Cochran's Q test - overview

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Cochran's Q test
One sample $z$ test for the mean
$z$ test for a single proportion
Sign test
Independent/grouping variableIndependent variableIndependent variableIndependent variable
One within subject factor ($\geq 2$ related groups)NoneNone2 paired groups
Dependent variableDependent variableDependent variableDependent variable
One categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with 2 independent groupsOne of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
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: $\mu = \mu_0$

Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the 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: P(first score of a pair exceeds second score of a pair) = P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H0: the population median of the difference scores is equal to zero
A difference score is the difference between the first score of a pair and the second score of a pair.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1: not all population proportions are equalH1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
  • H1 two sided: P(first score of a pair exceeds second score of a pair) $\neq$ P(second score of a pair exceeds first score of a pair)
  • H1 right sided: P(first score of a pair exceeds second score of a pair) > P(second score of a pair exceeds first score of a pair)
  • H1 left sided: P(first score of a pair exceeds second score of a pair) < P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H1 two sided: the population median of the difference scores is different from zero
  • H1 right sided: the population median of the difference scores is larger than zero
  • H1 left sided: the population median of the difference scores is smaller than zero
AssumptionsAssumptionsAssumptionsAssumptions
  • Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
  • Scores are normally distributed in the population
  • Population standard deviation $\sigma$ is known
  • Sample is a simple random sample from the population. That is, observations 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 statisticTest statistic
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.
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.

The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
$W = $ number of difference scores that is larger than 0
Sampling distribution of $Q$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $W$ if H0 were true
If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedomStandard normal distributionApproximately the standard normal distributionThe 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?Significant?Significant?
If the number of blocks 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: Two sided: Right sided: Left sided: If $n$ is small, the table for the binomial distribution should be used:
Two sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$

If $n$ is large, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
n.a.$C\%$ confidence interval for $\mu$Approximate $C\%$ confidence interval for $\pi$n.a.
-$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
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).

The confidence interval for $\mu$ can also be used as significance test.
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)
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n.a.Effect sizen.a.n.a.
-Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
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n.a.Visual representationn.a.n.a.
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One sample z test
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Equivalent ton.a.Equivalent toEquivalent to
Friedman test, with a categorical dependent variable consisting of two independent groups.-
  • 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.
Two sided sign test is equivalent to
Example contextExample contextExample contextExample context
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 average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$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.Do people tend to score higher on mental health after a mindfulness course?
SPSSn.a.SPSSSPSS
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 Cochran's Q 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 Sign test
Jamovin.a.JamoviJamovi
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
  • 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
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
  • Put the two paired variables in the box below Measures
Practice questionsPractice questionsPractice questionsPractice questions