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We have a single-case experimental design and we calculated Cohen's d for effect size using the mean difference between pre and post, divided by pooled standard deviation. However, for one of the outcomes, the variance for pre test is zero and the variance for post test is zero, even though there is clearly a change in the mean scores of pre and post. What should we do in this situation? Do we need to use a different equation?

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In your case --- because both samples have zero variance --- the denominator in the formula for Cohen's d will be zero. Reporting the result as "infinite" may be the only option.

As a side note, since you are looking at pre-post data, a variant of Cohen's d for paired data may be more appropriate. There appear to be some possible variants

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First, investigate why all observations in pregroup and one postgroup are equal. Maybe very low $n$? Are the data reals or integers? Since you use the pooled variance for the standard errors, calculating effect size should not be a problem.

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  • $\begingroup$ Because both samples have zero variance, there will be a zero in the denominator of the Cohen's d formula. $\endgroup$ Jul 12 '18 at 14:53
  • $\begingroup$ I thought about pooling among all the groups? $\endgroup$ Jul 12 '18 at 15:24
  • $\begingroup$ The pooled sd is a form of sqrt( (sd1 ^ 2 + sd2 ^ 2) / 2), usually with some adjustments for sample size $\endgroup$ Jul 12 '18 at 15:31

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