We show that the one-way ANOVA and Tukey–Kramer (TK) tests agree on any sample with two groups. This result is based on a simple identity connecting the Fisher–Snedecor and studentized probabilistic distributions and is proven without any additional assumptions; in particular, the standard ANOVA assumptions (independence, normality, and homoscedasticity (INAH)) are not needed. In contrast, it is known that for a sample with $k>2$ groups of observations, even under the INAH assumptions, with the same significance level $\alpha$, the above two tests may give opposite results: (i) ANOVA rejects its null hypothesis $H_0^{\mathit{A}}:{\mu }_1=\dots ={\mu }_k$, while the TK one, $H_0^{\mathit{TK}}(i,j):{\mu }_i={\mu }_j$, is not rejected for any pair $i,j\in \{1,\dots ,k\}$; (ii) the TK test rejects $H_0^{\mathit{TK}}(i,j)$ for a pair $(i,j)$ (with $i\ne j$), while ANOVA does not reject $H_0^{\mathit{A}}$. We construct two large infinite pseudo-random families of samples of both types satisfying INAH: in case (i) for any $k\ge 3$ and in case (ii) for some larger k. Furthermore, case (ii) ANOVA, being restricted to the pair of groups $(i,j)$, may reject equality ${\mu }_i={\mu }_j$ with the same $\alpha$. This is an obvious contradiction, since ${\mu }_1=\dots ={\mu }_k$ implies ${\mu }_i={\mu }_j$ for all $i,j\in \{1,\dots ,k\}.$ Such contradictions appear already in the symmetric case for $k=3$, or in other words, for three groups of $d,d$, and c observations with sample means $+1,-1$, and 0, respectively. We outline conditions necessary and sufficient for this phenomenon. Similar contradictory examples are constructed for the multivariable linear regression (MLR). However, for these constructions, it seems difficult to verify the Gauss–Markov assumptions, which are standardly required for MLR. Mathematics Subject Classification: 62 Statistics.

中文翻译：

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单向方差分析和 Tukey-Kramer 多重比较检验中的逻辑矛盾，具有两个以上的观察组

我们表明，单向方差分析和 Tukey-Kramer ( TK ) 检验在两组的任何样本上都一致。该结果基于连接 Fisher-Snedecor 和学生化概率分布的简单恒等式，无需任何额外假设即可证明；特别是，不需要标准方差分析假设（独立性、正态性和同方差性 ( INAH )）。相比之下，众所周知，对于具有$克>2$观察组，即使在INAH假设下，具有相同的显着性水平$\alpha$，上述两个检验可能会给出相反的结果：(i) 方差分析拒绝其原假设 $H_0^{\mathit{一种}}：{\mu }_1=\dots ={\mu }_{克}$，而传统知识之一，$H_0^{\mathit{传统知识}}(\mathrm{一世},j)：{\mu }_{\mathrm{一世}}={\mu }_j$, 不会被任何对拒绝 $\mathrm{一世},j\in \{1,\dots ,克\}$; (ii) TK测试拒绝$H_0^{\mathit{传统知识}}(\mathrm{一世},j)$ 一对 $(\mathrm{一世},j)$ （和 $\mathrm{一世}\ne j$)，而方差分析不拒绝 $H_0^{\mathit{一种}}$. 我们构造了两个满足INAH的两种类型的大型无限伪随机样本族：在情况（i）中，对于任何$克\ge 3$并且在情况（ii）对于一些更大的k。此外，情况 (ii) 方差分析，仅限于组对$(\mathrm{一世},j)$, 可能拒绝平等 ${\mu }_{\mathrm{一世}}={\mu }_j$ 与相同 $\alpha$. 这是一个明显的矛盾，因为${\mu }_1=\dots ={\mu }_{克}$ 暗示 ${\mu }_{\mathrm{一世}}={\mu }_j$ 对全部 $\mathrm{一世},j\in \{1,\dots ,克\}.$ 这种矛盾已经出现在对称情况下 $克=3$，或者换句话说，对于三组 $d,d$, 和c 个样本均值的观测值$+1,-1$, 和 0 分别。我们概述了这种现象的必要和充分条件。为多变量线性回归 (MLR) 构建了类似的矛盾示例。然而，对于这些构造，似乎很难验证高斯-马尔可夫假设，这是 MLR 的标准要求。数学学科分类：62个统计。