the p-value for the sign test.

By ties I mean ($x_{before} = x_{after}$, where $x$ represents the measurement).

According to wikipedia's example, this procedure is acceptable. However, there's no mention of whether or not removing ties is only acceptable when ties are rare. What about when there are many ties?

I was able to find a related post, but it didn't add much except recommend this paper, which suggests using "Fisher's principal of randomization"(?). My understanding from reading this is that: It's okay to subtract (i.e., remove/ignore) ties from the set. Is that correct?

  • 1
    $\begingroup$ You might look up how Mood's median test (for independent samples) handles ties. It includes them with either the > or < group. When it's likely that observations tie the median value, this can lead to some unusual results, where the result depends on if ties are included with > or <. You might also check out the Trinomial test mentioned in that Wikipedia article. $\endgroup$ Aug 22, 2019 at 10:08
  • $\begingroup$ You have misquoted the linked paper; they (correctly) use "principle" where you have "principal". Since it's an electronic document you can usually avoid such issues by copying and pasting what you want to quote (though some pdfs won't generate the selected text exactly, so you still need to check). $\endgroup$
    – Glen_b
    Aug 23, 2019 at 5:37

1 Answer 1


Thanks to Sal's comment, I think I was able to find a reasonably complete solution to my question. I'm sharing a summary, that I've put together based on this paper on Trinomial test, which I definitely recommend.

Note that $B(x,n,p)$ is binomial test for x successes and n trials with probability p. $N_+$ is the number of positives signs, $N_-$ is the number of negative signs, and $N_0$ is the number of ties.

History of treatment of ties in the sign test:

Dixon and Mood (1946): include half the number of ties to positive observations as a nonrandomized unconditional exact test ($B(N_+ + N_0/2, N, 1/2)$)

Dixon and Massey (1951) -- most popular: exclude ties $B(N_+, N - N_0, 1/2)$

Putter (1955): asymptotic uniformly most powerful nonrandomized test which uses: $S_{1/2} = \frac{N_+ - N_-}{\sqrt{N_+ + N_-}}$ Rejects the null hypothesis if $S_{1/2}$ is greater the the $100(1-\alpha)th$ percentile of a standard normal distribution. $N$ must be large.

Coakley and Heise (1996): Improved nonrandomized unconditional test: $S_{2/3} = N_+ + 2N_0/3$ where the null hypothesis is rejected if $S_{2/3} > \kappa (p_0)$. Disclosure: "I don't know what $\kappa$ means here, the paper doesn't mention, but I'm including it for the sake of being thorough".

Finally, and most importantly, according to Wittkowski (1989): If ties are due to the nature of the phenomenon, which is true in my case, they will not give valuable information. Therefore, I think they can be ignored. However, if they're due to rounding errors, their inclusion should be considered.


  • Dixon, W.J. and Mood, A.M. (1946). The Statistical Sign Test. Journal of the American Statistical Association 41, 557-566.
  • Dixon, W.J. and Massey, F.J.Jr. (1951). An Introduction to Statistical Analysis. New York: McGraw-Hill.
  • Putter, J. (1955). The Treatment of Ties in Some Nonparametric Tests. Annals of Math. Stat. 26, 368-386.
  • Coakley, C.W. and Heise, M.A. (1996). Versions of the Sign Test in the Presence of Ties. Biometrics 52(4), 1242-1251.
  • Wittkowski, K.M. (1989). An Asymptotic UMP Sign Test for Discretised Data. The Statistician 38, 93-96.
  • 1
    $\begingroup$ Can you give full references please? Many authors produce multiple papers per year and many have fairly common surnames making the effort involved in figuring out which papers you mean substantial. $\endgroup$
    – Glen_b
    Aug 23, 2019 at 5:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.