This page offers structured overviews of one or more selected methods. Add additional methods for comparisons 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
One of ordinal level
One quantitative of interval or ratio level
Null hypothesis
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:
H_{0}: the population medians for the $I$ groups are equal
Else:
Formulation 1:
H_{0}: 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:
H_{0}:
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.
H_{0}: $m = 0$
Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair.
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 hypothesis
Alternative 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:
H_{1}: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
H_{1}:
the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
H_{1}:
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 $g$)
H_{1} two sided: $m \neq 0$
H_{1} right sided: $m > 0$
H_{1} left sided: $m < 0$
Assumptions
Assumptions
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
The population distribution of the difference scores is symmetric
Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured on an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
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.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic.
In order to compute each of the test statistics, follow the steps below:
For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2 - \mbox{score}_1)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
$W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{-}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
$W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
Sampling distribution of $H$ if H_{0} were true
Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true
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.
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here
$$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$
$$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$
Hence, if $N_r$ is large, the standardized test statistic
$$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$
follows approximately the standard normal distribution if the null hypothesis were true.
Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here
$$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$
Hence, if $N_r$ is large, the standardized test statistic
$$z = \frac{W_2}{\sigma_{W_2}}$$
follows approximately the standard normal distribution if the null hypothesis were true.
If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.
Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
Significant?
Significant?
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 Wilcoxon test
Jamovi
Jamovi
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
T-Tests > Paired Samples T-Test
Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
Under Tests, select Wilcoxon rank
Under Hypothesis, select your alternative hypothesis