# Sign test - overview

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Sign test
Binomial test for a single proportion
Chi-squared test for the relationship between two categorical variables
Kruskal-Wallis test
Independent variableIndependent variableIndependent /column variableIndependent/grouping variable
2 paired groupsNoneOne categorical with $I$ independent groups ($I \geqslant 2$)One categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent /row variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)One of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
• H0: P(first score of a pair exceeds second score of a pair) = P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
• H0: the population median of the difference scores is equal to zero
A difference score is the difference between the first score of a pair and the second score of a pair.
H0: $\pi = \pi_0$

Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
H0: there is no association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
• H0: the distribution of the dependent variable is the same in each of the $I$ populations
If there is one random sample of size $N$ from the total population:
• H0: the row and column variables are independent
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
• H1 two sided: P(first score of a pair exceeds second score of a pair) $\neq$ P(second score of a pair exceeds first score of a pair)
• H1 right sided: P(first score of a pair exceeds second score of a pair) > P(second score of a pair exceeds first score of a pair)
• H1 left sided: P(first score of a pair exceeds second score of a pair) < P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
• H1 two sided: the population median of the difference scores is different from zero
• H1 right sided: the population median of the difference scores is larger than zero
• H1 left sided: the population median of the difference scores is smaller than zero
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1: there is an association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
• H1: the distribution of the dependent variable is not the same in all of the $I$ populations
If there is one random sample of size $N$ from the total population:
• H1: the row and column variables are dependent
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
AssumptionsAssumptionsAssumptionsAssumptions
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
• 2 $\times$ 2 table: all four expected cell counts are 5 or more
• Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
• There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
Test statisticTest statisticTest statisticTest statistic
$W =$ number of difference scores that is larger than 0$X$ = number of successes in the sample$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.

$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$

Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.

Note: if ties are present in the data, the formula for $H$ is more complicated.
Sampling distribution of $W$ if H0 were trueSampling distribution of $X$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $H$ if H0 were true
The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.

If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
Binomial($n$, $P$) distribution.

Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom

For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.

For small samples, the exact distribution of $H$ should be used.

Significant?Significant?Significant?Significant?
If $n$ is small, the table for the binomial distribution should be used:
Two sided:
• Check if $W$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $W$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $W$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$

If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
Two sided:
• Check if $X$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $X$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $X$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Equivalent ton.a.n.a.n.a.
Two sided sign test is equivalent to
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Example contextExample contextExample contextExample context
Do people tend to score higher on mental health after a mindfulness course?Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?Is there an association between economic class and gender? Is the distribution of economic class different between men and women?Do people from different religions tend to score differently on social economic status?
SPSSSPSSSPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
• Put the two paired variables in the boxes below Variable 1 and Variable 2
• Under Test Type, select the Sign test
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
Analyze > Descriptive Statistics > Crosstabs...
• Put one of your two categorical variables in the box below Row(s), and the other categorical variable in the box below Column(s)
• Click the Statistics... button, and click on the square in front of Chi-square
• Continue and click OK
Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples...
• Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum
• Continue and click OK
JamoviJamoviJamoviJamovi
Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:

ANOVA > Repeated Measures ANOVA - Friedman
• Put the two paired variables in the box below Measures
Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
Frequencies > Independent Samples - $\chi^2$ test of association
• Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
ANOVA > One Way ANOVA - Kruskal-Wallis
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
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