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I have a propensity score matched data after 1-to-2 matching. Each treatment case is matched to one or two control cases. May I use paired tests to compare the outcomes? The discussion https://www.researchgate.net/post/how_is_paired_t-test_performed_for_21_case_controlled_studies_on_SPSS suggests that if a treatment case is matched to two control cases, the average of the outcomes of these two cases be compared with the treatment case outcome. Are there any other approaches? For instance, instead of the averaging, we may "double" the treatment cases matched to only one control case (and double this control case as well)? Or may be just put into the analysis all matched pairs as different pairs (each treatment case may participate in one or two such pairs)?

Or are there some special tests for one-to-many matching?

Viktor
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    I think the key search term for which you are looking is stratification. Each matched set, in your case one case and two controls, corresponds to one stratum. If you had a binary outcome you would use conditional logistic regression. I do not use SPSS so have no idea what it calls stratified analyses for continuous outcomes. – mdewey Oct 02 '18 at 15:57
  • @mdewey I do not use SPSS as well. As far as I know, stratification is not appropriate here. Logistic regression is an alternative approach, but here I need matching, not a regression. – Viktor Oct 02 '18 at 16:04
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    Googling "matching with multiple controls" gives a lot of relevant hits – kjetil b halvorsen Oct 02 '18 at 19:54
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    Not all researchers thinking you need to do a paired t-test, and so a (weighted) t-test/regression of the outcome on treatment may be sufficient. Another possibility is to use a cluster-robust standard error, where pair membership is the clustering variable. – Noah Oct 02 '18 at 20:47
  • @Noah Thank you for your answer! Is there an analog of Mann-Whitney's test for the situation (my data has a certain number of outliers, who make t-test comparison inappropriate)? – Viktor Oct 02 '18 at 23:20
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    @Viktor the exact extension would be ordinal logistic regression with cluster-robust standard errors. – Heteroskedastic Jim Oct 03 '18 at 22:06
  • @HeteroskedasticJim Could you explain? Ordinal logistic regression is somewhat different from rank tests. There is proportional odds assumption for ordinal logistic regression. – Viktor Oct 04 '18 at 10:48
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    @Viktor ordinal regression with the proportional odds assumption is the regression extension of the rank test. See this answer and the comments beneath it https://stats.stackexchange.com/a/43897/162986 – Heteroskedastic Jim Oct 04 '18 at 11:29
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    @Victor: Noah's and HeteroskedasticJim's suggestions are solid (+1 to both); using an regression approach is fine and if anything will allow the treatment estimate to be efficient if certain control variates are associated with the outcome. It will effectively emulate a doubly-robust estimate. – usεr11852 Oct 04 '18 at 21:35
  • @HeteroskedasticJim There is a certain controversy concerning cluster robust standard errors for nonlinear regression models. Because nonlinear heteroskedastic models' estimates are usually inconsistent. https://davegiles.blogspot.com/2013/05/robust-standard-errors-for-nonlinear.html – Viktor Jun 29 '19 at 11:55
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    @Viktor I'm familiar with the problem. Standard logit link has unpredictable coefficient behavior under heteroskedasticity. The problem here is not heteroskedasticity but violation of independence. So instead, what you'll get is a downward bias in the coefficients due to neglected heterogeneity but testing should be fine. And I think with cluster robust SEs, inference should proceed without problem. Or one could just do multilevel (mixed effects) ordinal logistic. – Heteroskedastic Jim Jun 29 '19 at 13:33

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