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 $J$ independent groups ($J \geqslant 2$)
One categorical with 2 independent groups
One categorical with $J$ independent groups ($J \geqslant 2$)
One of ordinal level
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$,
the probability of drawing an observation from condition $J$ is $\pi_J$
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)
H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$,
the probability of drawing an observation from condition $J$ is $\pi_J$
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.
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
H1: the population proportions are not all as specified under the null hypothesis
or equivalently
H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
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)
H1: the population proportions are not all as specified under the null hypothesis
or equivalently
H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null 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
Assumptions
Assumptions
Assumptions
Assumptions
Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
Sample is a simple random sample from the population. That is, 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
Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
Sample is a simple random sample from the population. That is, 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
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$X^2 = \dfrac{(b - c)^2}{b + c}$
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.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$W = $ number of difference scores that is larger than 0
Approximately the chi-squared distribution with $J - 1$ degrees of freedom
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 $b$.
Approximately the chi-squared distribution with $J - 1$ degrees of freedom
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.
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?
Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?
Do people tend to score higher on mental health after a mindfulness course?
Put your categorical variable in the box below Test Variable List
Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
Put your categorical variable in the box below Test Variable List
Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the Sign test
Jamovi
Jamovi
Jamovi
Jamovi
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
Put your categorical variable in the box below Variable
Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
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
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
Put your categorical variable in the box below Variable
Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
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