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I wasn't able to find one on Google, and was wondering if one exists...

whuber
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Tal Galili
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4 Answers4

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Good (2005) defines the one-sample sign-test for the location parameter $\theta$ for a continuous symmetric variable $X$ as follows:

  1. Take the difference $D_i$ of each observation to the location parameter $\theta_0$ under the null hypothesis.

  2. Define an indicator variable $Z_i$ as $0$ when $D_i < 0$, and as $1$ when $D_i > 0$. Since $X$ is continuous, $P(D_i = 0) = 0$.

  3. Calculate test statistic $T=\sum_i Z_i$.

  4. The distribution of $T$ is is found by generating all $2^N$ possible outcomes of the $Z_i$ indicator variables (2 possibilities for each observation with equal probability $\frac{1}{2}$ under H0). This leads to the binomial distribution as in the sign test for 2 dependent samples.

The justification for step 4 is:

Suppose we had lost track of the signs of the deviations [...]. We could attach new signs at random [...]. If we are correct in our hypothesis that the variable has a symmetric distribution about $\theta_0$, the resulting values should have precisely the same distribution as the original observations. That is, the absolute values of the deviations are sufficient for regenerating the sample. (p34f)

I agree that this reasoning seems somewhat different from a 2-sample permutation test where you re-assign experimental conditions to observations with the justification of exchangeability under H0.

Good, P. 2005. Permutation, Parametric, and Bootstrap Tests of Hypotheses. New York: Springer.

caracal
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O.k, I found that there is an unpaired solution to a sign test (A test of medians). It is called "Median test" And you can read about it in Wikipedia.

Tal Galili
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The extension goes thorugh introducing rank to somewhat regulate the order of data and the result are Wilcoxon tests (Mann-Whitney in particular).

  • Thanks mbq, but that will give us a test on the rank, not on the median - wouldn't it? – Tal Galili Nov 23 '10 at 11:07
  • @Tal Yes, but as I wrote -- this is all one can get. Yet they both answer the same holistic question "Is there a significant difference?", so both are considered nonparametric t-test variants. –  Nov 23 '10 at 12:40
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    @mbq, @Tal The Wilxocon-Mann-Whitney test is an unpaired test based on ranks. It has two interpretations (the Wikipedia article has a good discussion). One of them is that it's a test of equality of *medians* relative to an alternative hypothesis that's essentially a shift of median. BTW, the Wilcoxon test (not to be confused with the Mann-Whitney test!) is a sign test for paired data. Is this the "Sign Test" referred to in the question? – whuber Nov 23 '10 at 16:01
  • Hi Whuber, this is the sign test: http://en.wikipedia.org/wiki/Sign_test – Tal Galili Nov 23 '10 at 16:03
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    Perhaps it's useful to distinguish the three Wilcoxon tests: 1. one-sample test: "Wilcoxon signed rank test" 2. two dependent samples -> can be reduced to 1. by taking the observation-wise difference and testing for location parameter of 0. 3. two independent samples: "Wilcoxon rank sum test". This is equivalent to the Mann-Whitney U-test: both test statistics are conceptually different but numerically identical. – caracal Nov 23 '10 at 16:10
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    @Tal Yep, your "sign test" is also known as the Wilcoxon test. The Mann-Whitney U test is its unpaired counterpart. The *Mann-Kendall* test is its extension to more than two sets of data. Although it doesn't seem to be commonly known, the U statistic can be obtained as the t-statistic for the ranks of the *combined* dataset (using midranks for tied groups) and the Mann-Kendall test is really just ANOVA on the ranks of the combined data. This insight allows you to avoid the complex formulas often cited and use simple software to do the calculations. – whuber Nov 23 '10 at 16:35
  • Whuber, if I understand correctly, it can be called wilcoxon-sign-test, but not wilcoxon-sign-rank-test. Other then that - thanks for your references - all are interesting to know. – Tal Galili Nov 23 '10 at 17:48
  • @Tal There is a total chaos with these nomenclature; I used Wilcoxon as a keyword, since it is most common. And the whole bundle of such tests is located under `wilcox.test` function in R. –  Nov 23 '10 at 22:53
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I'm not sure if such a test can exist conceptually. The sign test uses the pairing of the data to decide whether one value is bigger than the corresponding other value. But in an unpaired situation there is nothing like a corresponding other value (every value in the other group could be a potential counterpart for comparison). Correct me please, if I'm not getting the point...

psj
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  • Hi psj, your answer gives a good explanation as to why I asked my question :) - exactly because I am not aware of how to generalize the sign test to unpaired data. However, there might be a solution out there, thus my question :) – Tal Galili Nov 23 '10 at 12:05
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    You have just described the Mann-Whitney U test! Also, you use "every value in the other group [as] a potential counterpart for comparison" and compute the median of all such paired differences you obtain the Hodges-Lehman estimate of the difference in medians of the two datasets: http://en.wikipedia.org/wiki/Hodges%E2%80%93Lehmann_estimate – whuber Nov 23 '10 at 16:31