One sample Wilcoxon signed-rank test - overview
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One sample Wilcoxon signed-rank test | Goodness of fit test |
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Independent variable | Independent variable | |
None | None | |
Dependent variable | Dependent variable | |
One of ordinal level | One categorical with $J$ independent groups ($J \geqslant 2$) | |
Null hypothesis | Null hypothesis | |
H0: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis. |
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Alternative hypothesis | Alternative hypothesis | |
H1 two sided: $m \neq m_0$ H1 right sided: $m > m_0$ H1 left sided: $m < m_0$ |
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Assumptions | Assumptions | |
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Test statistic | Test statistic | |
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:
| $X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells. | |
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 $W_1$:
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. | Approximately the chi-squared distribution with $J - 1$ degrees of freedom | |
Significant? | Significant? | |
For large samples, the table for standard normal probabilities can be used: Two sided:
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Example context | Example context | |
Is the median mental health score of office workers different from $m_0 = 50$? | Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$? | |
SPSS | SPSS | |
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 > Chi-square...
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Jamovi | Jamovi | |
T-Tests > One Sample T-Test
| Frequencies > N Outcomes - $\chi^2$ Goodness of fit
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Practice questions | Practice questions | |