The sign test

We have seen the definition of a non-parametric hypothesis test. The sign test is an example. We want to test the effectiveness of a drug from 12 observations. The data is the number of hours that the drug relieves pain. The null hypothesis is that the difference between paired medians equals zero.

Case #	Drug A	Drug B
1	2.0	3.5
2	3.6	5.7
3	2.6	2.9
4	2.6	2.4
5	7.3	9.9
6	3.4	3.3
7	14.9	16.7
8	6.6	6.0
9	2.3	3.8
10	2	4.0
11	6.8	9.1
12	8.5	20.9

The paired differences (Drug B – Drug A) are:

data <- c(1.5, 2.1, 0.3, -0.2, 2.6, -0.1, 1.8, -0.6, 1.5, 2.0, 2.3, 12.4)

Let’s order them in the increasing magnitude.

sort(data) 

-0.6 -0.2 -0.1  0.3  1.5  1.5  1.8  2.0  2.1  2.3  2.6 12.4

Under the null hypothesis, we expect half the numbers to be above zero (median) and half below. Suppose r⁺ observations are > 0 and r⁻ < 0, then under the null hypothesis, r⁺ and r⁻ follow a binomial distribution with p = 1/2.

In our case, three cases are below zero (r^–), and nine are above (r⁺). So, we estimate the p-value in a binomial test with 9 successes out of 12, but the expected probability is 0.5 under the null hypothesis.

binom.test(9, 12,  p = 0.5, alternative = "two.sided") 

	Exact binomial test

data:  9 and 12
number of successes = 9, number of trials = 12, p-value = 0.146
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.4281415 0.9451394
sample estimates:
probability of success 
                  0.75 

Since the p-value is 0.156, we would conclude that there is no evidence of a difference between the two treatments.