Friedman test - overview
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Friedman test | Sign test |
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Independent/grouping variable | Independent variable | |
One within subject factor ($\geq 2$ related groups) | 2 paired groups | |
Dependent variable | Dependent variable | |
One of ordinal level | One of ordinal level | |
Null hypothesis | 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. |
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Alternative hypothesis | Alternative hypothesis | |
H1: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups |
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Assumptions | Assumptions | |
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Test statistic | Test statistic | |
$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
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. | $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of $Q$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
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. | 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. | |
Significant? | Significant? | |
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
| 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:
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n.a. | Equivalent to | |
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Two sided sign test is equivalent to
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Example context | Example context | |
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)? | Do people tend to score higher on mental health after a mindfulness course? | |
SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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Jamovi | Jamovi | |
ANOVA > Repeated Measures ANOVA - Friedman
| 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
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Practice questions | Practice questions | |