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$m = 0$
$m$ is the unknown population median of the difference scores
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesis
Alternative hypothesis
Model chisquared test for the complete regression model:
not all population regression coefficients are 0
Wald test for individual $\beta_k$:
$\beta_k \neq 0$
or in terms of odds ratio:
$e^{\beta_k} \neq 1$
If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
right sided: $\beta_k > 0$
left sided: $\beta_k < 0$
Likelihood ratio chisquared test for individual $\beta_k$:
$\beta_k \neq 0$
or in terms of odds ratio:
$e^{\beta_k} \neq 1$
Two sided: $m \neq 0$
Right sided: $m > 0$
Left sided: $m < 0$
Assumptions
Assumptions
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
The population distribution of the difference scores is symmetric
Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured at an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
Test statistic
Test statistic
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.
Two different types of test statistics can be used; 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 $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2  \mbox{score}_1)$. 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 $\mbox{score}_2  \mbox{score}_1$.
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.
Sampling distribution of $X^2$ and of the Wald statistic if H0 were true
Sampling distribution of $W_1$ and of $W_2$ if H0 were true
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 a chisquared distribution with 1 degree of freedom
If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: approximately a 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
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 a 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 a 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: the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated if ties are present in the data.
Significant?
Significant?
For the model chisquared test for the complete regression model and likelihood ratio chisquared test for individual $\beta_k$:
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 $z$.
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 $\alpha$
Waldtype approximate $C\%$ confidence interval for $\beta_k$
n.a.
$b_k \pm z^* \times SE_{b_k}$
where $z^*$ 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)

Goodness of fit measure $R^2_L$
n.a.
$R^2_L = \dfrac{D_{null}  D_K}{D_{null}}$
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.

Example context
Example context
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
Is the median of the differences between the mental health scores before and after an intervention different from 0?
SPSS
SPSS
Analyze > Regression > Binary Logistic...
Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the Wilcoxon test
Jamovi
Jamovi
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'
TTests > Paired Samples TTest
Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
Under Tests, select Wilcoxon rank
Under Hypothesis, select your alternative hypothesis