Sign test  overview
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Sign test  Binomial test for a single proportion  $z$ test for the difference between two proportions 


Independent variable  Independent variable  Independent/grouping variable  
2 paired groups  None  One categorical with 2 independent groups  
Dependent variable  Dependent variable  Dependent variable  
One of ordinal level  One categorical with 2 independent groups  One categorical with 2 independent groups  
Null hypothesis  Null hypothesis  Null hypothesis  
 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}: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.  
Alternative hypothesis  Alternative hypothesis  Alternative hypothesis  
 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: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$  
Assumptions  Assumptions  Assumptions  


 
Test statistic  Test statistic  Test statistic  
$W = $ number of difference scores that is larger than 0  $X$ = number of successes in the sample  $z = \dfrac{p_1  p_2}{\sqrt{p(1  p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$  
Sampling distribution of $W$ if H_{0} were true  Sampling distribution of $X$ if H0 were true  Sampling distribution of $z$ 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.  Binomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).  Approximately the standard normal distribution  
Significant?  Significant?  Significant?  
If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
 Two sided:
 Two sided:
 
n.a.  n.a.  Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  
    Regular (large sample):
 
Equivalent to  n.a.  Equivalent to  
Two sided sign test is equivalent to
   When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels.  
Example context  Example context  Example context  
Do people tend to score higher on mental health after a mindfulness course?  Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?  Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.  
SPSS  SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
 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...
 
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
 Frequencies > 2 Outcomes  Binomial test
 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
 
Practice questions  Practice questions  Practice questions  