One sample Wilcoxon signedrank test  overview
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One sample Wilcoxon signedrank test  $z$ test for the difference between two proportions  Paired sample $t$ test  Friedman test  Wilcoxon signedrank test  Two way ANOVA 


Independent variable  Independent/grouping variable  Independent variable  Independent/grouping variable  Independent variable  Independent/grouping variables  
None  One categorical with 2 independent groups  2 paired groups  One within subject factor ($\geq 2$ related groups)  2 paired groups  Two categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)  
Dependent variable  Dependent variable  Dependent variable  Dependent variable  Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  One quantitative of interval or ratio level  One of ordinal level  One quantitative of interval or ratio level  One quantitative of interval or ratio level  
Null hypothesis  Null hypothesis  Null hypothesis  Null hypothesis  Null hypothesis  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.  H_{0}: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.  H_{0}: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.  H_{0}: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups
Usually the related groups are the different measurement points. 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.  H_{0}: $m = 0$
Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair. 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.  ANOVA $F$ tests:
 
Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  
H_{1} two sided: $m \neq m_0$ H_{1} right sided: $m > m_0$ H_{1} left sided: $m < m_0$  H_{1} two sided: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$  H_{1} two sided: $\mu \neq \mu_0$ H_{1} right sided: $\mu > \mu_0$ H_{1} left sided: $\mu < \mu_0$  H_{1}: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups  H_{1} two sided: $m \neq 0$ H_{1} right sided: $m > 0$ H_{1} left sided: $m < 0$  ANOVA $F$ tests:
 
Assumptions  Assumptions  Assumptions  Assumptions  Assumptions  Assumptions  





 
Test statistic  Test statistic  Test statistic  Test statistic  Test statistic  Test statistic  
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:
 $z = \dfrac{p_1  p_2}{\sqrt{p(1  p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. Note: we could just as well compute $p_2  p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$  $t = \dfrac{\bar{y}  \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores). The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.  $Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i  3 \times N(k + 1)$
Here $N$ is the number of 'blocks' (usually the subjects  so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$. Note: if ties are present in the data, the formula for $Q$ is more complicated.  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 main and interaction effects together (model):
 
n.a.  n.a.  n.a.  n.a.  n.a.  Pooled standard deviation  
          $ \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  Sampling distribution of $z$ if H_{0} were true  Sampling distribution of $t$ if H_{0} were true  Sampling distribution of $Q$ if H_{0} were true  Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true  Sampling distribution of $F$ if H_{0} were true  
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.  Approximately the standard normal distribution  $t$ distribution with $N  1$ degrees of freedom  If the number of blocks $N$ is large, approximately the chisquared distribution with $k  1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used.  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.  For main and interaction effects together (model):
 
Significant?  Significant?  Significant?  Significant?  Significant?  Significant?  
For large samples, the table for standard normal probabilities can be used: Two sided:
 Two sided:
 Two sided:
 If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
 For large samples, the table for standard normal probabilities can be used: Two sided:

 
n.a.  Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  $C\%$ confidence interval for $\mu$  n.a.  n.a.  n.a.  
  Regular (large sample):
 $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N1}$ distribution with the area $C / 100$ between $t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu$ can also be used as significance test.        
n.a.  n.a.  Effect size  n.a.  n.a.  Effect size  
    Cohen's $d$: Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y}  \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0.$     
 
n.a.  n.a.  Visual representation  n.a.  n.a.  n.a.  
          
n.a.  n.a.  n.a.  n.a.  n.a.  ANOVA table  
          
n.a.  Equivalent to  Equivalent to  n.a.  n.a.  Equivalent to  
  When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels. 
     OLS regression with two categorical independent variables and the interaction term, transformed into $(I  1)$ + $(J  1)$ + $(I  1) \times (J  1)$ code variables.  
Example context  Example context  Example context  Example context  Example context  Example context  
Is the median mental health score of office workers different from $m_0 = 50$?  Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.  Is the average difference between the mental health scores before and after an intervention different from $\mu_0 = 0$?  Is there a difference in depression level between measurement point 1 (preintervention), measurement point 2 (1 week postintervention), and measurement point 3 (6 weeks postintervention)?  Is the median of the differences between the mental health scores before and after an intervention different from 0?  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?  
SPSS  SPSS  SPSS  SPSS  SPSS  SPSS  
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...
 SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
 Analyze > Compare Means > PairedSamples T Test...
 Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > General Linear Model > Univariate...
 
Jamovi  Jamovi  Jamovi  Jamovi  Jamovi  Jamovi  
TTests > One Sample TTest
 Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples  $\chi^2$ test of association
 TTests > Paired Samples TTest
 ANOVA > Repeated Measures ANOVA  Friedman
 TTests > Paired Samples TTest
 ANOVA > ANOVA
 
Practice questions  Practice questions  Practice questions  Practice questions  Practice questions  Practice questions  