sign test

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A nonparametric statistical test based on the comparison of the signs of the differences between
paired values.

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Wikipedia: Sign test
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In statistics, the sign test can be used to test the hypothesis that there is "no difference" between
the continuous distributions of two random variables X and Y. Formally:

Let p = Pr(X > Y), and then test the null hypothesis H0: p = 0.50. This hypothesis implies that
given a random pair of measurements (xi, yi), then xi and yi are equally likely to be larger than the
other.

Independent pairs of sample data are collected from the populations {(x1, y1), (x2, y2), . . ., (xn,
yn)}. Pairs are omitted for which there is no difference so that there is a possibility of a reduced
sample of m pairs.

Then let w be the number of pairs for which yi − xi > 0. Assuming that H0 is true, then W follows
a binomial distribution W ~ b(m, 0.5).

The left-tail value is computed by Pr(W ≤ w), which is the p-value for the alternative H1: p <
0.50. This alternative means that the X measurements tend to be higher.

The right-tail value is computed by Pr(W ≥ w), which is the p-value for the alternative H1: p >
0.50. This alternative means that the Y measurements tend to be higher.

For a two-sided alternative H1 the p-value is twice the smaller tail-value.

This is the html version of the file http://mlsc.lboro.ac.uk/resources/statistics/signtest.pdf.

Page 1
Statistics: 2.1 The sign test
Rosie Shier. 2004.
1 Introduction
Many statistical tests require that your data follow a normal distribution. Sometimes
this is not the case. In some instances it is possible to transform the data to make them
follow a normal distribution; in others this is not possible or the sample size might be so
small that it is difficult to ascertain whether or not the data a normally distributed. In
such cases it is necessary to use a statistical test that does not require the data to follow
a particular distribution. Such a test is called a non-parametric or distribution free
test. The sign test is an example of one of these.
The sign test is used to test the null hypothesis that the median of a distribution is equal
to some value. It can be used a) in place of a one-sample t-test b) in place of a paired
t-test or c) for ordered categorial data where a numerical scale is inappropriate but where
it is possible to rank the observations. (Note that the Wilcoxon Signed Rank Sum Test
is also appropriate in these situations and is a more powerful test than the sign test.)
2 Procedure for carrying out the sign test
The observations in a sample of size n are x
1
,x
2
,...,x
n
(these observations could be
paired differences); the null hypothesis is that the population median is equal to some
value M. Suppose that r
+
of the observations are greater than M and r
−
are smaller than
M (in the case where the sign test is being used in place of a paired t-test, M would
be zero). Values of x which are exactly equal to M are ignored; the sum r
+
+r
−
may
therefore be less than n — we will denote it by n .
Under the null hypothesis we would expect half the x’s to be above the median and half
below. Therefore, under the null hypothesis both r
+
and r
−
follow a binomial distribution
with p =
1
2
and n = n .
The test procedure is as follows:
1. Choose r = max(r
−
,r
+
).
2. Use tables of the binomial distribution to find the probability of observing a value of
r or higher assuming p =
1
2
and n = n . If the test is one-sided, this is your p-value.
3. If the test is a two-sided test, double the probabililty obtained in (2) to obtain the
p-value.
1

Page 2
Example:
The table below shows the hours of relief provided by two analgesic drugs in 12 patients
suffering from arthritis. Is there any evidence that one drug provides longer relief than
the other?
Case Drug A Drug B Case Drug A Drug B
1
2.0
3.5
7
14.9
16.7
2
3.6
5.7
8
6.6
6.0
3
2.6
2.9
9
2.3
3.8
4
2.6
2.4
10
2.0
4.0
5
7.3
9.9
11
6.8
9.1
6
3.4
3.3
12
8.5
20.9
Solution:
In this case our null hypothesis is that the median difference is zero. Our actual differ-
ences (Drug B - Drug A) are:
+1.5,+2.1,+0.3,−0.2,+2.6,−0.1,+1.8,−0.6,+1.5,+2.0,+2.3,+12.4
Our actual median difference is 1.65 hours. We have r
+
= 9,r
−
= 3,n = 12, r =
max(r
−
,r
+
) = 9. Therefore our two-sided p-value (from binomial tables) is p = 0.146.
We would conclude that there is no evidence for a difference between the two treatments.
Note that the Wilcoxon Signed Rank Sum Test would also be appropriate in this case
and is a more powerful test because it takes account of the magnitude of the differences
as well as the sign.
3 Carrying out the sign test in SPSS
Case 1: Paired data
— Choose Analyze
— Select Nonparametric Tests
— Select 2 Related Samples
— Specify which two variables comprise your pairs of observation by clicking on them
both then clicking on the arrow to put them under Test Pair(s) List.
— Under Test Type select Sign
— Click on OK
The output will look like this:
Frequencies
N
Drug B - Drug A Negative Differences
a
3
Positive Differences
b
9
Ties
c
0
Total
12
a. DRUGB < DRUGA
b. DRUGB > DRUGA
c. DRUGA = DRUGB
2

Page 3
Test Statistics
b
DRUGB - DRUGA
Exact Sig. (2-tailed) 0.146
a. Binomial distribution used.
b. Sign Test
If you were carrying out a one-sided test, you would need to divide the p-value by 2.
Case 2: Single set of observations
Note that in this case SPSS includes values that are exactly equal to the median — it
takes r
+
= the number of observations that are greater than M and r
−
= the number
of observations that are less than or equal to M. Any ties should therefore be excluded
manually as follows: choose Data then Select Cases; choose If condition is satisfied
and click on If; in the box type in your variable name then = M, where M is the median
value specified in your null hypothesis; click on Continue then OK and proceed as
detailed below.
— Choose Analyze
— Select Nonparametric Tests
— Select Binomial
— Choose the relevant variable as the Test Variable
— Under Define Dichotomy in the lower left hand corner of the pop-up screen, choose
Cut Point and specify your null value (i.e. 0 in the case of paired data, etc.)
— Click on OK
The output will look like this:
Binomial Test
Category N Observed Test Prop. Exact Sig.
Prop.
(2-tailed)
Drug B - Drug A Group 1
≤0
3
0.25
0.50
0.146
Group 2
>0
9
0.75
Total
12
1
If you were carrying out a one-sided test, you would need to divide the p-value by 2.
3

SIGN TEST
Name:
SIGN TEST
Type:
 Analysis Command
Purpose:
Perform a one sample or a paired two sample sign test.
Description:
The t-test is the standard test for testing that the difference between population
means for two paired samples are equal. If the populations are non-normal,
particularly for small samples, then the t-test may not be valid. The sign test is an
alternative that can be applied when distributional assumptions are suspect.
However, it is not as powerful as the t-test when the distributional assumptions
are in fact valid. It can also be applied in the case where there is no quantitative
scale, but it is possible to order the data (i.e., an ordinal scale). Dataplot states
the sign test in terms of medians, but it can also be expressed in terms of means.

To form the sign test, compute di = Xi - Yi where X and Y are the two samples.
Count the number of times di is positive, R+, and the number of times it is
negative, R-. If the samples have equal medians and the populations are
symmetric, then R+ and R- should be similar. If there are too many positives (R+)
or negatives (R-), then we reject the hypothesis of equality. Ties are excluded
from the analysis. Since there are only two choices (+ or -) for di the test statistic
for the sign test follows a binomial distribution with p=0.5.