Friedman test overview

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Friedman test	$z$ test for a single proportion	One sample Wilcoxon signed-rank test
Independent/grouping variable	Independent variable	Independent variable
One within subject factor ($\geq 2$ related groups)	None	None
Dependent variable	Dependent variable	Dependent variable
One of ordinal level	One categorical with 2 independent groups	One of ordinal level
Null hypothesis	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.	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.
Alternative hypothesis	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	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: $m < m_0$
Assumptions	Assumptions	Assumptions
Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another	Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb: Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10 Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more Sample is a simple random sample from the population. That is, observations are independent of one another If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.	The population distribution of the scores is symmetric Sample is a simple random sample from the population. That is, observations are independent of one another
Test statistic	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.	$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$ 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.	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: For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero. For each subject, compute the absolute value of the difference score $\|\mbox{score} - m_0\|$. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy. Then compute the test statistic: $W_1 = \sum\, R_d^{+}$ or $W_1 = \sum\, R_d^{-}$ That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However: Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one. If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip'). $W_2 = \sum\, \mbox{sign}_d \times R_d$ That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
Sampling distribution of $Q$ if H₀ were true	Sampling distribution of $z$ if H₀ were true	Sampling distribution of $W_1$ and of $W_2$ 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.	Approximately the standard normal distribution	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.
Significant?	Significant?	Significant?
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$: Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$	Two sided: Check if $z$ observed in sample is at least as extreme as critical value $z^$ or Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $z$ observed in sample is equal to or larger than critical value $z^$ or Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$	For large samples, the table for standard normal probabilities can be used: Two sided: Check if $z$ observed in sample is at least as extreme as critical value $z^$ or Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $z$ observed in sample is equal to or larger than critical value $z^$ or Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
n.a.	Approximate $C\%$ confidence interval for $\pi$	n.a.
-	Regular (large sample): $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$ 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) With plus four method: $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$ where $p_{plus} = \dfrac{X + 2}{N + 4}$ and 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)	-
n.a.	Equivalent to	n.a.
-	When testing two sided: goodness of fit test, with a categorical variable with 2 levels. When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.	-
Example context	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)?	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 the median mental health score of office workers different from $m_0 = 50$?
SPSS	SPSS	SPSS
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples... Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables Under Test Type, select the Friedman test	Analyze > Nonparametric Tests > Legacy Dialogs > Binomial... Put your dichotomous variable in the box below Test Variable List Fill in the value for $\pi_0$ in the box next to Test Proportion If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution	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... On the Objective tab, choose Customize Analysis On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your $m_0$ in the box next to Hypothesized median Click Run Double click on the output table to see the full results
Jamovi	Jamovi	Jamovi
ANOVA > Repeated Measures ANOVA - Friedman Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures	Frequencies > 2 Outcomes - Binomial test Put your dichotomous variable in the white box at the right Fill in the value for $\pi_0$ in the box next to Test value Under Hypothesis, select your alternative hypothesis Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution	T-Tests > One Sample T-Test Put your variable in the box below Dependent Variables Under Tests, select Wilcoxon rank Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis
Practice questions	Practice questions	Practice questions

Friedman test - overview