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
Marginal Homogeneity test / Stuart-Maxwell test
$z$ test for a single proportion
Independent variableIndependent variableIndependent /column variableIndependent variableIndependent variable
2 paired groupsNoneOne categorical with $I$ independent groups ($I \geqslant 2$)2 paired groupsNone
Dependent variableDependent variableDependent /row variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)One categorical with $J$ independent groups ($J \geqslant 2$)One categorical with 2 independent groups
Null hypothesisNull 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
H0: for each category $j$ of the dependent variable, $\pi_j$ for the first paired group = $\pi_j$ for the second paired group.

Here $\pi_j$ is the population proportion in category $j.$
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.
Alternative hypothesisAlternative 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
H1: for some categories of the dependent variable, $\pi_j$ for the first paired group $\neq$ $\pi_j$ for the second paired group.H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptions
  • 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
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
    • Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
  • Sample is a simple random sample from the population. That is, observations are independent of one another
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
Test statisticTest 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.
Computing the test statistic is a bit complicated and involves matrix algebra. Unless you are following a technical course, you probably won't need to calculate it by hand.$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
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 the test statistic if H0 were trueSampling distribution of $z$ 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 freedomApproximately the chi-squared distribution with $J - 1$ degrees of freedomApproximately the standard normal distribution
Significant?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 $\alpha$
If we denote the test statistic as $X^2$:
  • 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 $\alpha$
Two sided: Right sided: Left sided:
n.a.n.a.n.a.n.a.Approximate $C\%$ confidence interval for $\pi$
----Regular (large sample):
  • $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
    where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
  • $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
    where $p_{plus} = \dfrac{X + 2}{N + 4}$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
Equivalent ton.a.n.a.n.a.Equivalent to
Two sided sign test is equivalent to ---
  • When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
  • When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example contextExample 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?Subjects are asked to taste three different types of mayonnaise, and to indicate which of the three types of mayonnaise they like best. They then have to drink a glass of beer, and taste and rate the three types of mayonnaise again. Does drinking a beer change which type of mayonnaise people like best?Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
SPSSSPSSSPSSSPSSSPSS
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 > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Marginal Homogeneity 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
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
JamoviJamoviJamovin.a.Jamovi
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
-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
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
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