K-W是否需要满足各组分的分布是相同的才可以使用。
疑问来源：
The other assumption of one-way anova is that the variation within the groups is equal (homoscedasticity). While Kruskal-Wallis does not assume that the data are normal, it does assume that the different groups have the same distribution, and groups with different standard deviations have different distributions. If your data are heteroscedastic, Kruskal–Wallis is no better than one-way anova, and may be worse. Instead, you should use Welch's anova for heteoscedastic data.
(http://www.biostathandbook.com/kruskalwallis.html
)
wiki百科：Since it is a non-parametric method, the Kruskal–Wallis test does not assume a normal distribution of the residuals, unlike the analogous one-way analysis of variance. If the researcher can make the more stringent assumptions of an identically shaped and scaled distribution for all groups, except for any difference in medians, then the null hypothesis is that the medians of all groups are equal, and the alternative hypothesis is that at least one population median of one group is different from the population median of at least one other group.
只是说如果可以满足，就怎样怎样，看似非强制要求。
然而 K-W基于 Mann–Whitney U test 所以继续wiki：
Although Mann and Whitney[1] developed the MWW test under the assumption of continuous responses with the alternative hypothesis being that one distribution is stochastically greater than the other, there are many other ways to formulate the null and alternative hypotheses such that the MWW test will give a valid test.[2]
A very general formulation is to assume that:
- All the observations from both groups are independent of each other,
- The responses are ordinal (i.e. one can at least say, of any two observations, which is the greater),
- The distributions of both groups are equal under the null hypothesis, so that the probability of an observation from one population (X) exceeding an observation from the second population (Y) equals the probability of an observation from Y exceeding an observation from X. That is, there is a symmetry between populations with respect to probability of random drawing of a larger observation.
- Under the alternative hypothesis, the probability of an observation from one population (X) exceeding an observation from the second population (Y) (after exclusion of ties) is not equal to 0.5. The alternative may also be stated in terms of a one-sided test, for example: P(X > Y) + 0.5 P(X = Y) > 0.5.
Under more strict assumptions than those above, e.g., if the responses are assumed to be continuous and the alternative is restricted to a shift in location (i.e. F1(x) = F2(x + δ)), we can interpret a significant MWW test as showing a difference in medians. Under this location shift assumption, we can also interpret the MWW as assessing whether the Hodges–Lehmann estimate of the difference in central tendency between the two populations differs from zero. The Hodges–Lehmann estimate for this two-sample problem is the median of all possible differences between an observation in the first sample and an observation in the second sample.
其中第三条的确说：在零假设下两组的分布相同。但是因为我才学习统计分很短的时间，我也不敢确认这句话是否意味着M-W必须满足两组分布相同，而且虽然K-W基于M-W，但m-w分析两个组分，K-W分析3个以上，也不清楚是否k-w 必须满足 M-W 应该满足的。
所以希望大神们帮我解答或分析一下：我的数据在对数处理后是正态的，但方差不是齐性的，我该选择one-way anova非齐性下的事后检验方法，还是选择非参数性方法中的K-W呢？万分感谢！
扫码加好友，拉您进群
全部版块
我的主页
收藏
