Marginal Homogeneity test / Stuart-Maxwell test - overview
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
Marginal Homogeneity test / Stuart-Maxwell test | Wilcoxon signed-rank test | Sign test |
You cannot compare more than 3 methods |
---|---|---|---|
Independent variable | Independent variable | Independent variable | |
2 paired groups | 2 paired groups | 2 paired groups | |
Dependent variable | Dependent variable | Dependent variable | |
One categorical with $J$ independent groups ($J \geqslant 2$) | One quantitative of interval or ratio level | One of ordinal level | |
Null hypothesis | Null hypothesis | Null hypothesis | |
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: $m = 0$
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 | Alternative hypothesis | |
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: $m \neq 0$ H1 right sided: $m > 0$ H1 left sided: $m < 0$ |
| |
Assumptions | Assumptions | Assumptions | |
|
|
| |
Test statistic | Test statistic | Test statistic | |
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. | 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:
| $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of the test statistic if H0 were true | Sampling distribution of $W_1$ and of $W_2$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
Approximately the chi-squared distribution with $J - 1$ degrees of freedom | 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. | 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. | |
Significant? | Significant? | Significant? | |
If we denote the test statistic as $X^2$:
| For large samples, the table for standard normal probabilities can be used: Two sided:
| If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
| |
n.a. | n.a. | Equivalent to | |
- | - |
Two sided sign test is equivalent to
| |
Example context | Example context | Example context | |
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 median of the differences between the mental health scores before and after an intervention different from 0? | Do people tend to score higher on mental health after a mindfulness course? | |
SPSS | SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| |
n.a. | Jamovi | Jamovi | |
- | T-Tests > Paired Samples T-Test
| 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
| |
Practice questions | Practice questions | Practice questions | |