1 Test Statistic for Sign Test for Paired Samples in R
2 Simple Sign Test for Paired Samples in R
3 Sign Test for Paired Samples Critical Value in R
4 Two-tailed Sign Test for Paired Samples in R
5 One-tailed Sign Test for Paired Samples in R
6 Sign Test for Paired Samples: Test Statistics & P-values in R

Here, we discuss the sign test for paired samples in R with interpretations, including, test statistics, p-values, and confidence intervals.

The sign test for paired samples in R can be performed with the SIGN.test() function from the "BSDA" package.

The sign test for paired samples can be used to test whether the median of the differences between paired values from the two dependent populations where two dependent random samples come from is equal to a certain value (which is stated in the null hypothesis) or not. It is a non-parametric alternative to the paired two sample t-test.

In the sign test for paired samples, the test statistic is based on the signs of the observed paired differences minus the null hypothesis median. It is the number of observed paired differences greater than the null hypothesis median of paired difference, and inference is based on binomial distribution.

Sign Test for Paired Samples Tests & Hypotheses
Question	Is the median of paired x and y differences equal to $m_0$?	Is the median of paired x and y differences greater than $m_0$?	Is the median of paired x and y differences less than $m_0$?
Form of Test	Two-tailed	Right-tailed test	Left-tailed test
Null Hypothesis, $H_0$	$m_d = m_0$	$m_d = m_0$	$m_d = m_0$
Alternate Hypothesis, $H_1$	$m_d \neq m_0$	$m_d > m_0$	$m_d < m_0$

Install and Load "BSDA" Package:

To use the SIGN.test() function form the "BSDA" package, first install the package, then load it as follows:

install.packages("BSDA")

library(BSDA)

Sample Steps to Run a Sign Test for Paired Samples Test:

# Create the data samples for the sign test for paired samples
# Values are paired based on matching position in each sample

data_x = c(5.3, 8.2, 6.8, 6.3, 5.6)
data_y = c(7.8, 6.2, 9.2, 7.1, 5.6)

# Run the sign test for paired samples with specifications

SIGN.test(data_x, data_y,
          alternative = "two.sided",
          md = 0, conf.level = 0.95)


    Dependent-samples Sign-Test

data:  data_x and data_y
S = 1, p-value = 0.625
alternative hypothesis: true median difference is not equal to 0
93.75 percent confidence interval:
 -2.5  2.0
sample estimates:
median of x-y 
         -0.8

Table of Some Sign Test for Paired Samples Arguments in R
Argument	Usage
x, y	x is the sample data values for the first sample, and y is for the second sample
md	Population median value of paired differences in null hypothesis
alternative	Set alternate hypothesis as "greater", "less", or the default "two.sided"
conf.level	Level of confidence for the test and confidence interval (default = 0.95)

Creating a Sign Test for Paired Samples Object:

# Create data
data_x = rnorm(100); data_y = rnorm(100)

# Create object
sgnt_object = SIGN.test(data_x, data_y,
                        alternative = "two.sided",
                        md = 0, conf.level = 0.95)

# Extract a component
sgnt_object$statistic

 S 
44

Table of Some Sign Test for Paired Samples Object Outputs in R
Test Component	Usage
sgnt_object$statistic	Test statistic; number of paired differences greater than the median
sgnt_object$p.value	P-value
sgnt_object$estimate	Point estimate or sample median of paired differences
sgnt_object$conf.int	Confidence interval

1 Test Statistic for Sign Test for Paired Samples in R

The sign test for paired samples has test statistics, $S$, of the form:

\[S = \sum_{i=1}^n I_{([x_i-y_i]-m_0)>0},\]

which is the number of observed paired differences greater than the null hypothesis median.

For $n_d$ which is the number of non-zero $((x_i-y_i) - m_0)$, inference on $S$ is based on the binomial distribution with $\text{size} = n_d$ and $\text{prob} = 0.5$.

$x_i's$ and $y_i's$ are the sample values,

$m_0$ is the population median of paired differences to be tested and set in the null hypothesis,

$I_{([x_i-y_i]-m_0)>0}$ is $1$ when $([x_i-y_i]-m_0)>0$ and $0$ otherwise, and

$n$ is the sample size of paired differences.

For unpaired (or independent) samples, see the Wilcoxon rank sum test.

2 Simple Sign Test for Paired Samples in R

Enter the data by hand.

data_x = c(20.9, 22.3, 18.1, 19.5, 16.9,
           20.4, 17.2, 18.5, 19.2, 25.5,
           23.0, 20.7, 20.2, 17.1, 19.8)
data_y = c(24.3, 21.4, 20.1, 18.7, 13.8,
           23.4, 15.6, 22.2, 25.7, 15.7,
           22.1, 19.2, 15.3, 15.5, 15.2)

For the following null hypothesis $H_0$, and alternative hypothesis $H_1$, with the level of significance $\alpha=0.05$.

$H_0:$ the population median of the paired differences is equal to 0 ($m_d = 0$).

$H_1:$ the population median of the paired differences is not equal to 0 ($m_d \neq 0$, hence the default two-sided).

Because the level of significance is $\alpha=0.05$, the level of confidence is $1 - \alpha = 0.95$.

The SIGN.test() function has the default alternative as "two.sided", the default median of the paired differences as 0, and the default level of confidence as 0.95, hence, you do not need to specify the "alternative", "md", and "conf.level" arguments in this case.

SIGN.test(data_x, data_y,
          alternative = "two.sided",
          md = 0, conf.level = 0.95)

Or:

SIGN.test(data_x, data_y)


    Dependent-samples Sign-Test

data:  data_x and data_y
S = 10, p-value = 0.3018
alternative hypothesis: true median difference is not equal to 0
95 percent confidence interval:
 -2.821832  2.832747
sample estimates:
median of x-y 
          0.9 

Achieved and Interpolated Confidence Intervals: 

                  Conf.Level  L.E.pt U.E.pt
Lower Achieved CI     0.8815 -2.0000 1.6000
Interpolated CI       0.9500 -2.8218 2.8327
Upper Achieved CI     0.9648 -3.0000 3.1000

The sample median of paired differences, $\tilde d$, is 0.9,

test statistic, $S$, is 10,

the p-value, $p$, is 0.3018,

the interpolated 95% confidence interval is [-2.821832, 2.832747].

Interpretation:

Note that for SIGN.test() in R, the two methods may disagree for some edge cases, as p-value is based on binomial distribution, and confidence interval is based on interpolation of binomial distribution.

P-value: With the p-value ($p = 0.3018$) being greater than the level of significance 0.05, we fail to reject the null hypothesis that the population median of the paired differences is equal to 0. This is the same as the "two.sided" test for $\text{Binomial($n_d$, 0.5)}$, that is, binom.test(10, 15, 0.5, "two.sided")$p.value $=0.3017578$.
Confidence Interval: With the null hypothesis median of paired differences ($m_d = 0$) being inside the confidence interval, $[-2.821832, 2.832747]$, we fail to reject the null hypothesis that the population median of the paired differences is equal to 0.

3 Sign Test for Paired Samples Critical Value in R

To get the critical value for a sign test for paired samples in R, you can use the qbinom() function for binomial distribution to derive the quantile associated with the given level of significance value $\alpha$.

With $n_d$ as the number of non-zero $((x_i-y_i) - m_0)$ as above.

For two-tailed test with level of significance $\alpha$. The critical values are: qbinom($\alpha/2$, $n_d$, 0.5) - 1, and qbinom($1-\alpha/2$, $n_d$, 0.5) + 1.

For one-tailed test with level of significance $\alpha$. The critical value is: for left-tailed, qbinom($\alpha$, $n_d$, 0.5) - 1; and for right-tailed, qbinom($1-\alpha$, $n_d$, 0.5) + 1.

Example:

For $\alpha = 0.1$, $n_d = 60$.

Two-tailed:

qbinom(0.05, 60, 0.5) - 1; qbinom(0.95, 60, 0.5) + 1

[1] 23

[1] 37

One-tailed:

#Left
qbinom(0.1, 60, 0.5) - 1

[1] 24

#Right
qbinom(0.9, 60, 0.5) + 1

[1] 36

4 Two-tailed Sign Test for Paired Samples in R

Using a subset of the AirPassengers data from the "datasets" package:

Air_data = data.frame(matrix(AirPassengers, ncol = 12, byrow = TRUE))[,7:12]
colnames(Air_data) = c("Jul", "Aug", "Sep", "Oct", "Nov", "Dec")
rownames(Air_data) = 1949:1960
Air_data

     Jul Aug Sep Oct Nov Dec
1949 148 148 136 119 104 118
1950 170 170 158 133 114 140
1951 199 199 184 162 146 166
1952 230 242 209 191 172 194
1953 264 272 237 211 180 201
1954 302 293 259 229 203 229
1955 364 347 312 274 237 278
1956 413 405 355 306 271 306
1957 465 467 404 347 305 336
1958 491 505 404 359 310 337
1959 548 559 463 407 362 405
1960 622 606 508 461 390 432

For "Sep" as the x group versus "Oct" as the y group.

For the following null hypothesis $H_0$, and alternative hypothesis $H_1$, with the level of significance $\alpha=0.1$.

$H_0:$ the population median of the paired differences is equal to 50 ($m_d = 50$).

$H_1:$ the population median of the paired differences is not equal to 50 ($m_d \neq 50$, hence the default two-sided).

Because the level of significance is $\alpha=0.1$, the level of confidence is $1 - \alpha = 0.9$.

SIGN.test(Air_data$Sep, Air_data$Oct,
          alternative = "two.sided",
          md = 50, conf.level = 0.9)


    Dependent-samples Sign-Test

data:  Air_data$Sep and Air_data$Oct
S = 2, p-value = 0.03857
alternative hypothesis: true median difference is not equal to 50
90 percent confidence interval:
 23.71545 47.85636
sample estimates:
median of x-y 
           34 

Achieved and Interpolated Confidence Intervals: 

                  Conf.Level  L.E.pt  U.E.pt
Lower Achieved CI     0.8540 25.0000 47.0000
Interpolated CI       0.9000 23.7155 47.8564
Upper Achieved CI     0.9614 22.0000 49.0000

Interpretation:

P-value: With the p-value ($p = 0.03857$) being less than the level of significance 0.1, we reject the null hypothesis that the population median of the paired differences is equal to 50.
$S$ T-statistic: With $n_d = 12$, and test statistics value ($S = 2$) being less than or equal to $\text{qbinom(0.05, 12, 0.5)-1}=2$, or being within $0 \text{ to } 2$, we reject the null hypothesis that the population median of the paired differences is equal to 50.
Confidence Interval: With the null hypothesis median of the paired differences ($m_d = 50$) being outside the confidence interval, $[23.71545, 47.85636]$, we reject the null hypothesis that the population median of the paired differences is equal to 50.

5 One-tailed Sign Test for Paired Samples in R

Right Tailed Test

Using a subset of the AirPassengers data from the "datasets" package:

Air_data = data.frame(matrix(AirPassengers, ncol = 12, byrow = TRUE))[,1:6]
colnames(Air_data) = c("Jan", "Feb", "Mar", "Apr", "May", "Jun")
rownames(Air_data) = 1949:1960
Air_data

     Jan Feb Mar Apr May Jun
1949 112 118 132 129 121 135
1950 115 126 141 135 125 149
1951 145 150 178 163 172 178
1952 171 180 193 181 183 218
1953 196 196 236 235 229 243
1954 204 188 235 227 234 264
1955 242 233 267 269 270 315
1956 284 277 317 313 318 374
1957 315 301 356 348 355 422
1958 340 318 362 348 363 435
1959 360 342 406 396 420 472
1960 417 391 419 461 472 535

For "Mar" as the x group versus "Apr" as the y group.

For the following null hypothesis $H_0$, and alternative hypothesis $H_1$, with the level of significance $\alpha=0.1$.

$H_0:$ the population median of the paired differences is equal to 4 ($m_d = 4$).

$H_1:$ the population median of the paired differences is greater than 4 ($m_d > 4$, hence one-sided).

Because the level of significance is $\alpha=0.1$, the level of confidence is $1 - \alpha = 0.9$.

SIGN.test(Air_data$Mar, Air_data$Apr,
          alternative = "greater",
          md = 4, conf.level = 0.9)


    Dependent-samples Sign-Test

data:  Air_data$Mar and Air_data$Apr
S = 7, p-value = 0.2744
alternative hypothesis: true median difference is greater than 4
90 percent confidence interval:
 3.223434      Inf
sample estimates:
median of x-y 
            7 

Achieved and Interpolated Confidence Intervals: 

                  Conf.Level L.E.pt U.E.pt
Lower Achieved CI     0.8062 4.0000    Inf
Interpolated CI       0.9000 3.2234    Inf
Upper Achieved CI     0.9270 3.0000    Inf

Interpretation:

P-value: With the p-value ($p = 0.2744$) being greater than the level of significance 0.1, we fail to reject the null hypothesis that the population median of the paired differences is equal to 4.
$S$ T-statistic: $n_d = 12 - 1 = 11$ because one paired difference is $4$. With $n_d = 11$, and test statistics value ($S = 7$) being less than $\text{qbinom(0.9, 11, 0.5)+1}=9$, or being outside $9 \text{ to } 11$, we fail to reject the null hypothesis that the population median of the paired differences is equal to 4.
Confidence Interval: With the null hypothesis median of the paired differences ($m_d = 4$) being inside the confidence interval, $[3.223434, \infty)$, we fail to reject the null hypothesis that the population median of the paired differences is equal to 4.

Left Tailed Test

For "Jun" as the x group versus "May" as the y group.

For the following null hypothesis $H_0$, and alternative hypothesis $H_1$, with the level of significance $\alpha=0.05$.

$H_0:$ the population median of the paired differences is equal to 65 ($m_d = 65$).

$H_1:$ the population median of the paired differences is less than 65 ($m_d < 65$, hence one-sided).

Because the level of significance is $\alpha=0.05$, the level of confidence is $1 - \alpha = 0.95$.

SIGN.test(Air_data$Jun, Air_data$May,
          alternative = "less",
          md = 65, conf.level = 0.95)


    Dependent-samples Sign-Test

data:  Air_data$Jun and Air_data$May
S = 2, p-value = 0.01929
alternative hypothesis: true median difference is less than 65
95 percent confidence interval:
     -Inf 58.99727
sample estimates:
median of x-y 
           40 

Achieved and Interpolated Confidence Intervals: 

                  Conf.Level L.E.pt  U.E.pt
Lower Achieved CI     0.9270   -Inf 56.0000
Interpolated CI       0.9500   -Inf 58.9973
Upper Achieved CI     0.9807   -Inf 63.0000

Interpretation:

P-value: With the p-value ($p = 0.01929$) being less than the level of significance 0.05, we reject the null hypothesis that the population median of the paired differences is equal to 65.
$S$ T-statistic: With $n_d = 12$, and the test statistics value ($S = 2$) being less than or equal to $\text{qbinom(0.05, 12, 0.5)-1}=2$, or being within $0 \text{ to } 2$, we reject the null hypothesis that the population median of the paired differences is equal to 65.
Confidence Interval: With the null hypothesis median value ($m_d = 65$) being outside the confidence interval, $(-\infty, 58.99727]$, we reject the null hypothesis that the population median of the paired differences is equal to 65.

6 Sign Test for Paired Samples: Test Statistics & P-values in R

Here for a sign test for paired samples, we show how to get the test statistics (or S-value), and p-values from the SIGN.test() function in R, or by written code.

data_x = Air_data$Jun; data_y = Air_data$May
sgnt_object = SIGN.test(data_x, data_y,
                        alternative = "two.sided",
                        md = 60, conf.level = 0.95)
sgnt_object


    Dependent-samples Sign-Test

data:  data_x and data_y
S = 3, p-value = 0.146
alternative hypothesis: true median difference is not equal to 60
95 percent confidence interval:
 15.06364 62.25545
sample estimates:
median of x-y 
           40 

Achieved and Interpolated Confidence Intervals: 

                  Conf.Level  L.E.pt  U.E.pt
Lower Achieved CI     0.8540 24.0000 56.0000
Interpolated CI       0.9500 15.0636 62.2555
Upper Achieved CI     0.9614 14.0000 63.0000

To get the test statistic or S-value:

\[S = \sum_{i=1}^n I_{([x_i-y_i]-m_0)>0},\]

which is the number of observed paired differences greater than the null hypothesis median.

sgnt_object$statistic

S 
3

# to remove name S
unname(sgnt_object$statistic)

[1] 3

Same as:

md = 60; diffs = (data_x - data_y) - md
sum(diffs>0)

[1] 3

To get the p-value:

with $n_d$ as the number of non-zero $((x_i-y_i) - m_0)$.

For $X \sim Binomial(n_d, 0.5)$

Two-tailed:

For $S = n_d/2$, $Pvalue = 1$.

For $S > n_d/2$, $Pvalue = 2 * P(X \geq S) = 2 * \sum_{x=S}^{n_d} P(X=x)$.

For $S < n_d/2$, $Pvalue = 2 * P(X \leq S) = 2 * \sum_{x=0}^{S} P(X=x)$.

One-tailed:

For right-tail, $Pvalue = P(X \geq S) = \sum_{x=S}^{n_d} P(X=x)$ or for left-tail, $Pvalue = P(X \leq S) = \sum_{x=0}^{S} P(X=x)$.

sgnt_object$p.value

[1] 0.1459961

Same as:

md = 60; diffs = (data_x - data_y) - md
nd = sum(diffs!=0); nd

[1] 12

Note that the p-value depends on the $\text{test statistics}$ ($S = 3 < n_d/2 = 6$). We also use the distribution function pbinom() for the binomial distribution in R.

2*sum(dbinom(0:3, 12, 0.5))

[1] 0.1459961

2*pbinom(3, 12, 0.5)

[1] 0.1459961

One-tailed example:

# Right tailed
1-pbinom(2, 12, 0.5)
# Left tailed
pbinom(3, 12, 0.5)

Question	Is the median of paired x and y differences equal to \(m_0\)?	Is the median of paired x and y differences greater than \(m_0\)?	Is the median of paired x and y differences less than \(m_0\)?
Form of Test	Two-tailed	Right-tailed test	Left-tailed test
Null Hypothesis, \(H_0\)	\(m_d = m_0\)	\(m_d = m_0\)	\(m_d = m_0\)
Alternate Hypothesis, \(H_1\)	\(m_d \neq m_0\)	\(m_d > m_0\)	\(m_d < m_0\)

Sign Tests for Paired Samples in R

Install and Load "BSDA" Package:

Sample Steps to Run a Sign Test for Paired Samples Test:

Creating a Sign Test for Paired Samples Object:

1 Test Statistic for Sign Test for Paired Samples in R

2 Simple Sign Test for Paired Samples in R

Interpretation:

3 Sign Test for Paired Samples Critical Value in R

4 Two-tailed Sign Test for Paired Samples in R

Interpretation:

5 One-tailed Sign Test for Paired Samples in R

Right Tailed Test

Interpretation:

Left Tailed Test

Interpretation:

6 Sign Test for Paired Samples: Test Statistics & P-values in R

To get the test statistic or S-value:

To get the p-value: