Wilcoxon 符号秩检验

先决条件：参数和非参数方法
假设检验

Wilcoxon 符号秩检验，也称为 Wilcoxon 配对检验，是一种非参数假设检验，它比较两个配对组的中位数并判断它们是否相同分布。

我们可以在以下情况下使用它：

数据对之间的差异是非正态分布的。
独立的数据对是相同的。（或匹配）例如。（数学、英语：两个科目）；（六月，七月：两个月）

涉及的步骤：

Step 1 - Determine the null (h0) and alternate (ha) hypothesis. 
Step 2 - Find the difference (D) between the two columns. [D = B-A]
Step 3 - Find absolute difference (Abs-D). [Abs-D = |D|]
Step 4 - Assign ranks to Abs-D from lowest (1) to highest (n).

分配等级：

如果任意两个或多个 Abs-D 值相同，则为它们分配连续的等级，然后找到每组重复值的等级平均值。考虑以下场景：

Abs-D	1	2	3	3	4	4	4	5
Rank	1	2	3	4	5	6	7	8
Modified Ranks	1	2	3.5	3.5	6	6	6	7

Case I - For Abs-D = 3 
       -> Assign them consecutive possible ranks. (3,4)
       -> Find average of 3,4 => (3+4)/2 = 3.5
       -> Assign the rank = 3.5 to both the 3's present in the table.

Case II - For Abs-D = 4
        -> Assign them consecutive possible ranks. (5,6,7)
        -> Find average of 5,6,7 => (5+6+7)/3 = 6
        -> Assign the rank = 6 to all the 4's present in the table.

Step 5 - Find the sum of the ranks assigned to positive (T+) and negative (T-) Abs-D values. 
Step 6 - Find the Wilcoxon Rank. (Wcalc = minimum(T+,T-))  
Step 7 - Use the value of n and α and find Wtable in two-tailed section of 
      'Critical values of wilcoxon signed rank test'. 
     (take α = 0.05, if not given)
Step 8 - Interpretation of result.

NOTE : We use two-tailed test when we are dealing with two hypothesis. (null and alternate)

编程需要懂一点英语

结果的解释

When Wcalc < Wtable :
    -> Reject H0 (null hypothesis) 
     -> The two groups are not identically distributed.

When Wcalc > Wtable :
    -> Accept H0 (null hypothesis)
     -> The two groups are identically distributed.

示例问题（循序渐进）：

考虑以下示例。测量了印度13个邦的烟雾浓度数据。执行 Wilcoxon 符号秩检验并确定 5 月记录的浓度与 12 月记录的浓度是否存在显着差异。 [取α = 0.05]

States	Smog concentration in May (A)	Smog concentration in December (B)	Difference [D] (B-A) (Step-2)	Absolute Difference [Abs-D] (Step-3)	Rank (Step-4)
Delhi	13.3	11.1	-2.2	2.2	5
Mumbai	10.0	16.2	6.2	6.2	9
Chennai	16.5	15.3	-1.2	1.2	3
Kerala	7.9	19.9	12.0	12.0	11
Karnataka	9.5	10.5	1.0	1.0	2
Tamil Nadu	8.3	15.5	7.2	7.2	10
Orissa	12.6	12.7	0.1	0.1	1
UP	8.9	14.2	5.3	5.3	7
MP	13.6	15.6	2.0	2.0	4
Rajasthan	8.1	20.4	12.3	12.3	12
Gujarat	18.3	12.7	-5.6	5.6	8
West Bengal	8.1	11.2	3.1	3.1	6
Jammu	13.4	36.8	23.4	23.4	13

n = 13
α = 0.05

Step 1 - h0 : Cmay = Cdecember (no change in the smog concentration)
     h1 : Cmay ≠ Cdecember (smog concentration changed)

Step 2,3,4 - Refer the table given above.

Step 5 - T+ marked as [ ] in table.  
     T- marked as [ ] in table.
∑T+ = 75
∑T- = 16

Step 6 - Wcalc = minimum(75,16) 
           = 16

Step 7 - Using n = 13 and α = 0.05 in table (click here)
     Wtable = 17

Step 8 - Wcalc < Wtable :
      Rejecting H0. 
     i.e smog concentration have changed from before.

结论：

Wilcoxon 符号秩检验是制药领域中非常常见的检验，尤其是在药物研究人员中，用于找出各种药物对人体的主要症状。作为一种非参数检验，它可以替代本质上是参数的 T 检验。如有任何疑问/疑问，请在下方评论。