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I have data from a 1 group repeated measures (Time 1 and Time 2) design to evaluate the impact of a psychological intervention for a group of participants. It is the same group of participants pre and post intervention but they are anonymous therefore, I cannot pair the individual data points for Time 1 and Time 2. So, I don't think I can use the paired-sample t-test because I can't calculate the differences for each individual.

I also don't think I can use the independent samples t-test as the sample is the same violating a key assumption.

Is my reasoning on t-tests correct and is there an alternative statistical method I can use?

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Being unable to know the values of the same individual at Time 1 and Time 2 because of "anonymization" is quite a nonsense. Anonymization refers to the impossibility of identify a person (name, address, e-mail...). That's why is common the use of identifiers (numerical or alphanumerical). If you have a dataset of paired data, the values of one row should represent the same individual (without knowing who is it). Otherwise, this dataset is corrupted.

Nevertheless, it could happen. So, if you can't help it, exceptionally you could apply an independent t-test, knowing that you can't apply a paired test because of the problem that you exposed. But you have to be carefull with your conclusions because you know that there's variability included that you are not considering (data is paired). Check what Jeremy Miles has pointed out in the comments of this answer for more information about what happens with the p-value.

Furthermore, remember that t-test has assumptions that should be accomplished:

  • Variance homogeneity
  • Normal distribution of residuals

In addition, I found another post with the same question: Can i do a paired t-test on a pre- and post questionnaire but without pairing the same student for the pre and post?

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    $\begingroup$ Failing to take into account pairing will (except in slightly unusual circumstances) mean your standard errors are inflated (and hence p-value is higher than it could be). $\endgroup$
    – Jeremy Miles
    Commented Jan 13, 2020 at 16:55
  • $\begingroup$ Thanks for this clarification, I will add it to my answer. $\endgroup$
    – Dave
    Commented Jan 13, 2020 at 16:56
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    $\begingroup$ Homogeneity of variance is not an issue, see for example: ncbi.nlm.nih.gov/pubmed/15171807 $\endgroup$
    – Jeremy Miles
    Commented Jan 13, 2020 at 16:56

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