This post discusses non parametric testing methods: sign test, signed rank test and rank sum test.

### Sign Test

Sign test tests if the median of data is md. For example, given:

> x <- c(1, 3, 2, 7, 4, 6, 8, 2)

If , then the number of positive signs of x – 5 should be about 4 (= the number of data / 2 = 8 / 2):

> sign(x - 5) [1] -1 -1 -1 1 -1 1 1 -1

If holds, number of positive sign is where 8 is the number of data. Thus, having observed 3 positive signs, we set in and .

Note that we should remove values with x = 5 as it does not fit binomial distribution we’re considering(plus or minus). But we don’t have 5 in this example.

Then, the p value is:

> 2 * pbinom(3, 8, .5) [1] 0.7265625

Thus, we can not reject .

We can do the same thing with binom.test():

> binom.test(3, 8) Exact binomial test data: 3 and 8 number of successes = 3, number of trials = 8, p-value = 0.7266 alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval: 0.08523341 0.75513678 sample estimates: probability of success 0.375

See that the p-value is the same as we calculated manually. BSDA package has a convenient function SIGN.test:

> install.packages("BSDA") > library(BSDA) > SIGN.test(x, md=5) One-sample Sign-Test data: x s = 3, p-value = 0.7266 alternative hypothesis: true median is not equal to 5 95 percent confidence interval: 1.675 7.325 sample estimates: median of x 3.5 Conf.Level L.E.pt U.E.pt Lower Achieved CI 0.9297 2.000 7.000 Interpolated CI 0.9500 1.675 7.325 Upper Achieved CI 0.9922 1.000 8.000

Based on the same idea, we can test if x and y has the median difference of 0:

> SIGN.test(c(1, 3, 2, 5, 6), c(2, 5, 3, 2, 1), md=0) Dependent-samples Sign-Test data: c(1, 3, 2, 5, 6) and c(2, 5, 3, 2, 1) S = 2, p-value = 1 alternative hypothesis: true median difference is not equal to 0 93.75 percent confidence interval: -2 5 sample estimates: median of x-y -1

### Signed Rank Test

Wilcoxon’s signed rank test uses the magnitude of difference in addition to the sign. In other words , sign test looked at sign (+1 or -1) only. But in signed rank test, we look at difference magnitude and their rank as well. Also, signed rank test assumes the difference, i.e., x – 5, is symmetric for two-sided test. (See assumptions section in the wiki.)

> m <- rbind(abs(x-5), rank(abs(x-5)), sign(x-5)) > rownames(m) <- c("abs", "rank", "sign") > m [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] abs 4 2.0 3 2.0 1.0 1.0 3 3 rank 8 3.5 6 3.5 1.5 1.5 6 6 sign -1 -1.0 -1 1.0 -1.0 1.0 1 -1

Then, we get the sum of rank for sign = -1 (T_minus) and sum of rank for sign = +1 (T_plus):

> T_plus <- sum(m["rank", m["sign",] > 0]) > T_minus <- sum(m["rank", m["sign",] < 0])

Finally, we get min of T_plus and T_minus:

> T_test <- min(T_plus, T_minus) > T_test [1] 11

If median were md, then T_test should not be too different from 18 (= (1+2+…+8)/2) as half of x-5 should be larger than 0 and half of them should be smaller than 0.

Package exactRankTests has wilcox.test() and wilcox.exact(). Among them, wilcox.exact() considers tie in ranking properly:

> wilcox.exact(x, mu=5, exact=FALSE) Asymptotic Wilcoxon signed rank test data: x V = 11, p-value = 0.3234 alternative hypothesis: true mu is not equal to 5

We can’t reject that mu equals to 5. Please note that I’ve used exact=FALSE. It’s necessary when there’s tie in the data. If there’s tie, we can’t compute p-value exactly, and need to use approximation to gaussian by specifying exact=FALSE.

For paired test(test if median of difference is 5):

> x [1] 1 3 2 7 4 6 8 2 > y [1] 1 3 2 4 5 3 2 7 > wilcox.exact(x, y, mu=5, exact=FALSE, paired=T) Asymptotic Wilcoxon signed rank test data: x and y V = 1, p-value = 0.01661 alternative hypothesis: true mu is not equal to 5

### Rank Sum Test

Rank sum test is for samples taken independently from x and y. Basic idea is to combine x and y, then sort them to give them ranks. If x and y are not too different, rank sum of x and y shouldn’t be too different:

> m <- rbind(c(x,y), c(rep(0, 8), rep(1, 8)), rank(c(x, y))) > rownames(m) <- c("data", "source", "rank") > m [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] data 1.0 3 2.0 7.0 4.0 6 8 2.0 1.0 3 2.0 4.0 5 source 0.0 0 0.0 0.0 0.0 0 0 0.0 1.0 1 1.0 1.0 1 rank 1.5 8 4.5 14.5 10.5 13 16 4.5 1.5 8 4.5 10.5 12 [,14] [,15] [,16] data 3 2.0 7.0 source 1 1.0 1.0 rank 8 4.5 14.5 > sum(m["rank", m["source",]==0]) [1] 72.5 > sum(m["rank", m["source",]==1]) [1] 63.5

More precisely, the sum of rank should be proportional to the number of data belongs to each source. But in our case, as the number of data in x and y is the same, they should be very similar.

wilcox.test() and wilcox.exact() performs rank sum test if paired = FALSE (which is default):

> wilcox.exact(x, y, exact=FALSE) Asymptotic Wilcoxon rank sum test data: x and y W = 36.5, p-value = 0.6322 alternative hypothesis: true mu is not equal to 0

As p > 0.05, if can not reject . Thus, mu is equal to zero.

### Efficiency

According to [4], signed rank test and rank sum test is efficient enough compared to t-test for data from normal distribution. Compared to signed rank test and rank sum test, t-test needed 90% of data, meaning that those non parametric methods are not too inferior to the parametric testing.

References)

1. r-tutor.com: sign test

2. rtutor.com: wilcoxon signed rank test

3. 임동훈, R을 이용한 비모수 통계학, 자유아카데미.

4. 배도선 외, 통계학 이론과 응용, 청문각.

5. http://en.wikipedia.org/wiki/Sign_test

6. http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test

7. http://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U: This is rank sum test called also as Mann–Whitney U test or Mann–Whitney–Wilcoxon (MWW) or Wilcoxon rank-sum test or Mann-Whitney test.

8. 안재형, R을 이용한 누구나하는 통계분석, 한나래.

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