建议将64个组，编码成63个伪变量，然后做回归分析，设置对照组为参照组，能够完美达到你的要求。spss教程：回归分析：[1]含虚拟自变量_百度经验  http://jingyan.baidu.com/article/851fbc37efdc463e1e15ab69.html

   直接做63次独立样本t检验就行，即分别将63组给药组中的每一组都与对照组做一次独立样本t检验。祝好运~

感谢，但是编辑告知我如下：

因为有数个组,与对照组进行比较(以及潜在的相互关系),应做一个方差分析以及相关的因果分析。因果分析不可使用独立样本t –检验单独与对照组比较。应该使用多重比较因果t-检验并执行使用某种形式的修正（例如Bonferroni）,用于比较不同的条件。

英文原文


Since there are several groups, that are being compared to a positive and negative control (as well as potentially to each other), an ANOVA is required as well as a relevant post-hoc analysis. The post-hoc analysis cannot be independent sample t- tests compared to the negative control alone. They should be multiple comparison post-hoc t-tests that are performed using some form of correction (such as Bonferroni) to compare the different conditions to each other.

盼回复

跟我想的，做的一模一样，

但是编辑告知我如下：

因为有数个组,与对照组进行比较(以及潜在的相互关系),应做一个方差分析以及相关的因果分析。因果分析不可使用独立样本t –检验单独与对照组比较。应该使用多重比较因果t-检验并执行使用某种形式的修正（例如Bonferroni）,用于比较不同的条件。

Since there are several groups, that are being compared to a positive and negative control (as well as potentially to each other), an ANOVA is required as well as a relevant post-hoc analysis . The post-hoc analysis cannot be independent sample t- tests compared to the negative control alone. They should be multiple comparison post-hoc t-tests that are performed using some form of correction (such as Bonferroni) to compare the different conditions to each other.

也就是针对我的问题要做一个  ANOVA post-hoc analysis

盼望回复

那你就按照post hoc的方法检验就可以了，它会将两两之间都进行比较，但是你可以选择性的报告。