Cochran's Q test - overview
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Cochran's Q 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 categorical with 2 independent groups | One of ordinal level | |
Null hypothesis | Null 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 $I.$ |
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Alternative hypothesis | Alternative hypothesis | |
H1: not all population proportions are equal |
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Assumptions | Assumptions | |
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Test statistic | Test 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. | $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 (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom | 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 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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Equivalent to | Equivalent to | |
Friedman test, with a categorical dependent variable consisting of two independent groups. |
Two sided sign test is equivalent to
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Example context | Example 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? | 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 | |
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
| 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 | |