Marginal Homogeneity test / Stuart-Maxwell test - overview
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Marginal Homogeneity test / Stuart-Maxwell test | Binomial test for a single proportion | Sign test |
You cannot compare more than 3 methods |
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Independent variable | Independent variable | Independent variable | |
2 paired groups | None | 2 paired groups | |
Dependent variable | Dependent variable | Dependent variable | |
One categorical with $J$ independent groups ($J \geqslant 2$) | One categorical with 2 independent groups | 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: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis. |
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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: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ |
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Assumptions | Assumptions | Assumptions | |
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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. | $X$ = number of successes in the sample | $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of the test statistic if H0 were true | Sampling distribution of $X$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
Approximately the chi-squared distribution with $J - 1$ degrees of freedom | Binomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis). | 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 $X^2$:
| 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:
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n.a. | n.a. | Equivalent to | |
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Two sided sign test is equivalent to
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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 proportion of smokers amongst office workers different from $\pi_0 = 0.2$? | 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 > Binomial...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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n.a. | Jamovi | Jamovi | |
- | Frequencies > 2 Outcomes - Binomial 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
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Practice questions | Practice questions | Practice questions | |