Sign test - overview
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Sign test | Binomial test for a single proportion | Wilcoxon signed-rank test | Cochran's Q test | $z$ test for a single proportion | Friedman test |
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Independent variable | Independent variable | Independent variable | Independent/grouping variable | Independent variable | Independent/grouping variable | |
2 paired groups | None | 2 paired groups | One within subject factor ($\geq 2$ related groups) | None | One within subject factor ($\geq 2$ related groups) | |
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 categorical with 2 independent groups | One categorical with 2 independent groups | One of ordinal level | |
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: $m = 0$
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. | H0: $\pi_1 = \pi_2 = \ldots = \pi_I$
Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$ | 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: 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 | 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: $m \neq 0$ H1 right sided: $m > 0$ H1 left sided: $m < 0$ | H1: not all population proportions are equal | H1 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ | 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 | 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:
| If a failure is scored as 0 and a success is scored as 1:
$Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$ Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores. Before computing $Q$, first exclude blocks with equal scores in all $k$ groups. | $z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis. | $Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
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 $W_1$ and of $W_2$ if H0 were true | Sampling distribution of $Q$ if H0 were true | Sampling distribution of $z$ 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). | 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 the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom | Approximately the standard normal distribution | 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? | 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:
| If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
| 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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n.a. | n.a. | n.a. | n.a. | Approximate $C\%$ confidence interval for $\pi$ | n.a. | |
- | - | - | - | Regular (large sample):
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Equivalent to | n.a. | n.a. | Equivalent to | Equivalent to | n.a. | |
Two sided sign test is equivalent to
| - | - | Friedman test, with a categorical dependent variable consisting of two independent groups. |
| - | |
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 of the differences between the mental health scores before and after an intervention different from 0? | Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks? | Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic. | 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 | SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
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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 > Paired Samples T-Test
| Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
| Frequencies > 2 Outcomes - Binomial test
| ANOVA > Repeated Measures ANOVA - Friedman
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Practice questions | Practice questions | Practice questions | Practice questions | Practice questions | Practice questions | |