Marginal Homogeneity test / Stuart-Maxwell test - overview
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Marginal Homogeneity test / Stuart-Maxwell test
One way ANOVA
$z$ test for a single proportion
You cannot compare more than 3 methods
Independent variable
Independent/grouping variable
Independent variable
2 paired groups
One categorical with $I$ independent groups ($I \geqslant 2$)
None
Dependent variable
Dependent variable
Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)
One quantitative of interval or ratio level
One categorical with 2 independent groups
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 $j.$
ANOVA $F$ test:
H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
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 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.
ANOVA $F$ test:
H1: not all population means are equal
$t$ Test for contrast:
H1 two sided: $\Psi \neq 0$
H1 right sided: $\Psi > 0$
H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
H1 - usually two sided: $\mu_g \neq \mu_h$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
Assumptions
Assumptions
Assumptions
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
The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
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
Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
Sample is a simple random sample from the population. That is, observations are independent of one another
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.
ANOVA $F$ test:
$\begin{aligned}[t]
F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\
&= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\
&= \dfrac{\mbox{mean square between}}{\mbox{mean square error}}
\end{aligned}
$
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
$t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
$t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$,
$s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA,
$n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$F$ test:
Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)
$t$ Test for contrast two sided:
Check if $t$ observed in sample is at least as extreme as critical value $t^*$ or
Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$
$t$ Test for contrast right sided:
Check if $t$ observed in sample is equal to or larger than critical value $t^*$ or
Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$
$t$ Test for contrast left sided:
Check if $t$ observed in sample is equal to or smaller than critical value $t^*$ or
Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$
$t$ Test multiple comparisons two sided:
Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
Two sided:
Check if $z$ observed in sample is at least as extreme as critical value $z^*$ or
Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Right sided:
Check if $z$ observed in sample is equal to or larger than critical value $z^*$ or
Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Left sided:
Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or
Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
n.a.
$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$
Approximate $C\%$ confidence interval for $\pi$
-
Confidence interval for $\Psi$ (contrast):
$c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
$(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* = $ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
$\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Regular (large sample):
$p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
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)
With plus four method:
$p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
where $p_{plus} = \dfrac{X + 2}{N + 4}$ and 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)
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Effect size
n.a.
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Proportion variance explained $\eta^2$ and $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variable:
$$
\begin{align}
\eta^2 = R^2
&= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}}
\end{align}
$$
Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to:
$$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$
$\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$
Cohen's $d$:
Standardized difference between the mean in group $g$ and in group $h$:
$$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$
Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
$F$ test ANOVA is equivalent to $F$ test regression model
$t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
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 average mental health score different between people from a low, moderate, and high economic class?
Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the Marginal Homogeneity test
Analyze > Compare Means > One-Way ANOVA...
Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
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
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
n.a.
Jamovi
Jamovi
-
ANOVA > ANOVA
Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
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
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
