Sign test - overview
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Sign test | Binomial test for a single proportion | One sample Wilcoxon signed-rank test | McNemar's test | Sign test | $z$ test for the difference between two proportions |
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Independent variable | Independent variable | Independent variable | Independent variable | Independent variable | Independent/grouping variable | |
2 paired groups | None | None | 2 paired groups | 2 paired groups | One categorical with 2 independent groups | |
Dependent variable | 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 of ordinal level | One categorical with 2 independent groups | |
Null hypothesis | Null hypothesis | 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: $m = m_0$
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:
Other formulations of the null hypothesis are:
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| H0: $\pi_1 = \pi_2$
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 | 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: $m \neq m_0$ H1 right sided: $m > m_0$ H1 left sided: $m < m_0$ | The alternative hypothesis H1 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:
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| H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | |
Assumptions | Assumptions | Assumptions | Assumptions | Assumptions | Assumptions | |
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Test statistic | 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:
| $X^2 = \dfrac{(b - c)^2}{b + c}$
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. | $W = $ number of difference scores that is larger than 0 | $z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 H0 were true | Sampling distribution of $X$ if H0 were true | Sampling distribution of $W_1$ and of $W_2$ if H0 were true | Sampling distribution of $X^2$ if H0 were true | Sampling distribution of $W$ if H0 were true | Sampling distribution of $z$ 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). | 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 chi-squared 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 $b$. | 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. | Approximately the standard normal distribution | |
Significant? | Significant? | 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:
| For large samples, the table for standard normal probabilities can be used: Two sided:
| For test statistic $X^2$:
| 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:
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n.a. | n.a. | n.a. | n.a. | n.a. | Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | |
- | - | - | - | - | Regular (large sample):
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Equivalent to | n.a. | n.a. | Equivalent to | Equivalent to | Equivalent to | |
Two sided sign test is equivalent to
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Two sided sign test is equivalent to
| When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. | |
Example context | Example context | 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 the median mental health score of office workers different from $m_0 = 50$? | Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders? | Do people tend to score higher on mental health after a mindfulness course? | 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 | SPSS | SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| 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...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
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Jamovi | 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
| Frequencies > 2 Outcomes - Binomial test
| T-Tests > One Sample T-Test
| Frequencies > Paired Samples - McNemar test
| 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
| Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared 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
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Practice questions | Practice questions | Practice questions | Practice questions | Practice questions | Practice questions | |