英文:
Why does Dunn's post hoc test & CDL output shows different CLD for two treatments with the same mean?
问题
我一直在使用R中的Kruskal-Wallis非参数检验,并随后进行Dunn's多重比较检验。从glyphosate H&S(abc)与其他组的比较结果来看,它显示与任何其他处理方式都没有显著差异。然而,triclopyr H&S(bc)与Imazapyr H&S(a)、Hexazinone Soil(a)以及未处理的对照组显示出显著差异。Triclopyr和Glyphosate具有相同的均值,但Triclopyr的标准误较大。这些多重比较结果和后续的CLD是否因中位数和标准误的差异而有所不同,或者是否存在其他问题?
我需要在可视化方面处理箱线图。显著性信息占用了太多空间,y轴延伸得太长 - 我们只能有100%的树木死亡。我不确定箱线图的这种可视化是否直观。也许没有显著性信息更好吗?
英文:
I've been working with a Kruskal wallis non-parametric test in R and following it up with Dunn's multiple comparison's test. Looking at the results of glyphosate H&S(abc) comparisons with others it is showing as not significantly different from any other treatment. However, triclopyr H&S(bc) shows significantly different from Imazapyr H&S(a), Hexazinone Soil(a), and the untreated control. Triclopyr and Glyphosate have the same mean with triclopyr having a greater SE. Is this difference in multiple comparisons results and subsequent CLD from the differences in median and SE, or is there something wrong?
Kt<-kruskal.test(mortality~treatment, data=callerydata)
> Kt
Kruskal-Wallis rank sum test
data: mortality by treatment
Kruskal-Wallis chi-squared = 40.897, df = 7, p-value = 8.473e-07
> library(dunn.test)
> #compact letter display
> library(rcompanion)
> ##https://rcompanion.org/handbook/F_08.html
> CLD1<-DT1$res
> CLD1
Comparison Z P.unadj P.adj
1 Control - Glyphosate HS -2.05136182 4.023172e-02 0.0804634485
2 Control - Glyphosate Stump -4.21069005 2.545918e-05 0.0002376191
3 Glyphosate HS - Glyphosate Stump -2.15932823 3.082471e-02 0.0663916861
4 Control - Hexazinone -0.87452793 3.818308e-01 0.4648375135
5 Glyphosate HS - Hexazinone 1.17683389 2.392618e-01 0.3349665403
6 Glyphosate Stump - Hexazinone 3.33616212 8.494361e-04 0.0039640353
7 Control - Imazapyr HS -0.88532458 3.759816e-01 0.4785220354
8 Glyphosate HS - Imazapyr HS 1.16603725 2.435994e-01 0.3247991829
9 Glyphosate Stump - Imazapyr HS 3.32536548 8.830270e-04 0.0027471951
10 Hexazinone - Imazapyr HS -0.01079664 9.913857e-01 1.0000000000
11 Control - Imazapyr Stump -4.21069005 2.545918e-05 0.0003564286
12 Glyphosate HS - Imazapyr Stump -2.15932823 3.082471e-02 0.0719243266
13 Glyphosate Stump - Imazapyr Stump 0.00000000 1.000000e+00 1.0000000000
14 Hexazinone - Imazapyr Stump -3.33616212 8.494361e-04 0.0047568424
15 Imazapyr HS - Imazapyr Stump -3.32536548 8.830270e-04 0.0030905945
16 Control - Triclopyr HS -2.73155021 6.303713e-03 0.0176503976
17 Glyphosate HS - Triclopyr HS -0.68018839 4.963852e-01 0.5791160442
18 Glyphosate Stump - Triclopyr HS 1.47913984 1.391029e-01 0.2049938130
19 Hexazinone - Triclopyr HS -1.85702228 6.330798e-02 0.1181749043
20 Imazapyr HS - Triclopyr HS -1.84622564 6.485945e-02 0.1135040371
21 Imazapyr Stump - Triclopyr HS 1.47913984 1.391029e-01 0.2163823582
22 Control - Triclopyr Stump -4.21069005 2.545918e-05 0.0007128572
23 Glyphosate HS - Triclopyr Stump -2.15932823 3.082471e-02 0.0784629017
24 Glyphosate Stump - Triclopyr Stump 0.00000000 1.000000e+00 1.0000000000
25 Hexazinone - Triclopyr Stump -3.33616212 8.494361e-04 0.0059460530
26 Imazapyr HS - Triclopyr Stump -3.32536548 8.830270e-04 0.0035321080
27 Imazapyr Stump - Triclopyr Stump 0.00000000 1.000000e+00 1.0000000000
28 Triclopyr HS - Triclopyr Stump -1.47913984 1.391029e-01 0.2291107322
> cldList(P.adj~Comparison, data=CLD1, threshold = 0.05)
Group Letter MonoLetter
1 Control a a
2 GlyphosateHS abc abc
3 GlyphosateStump b b
4 Hexazinone ac a c
5 ImazapyrHS ac a c
6 ImazapyrStump b b
7 TriclopyrHS bc bc
8 TriclopyrStump b b
I've checked my data set to be certain the data has been input correctly. Graph attached is mean+/-SE % mortality one year post treatment with compact letter display from the output. Thanks for your thoughts in advance. I'm no means a statistician and always assume there's user error here.
I need to work on the box plot for visualizing. The significance takes up so much space that the y axis extends too much- we can only have 100% of trees die. I'm not sure if this visualization with box plot is intuitive. Perhaps better without the significance?
答案1
得分: 1
K-W检验是一种非参数检验,不测试均值。它测试的H0假设是两个样本是否来自相同的分布,而不是样本的均值是否相同。如果你进行了一个参数检验来比较均值,那么你的担忧是正确的。然而,两个样本在非参数检验中可能具有相同的均值,但由于不同的分布而被检测出来。你的柱状图也并没有真正帮助你区分假设之间的差异。也许可以尝试使用箱线图。
英文:
The K-W test is a nonparametric test that doesn't test means. The H0 hypothesis that it is testing is whether two samples come from the same distribution, not that the means of the samples are the same. If you did a parametric test for means, then you'd be right to be concerned. However, it's entirely possible for two samples to have the same mean but different distributions as detected by nonparametric tests. Your barplots aren't really helping to distinguish the difference between hypotheses either. Maybe try a box plot instead.
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