Sign test - overview
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Sign test | Binomial test for a single proportion | Spearman's rho | Friedman test |
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Independent variable | Independent variable | Variable 1 | Independent/grouping variable | |
2 paired groups | None | One of ordinal level | One within subject factor ($\geq 2$ related groups) | |
Dependent variable | Dependent variable | Variable 2 | Dependent variable | |
One of ordinal level | One categorical with 2 independent groups | One of ordinal level | One of ordinal level | |
Null hypothesis | Null hypothesis | Null hypothesis | Null hypothesis | |
| H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis. | H0: $\rho_s = 0$
Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level. In words, the null hypothesis would be: H0: there is no monotonic relationship between the two variables in the population. | H0: 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. | |
Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | |
| H1 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ | H1 two sided: $\rho_s \neq 0$ H1 right sided: $\rho_s > 0$ H1 left sided: $\rho_s < 0$ | H1: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups | |
Assumptions | Assumptions | Assumptions | Assumptions | |
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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 | $t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}} $ Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores. | $Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
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. | |
Sampling distribution of $W$ if H0 were true | Sampling distribution of $X$ if H0 were true | Sampling distribution of $t$ if H0 were true | Sampling distribution of $Q$ if H0 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 $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. 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 $t$ distribution with $N - 2$ degrees of freedom | If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used. | |
Significant? | 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:
| If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
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Equivalent to | n.a. | n.a. | n.a. | |
Two sided sign test is equivalent to
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Example context | 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 there a monotonic relationship between physical health and mental health? | Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)? | |
SPSS | SPSS | SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Correlate > Bivariate...
| Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
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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
| Frequencies > 2 Outcomes - Binomial test
| Regression > Correlation Matrix
| ANOVA > Repeated Measures ANOVA - Friedman
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Practice questions | Practice questions | Practice questions | Practice questions | |