Sign test - overview
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Sign test | Binomial test for a single proportion | One sample Wilcoxon signed-rank test | One sample $z$ test for the mean | Spearman's rho |
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Independent variable | Independent variable | Independent variable | Independent variable | Variable 1 | |
2 paired groups | None | None | None | One of ordinal level | |
Dependent variable | Dependent variable | Dependent variable | Dependent variable | Variable 2 | |
One of ordinal level | One categorical with 2 independent groups | One of ordinal level | One quantitative of interval or ratio level | One of ordinal level | |
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. | H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean 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. | |
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: $m < m_0$ | H1 two sided: $\mu \neq \mu_0$ H1 right sided: $\mu > \mu_0$ H1 left sided: $\mu < \mu_0$ | H1 two sided: $\rho_s \neq 0$ H1 right sided: $\rho_s > 0$ H1 left sided: $\rho_s < 0$ | |
Assumptions | Assumptions | Assumptions | Assumptions | Assumptions | |
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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:
| $z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size. The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$. | $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. | |
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 $z$ if H0 were true | Sampling distribution of $t$ 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. | Standard normal distribution | Approximately the $t$ distribution with $N - 2$ degrees of freedom | |
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:
| Two sided:
| Two sided:
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n.a. | n.a. | n.a. | $C\%$ confidence interval for $\mu$ | n.a. | |
- | - | - | $\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ 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). The confidence interval for $\mu$ can also be used as significance test. | - | |
n.a. | n.a. | n.a. | Effect size | n.a. | |
- | - | - | Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$ | - | |
n.a. | n.a. | n.a. | Visual representation | n.a. | |
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Equivalent to | n.a. | 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 | 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 $m_0 = 50$? | Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$ | Is there a monotonic relationship between physical health and mental health? | |
SPSS | SPSS | SPSS | n.a. | 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 > Correlate > Bivariate...
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Jamovi | Jamovi | Jamovi | n.a. | 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
| - | Regression > Correlation Matrix
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Practice questions | Practice questions | Practice questions | Practice questions | Practice questions | |