This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
One categorical with $I$ independent groups ($I \geqslant 2$)
2 paired groups
Dependent variable
Dependent variable
Dependent variable
Dependent variable
One of ordinal level
One categorical with 2 independent groups
One of ordinal level
One categorical with 2 independent groups
Null hypothesis
Null hypothesis
Null hypothesis
Null 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.
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.
Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:
First score of pair is 0, second score of pair is 0
First score of pair is 0, second score of pair is 1 (switched)
First score of pair is 1, second score of pair is 0 (switched)
First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the null hypothesis are:
H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative 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: $\pi < \pi_0$
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 $g$)
The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the alternative hypothesis are:
H1: $\pi_1 \neq \pi_2$
H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
Assumptions
Assumptions
Assumptions
Assumptions
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
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
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Test statistic
Test statistic
Test statistic
Test statistic
$W = $ number of difference scores that is larger than 0
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.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
Sampling distribution of $W$ if H0 were true
Sampling distribution of $X$ if H0 were true
Sampling distribution of $H$ if H0 were true
Sampling distribution of $X^2$ 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.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
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.
If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.
If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.
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:
Check if $z$ observed in sample is at least as extreme as critical value $z^*$ or
Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Right sided:
Check if $z$ observed in sample is equal to or larger than critical value $z^*$ or
Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Left sided:
Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or
Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
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 $\alpha$
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
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
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the McNemar test
Jamovi
Jamovi
Jamovi
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
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
Frequencies > Paired Samples - McNemar test
Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns