Friedman test - overview

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Friedman test
$z$ test for a single proportion
McNemar's test
You cannot compare more than 3 methods
Independent/grouping variableIndependent variableIndependent variable
One within subject factor ($\geq 2$ related groups)None2 paired groups
Dependent variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull 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.

Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:

  1. First score of pair is 0, second score of pair is 0
  2. First score of pair is 0, second score of pair is 1 (switched)
  3. First score of pair is 1, second score of pair is 0 (switched)
  4. First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the null hypothesis are:

  • H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
  • H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)

Alternative hypothesisAlternative hypothesisAlternative 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: $\pi < \pi_0$

The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the alternative hypothesis are:

  • H1: $\pi_1 \neq \pi_2$
  • H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)

AssumptionsAssumptionsAssumptions
  • 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.
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Test statisticTest statisticTest 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.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
Sampling distribution of $Q$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ 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

If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.

If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.

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: Right sided: Left sided: For test statistic $X^2$:
  • 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 $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
  • Check if $b$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $b$ 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 toEquivalent to
-
  • 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 contextExample contextExample 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.Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
SPSSSPSSSPSS
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
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the McNemar test
JamoviJamoviJamovi
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
Frequencies > Paired Samples - McNemar test
  • Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
Practice questionsPractice questionsPractice questions