Marginal Homogeneity test / Stuart-Maxwell test - overview

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Marginal Homogeneity test / Stuart-Maxwell test
Two sample $z$ test
Binomial test for a single proportion
You cannot compare more than 3 methods
Independent variableIndependent/grouping variableIndependent variable
2 paired groupsOne categorical with 2 independent groupsNone
Dependent variableDependent variableDependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull 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: $\mu_1 = \mu_2$

Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
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.
Alternative hypothesisAlternative hypothesisAlternative 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: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptions
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Within each population, the scores on the dependent variable are normally distributed
  • Population standard deviations $\sigma_1$ and $\sigma_2$ are known
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
  • Sample is a simple random sample from the population. That is, observations are independent of one another
Test statisticTest statisticTest 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.$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.

The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0.

Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
$X$ = number of successes in the sample
Sampling distribution of the test statistic if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $X$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedomStandard normal distributionBinomial($n$, $P$) distribution.

Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
Significant?Significant?Significant?
If we denote the test statistic as $X^2$:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Two sided: Right sided: Left sided: 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 $\alpha$
n.a.$C\%$ confidence interval for $\mu_1 - \mu_2$n.a.
-$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).

The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
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n.a.Visual representationn.a.
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Two sample z test
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Example contextExample contextExample 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 average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women.Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?
SPSSn.a.SPSS
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Marginal Homogeneity test
-Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
  • Put your dichotomous variable in the box below Test Variable List
  • Fill in the value for $\pi_0$ in the box next to Test Proportion
n.a.n.a.Jamovi
--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
Practice questionsPractice questionsPractice questions