This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the righthand column. To practice with a specific method click the button at the bottom row of the table
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Dependent variable
Dependent variable
Dependent variable
Dependent variable
Dependent variable
One of ordinal level
One categorical with 2 independent groups
One of ordinal level
One categorical with 2 independent groups
One categorical with 2 independent groups
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
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}: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
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)
Model chisquared test for the complete regression model:
H_{0}: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
H_{0}: $\beta_k = 0$
or in terms of odds ratio:
H_{0}: $e^{\beta_k} = 1$
Likelihood ratio chisquared test for individual regression coefficient $\beta_k$:
H_{0}: $\beta_k = 0$
or in terms of odds ratio:
H_{0}: $e^{\beta_k} = 1$
in the regression equation
$
\ln \big(\frac{\pi_{y = 1}}{1  \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K
$. Here $ x_i$ represents independent variable $ i$, $\beta_i$ is the regression weight for independent variable $ x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $ y = 1$ (or equivalently, the proportion of $ y = 1$ in the population) given the scores on the independent variables.
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
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: $\pi \neq \pi_0$
H_{1} right sided: $\pi > \pi_0$
H_{1} left sided: $\pi < \pi_0$
H_{1} two sided: $m \neq m_0$
H_{1} right sided: $m > m_0$
H_{1} left sided: $m < m_0$
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)
Model chisquared test for the complete regression model:
H_{1}: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
H_{1}: $\beta_k \neq 0$
or in terms of odds ratio:
H_{1}: $e^{\beta_k} \neq 1$
If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
H_{1} right sided: $\beta_k > 0$
H_{1} left sided: $\beta_k < 0$
Likelihood ratio chisquared test for individual regression coefficient $\beta_k$:
H_{1}: $\beta_k \neq 0$
or in terms of odds ratio:
H_{1}: $e^{\beta_k} \neq 1$
Assumptions
Assumptions
Assumptions
Assumptions
Assumptions
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Sample is a simple random sample from the population. That is, observations 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
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1  \pi_{y=1}})$ is linear
The residuals are independent of one another
Often ignored additional assumption:
Variables are measured without error
Also pay attention to:
Multicollinearity
Outliers
Test statistic
Test statistic
Test statistic
Test statistic
Test statistic
$W = $ number of difference scores that is larger than 0
$X$ = number of successes in the sample
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.
$X^2 = \dfrac{(b  c)^2}{b + c}$
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.
Model chisquared test for the complete regression model:
$X^2 = D_{null}  D_K = \mbox{null deviance}  \mbox{model deviance} $
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$
Wald $ = \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.
Likelihood ratio chisquared test for individual $\beta_k$:
$X^2 = D_{K1}  D_K$
$D_{K1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
Sampling distribution of $W$ if H_{0} were true
Sampling distribution of $X$ if H0 were true
Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true
Sampling distribution of $X^2$ if H_{0} were true
Sampling distribution of $X^2$ and of the Wald statistic if H_{0} were true
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.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
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.
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 $X^2$, as computed in the model chisquared test for the complete model:
chisquared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chisquared distribution with 1 degree of freedom
If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chisquared test for individual $\beta_k$:
chisquared distribution with 1 degree of freedom
Significant?
Significant?
Significant?
Significant?
Significant?
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 $X$ 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 $\alpha$
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 $\alpha$
Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chisquared tests. Wald can be interpret as $X^2$
If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
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Waldtype approximate $C\%$ confidence interval for $\beta_k$




$b_k \pm z^* \times SE_{b_k}$
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).
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Goodness of fit measure $R^2_L$




$R^2_L = \dfrac{D_{null}  D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
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
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
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the McNemar test
Analyze > Regression > Binary Logistic...
Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Jamovi
Jamovi
Jamovi
Jamovi
Jamovi
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 > 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
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
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
Regression > 2 Outcomes  Binomial
Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'