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Two categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)
2 paired groups
One categorical with 2 independent groups
2 paired groups
None
Dependent variable
Dependent variable
Dependent variable
Dependent variable
Dependent variable
One quantitative of interval or ratio level
One of ordinal level
One quantitative of interval or ratio level
One categorical with 2 independent groups
One of ordinal level
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
ANOVA $F$ tests:
H_{0} for main and interaction effects together (model): no main effects and interaction effect
H_{0} for independent variable A: no main effect for A
H_{0} for independent variable B: no main effect for B
H_{0} for the interaction term: no interaction effect between A and B
Like in one way ANOVA, we can also perform $t$ tests for specific contrasts and multiple comparisons. This is more advanced stuff.
H_{0}: 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:
H_{0}: 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.
H_{0}: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
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 H_{0} 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:
H_{0}: $\pi_1 = \pi_2$, 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
H_{0}: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
ANOVA $F$ tests:
H_{1} for main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
H_{1} for independent variable A: there is a main effect for A
H_{1} for independent variable B: there is a main effect for B
H_{1} for the interaction term: there is an interaction effect between A and B
H_{1} 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)
H_{1} 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)
H_{1} 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:
H_{1} two sided: the population median of the difference scores is different from zero
H_{1} right sided: the population median of the difference scores is larger than zero
H_{1} left sided: the population median of the difference scores is smaller than zero
H_{1} two sided: $\mu_1 \neq \mu_2$
H_{1} right sided: $\mu_1 > \mu_2$
H_{1} left sided: $\mu_1 < \mu_2$
The alternative hypothesis H_{1} 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:
H_{1}: $\pi_1 \neq \pi_2$
H_{1}: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
H_{1} two sided: $m \neq m_0$
H_{1} right sided: $m > m_0$
H_{1} left sided: $m < m_0$
Assumptions
Assumptions
Assumptions
Assumptions
Assumptions
Within each of the $I \times J$ populations, 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 $I \times J$ populations
For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
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 of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
The population distribution of the scores is symmetric
Sample is a simple random sample from the population. That is, observations are independent of one another
Test statistic
Test statistic
Test statistic
Test statistic
Test statistic
For main and interaction effects together (model):
Note: mean square error is also known as mean square residual or mean square within.
$W = $ number of difference scores that is larger than 0
$z = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
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.
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 $\mu_2 > \mu_1$.
$X^2 = \dfrac{(b  c)^2}{b + c}$
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.
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:
For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}  m_0)$. The sign is 1 if the difference is larger than zero, 1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
For each subject, compute the absolute value of the difference score $\mbox{score}  m_0$.
Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
$W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{}$. So if you are using such a table, pick the smaller one.
If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{}$, the right and left sided alternative hypotheses 'flip').
$W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
Pooled standard deviation
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$
\begin{aligned}
s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score}  \mbox{its group mean})^2}{N  (I \times J)}}\\
&= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\
&= \sqrt{\mbox{mean square error}}
\end{aligned}
$
Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true
For main and interaction effects together (model):
$F$ distribution with $(I  1) + (J  1) + (I  1) \times (J  1)$ (df model, numerator) and $N  (I \times J)$ (df error, denominator) degrees of freedom
For independent variable A:
$F$ distribution with $I  1$ (df A, numerator) and $N  (I \times J)$ (df error, denominator) degrees of freedom
For independent variable B:
$F$ distribution with $J  1$ (df B, numerator) and $N  (I \times J)$ (df error, denominator) degrees of freedom
For the interaction term:
$F$ distribution with $(I  1) \times (J  1)$ (df interaction, numerator) and $N  (I \times J)$ (df error, denominator) degrees of freedom
Here $N$ is the total sample size.
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 $\sqrt{nP(1P)} = \sqrt{n \times 0.5(1  0.5)}$. 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.
Standard normal distribution
If $b + c$ is large enough (say, > 20), approximately the chisquared 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 $b$.
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.
Significant?
Significant?
Significant?
Significant?
Significant?
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 $\alpha$
If $n$ is small, the table for the binomial distribution should be used:
Two sided:
Check if $W$ observed in sample is in the rejection region or
Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
Check if $W$ observed in sample is in the rejection region or
Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
Check if $W$ observed in sample is in the rejection region or
Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
If $n$ is large, the table for standard normal probabilities can be used:
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 $\alpha$
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 $\alpha$
Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
Check if $b$ observed in sample is in the rejection region or
Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
For large samples, the table for standard normal probabilities can be used:
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 $\alpha$
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$C\%$ confidence interval for $\mu_1  \mu_2$
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$(\bar{y}_1  \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
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).
Proportion variance explained $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variables and the interaction effect together:
$$
\begin{align}
R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}
\end{align}
$$
$R^2$ 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 $\eta^2$:
Proportion variance of the dependent variable $y$ explained by an independent variable or interaction effect:
$$
\begin{align}
\eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\
\\
\eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\
\\
\eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}}
\end{align}
$$
$\eta^2$ 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:
$$
\begin{align}
\omega^2_A &= \dfrac{\mbox{sum of squares A}  \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\
\\
\omega^2_B &= \dfrac{\mbox{sum of squares B}  \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\
\\
\omega^2_{int} &= \dfrac{\mbox{sum of squares int}  \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\
\end{align}
$$
$\omega^2$ is a better estimate of the explained variance in the population than
$\eta^2$. Only for balanced designs (equal sample sizes).
Proportion variance explained $\eta^2_{partial}$:
$$
\begin{align}
\eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\
\\
\eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\
\\
\eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}}
\end{align}
$$




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Visual representation
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ANOVA table
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Equivalent to
Equivalent to
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OLS regression with two categorical independent variables and the interaction term, transformed into $(I  1)$ + $(J  1)$ + $(I  1) \times (J  1)$ code variables.
Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender?
Do people tend to score higher on mental health after a mindfulness course?
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.
Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
Is the median mental health score of office workers different from $m_0 = 50$?
SPSS
SPSS
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SPSS
SPSS
Analyze > General Linear Model > Univariate...
Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s)
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the McNemar test
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
On the Objective tab, choose Customize Analysis
On the Fields tab, specify the variable for which you want to compute the Wilcoxon signedrank test
On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signedrank test)'. Fill in your $m_0$ in the box next to Hypothesized median
Click Run
Double click on the output table to see the full results
Jamovi
Jamovi
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Jamovi
Jamovi
ANOVA > ANOVA
Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
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

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
TTests > One Sample TTest
Put your variable in the box below Dependent Variables
Under Tests, select Wilcoxon rank
Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis