Question

这是我的数据

> longnacl
   essays Ionic_Strength Water_Absorbancy
1   NaCl1           0.01            52.70
2   NaCl2           0.01            52.02
3   NaCl3           0.01            52.98
4   NaCl1           0.02            43.90
5   NaCl2           0.02            46.54
6   NaCl3           0.02            43.60
7   NaCl1           0.03            30.40
8   NaCl2           0.03            31.70
9   NaCl3           0.03            31.80
10  NaCl1           0.04            28.00
11  NaCl2           0.04            27.75
12  NaCl3           0.04            26.80

> anova = ezANOVA(data = longnacl , 
+                      dv = Water_Absorbancy,
+                      wid= essays,
+                      within = Ionic_Strength,
+                      
+                      detailed = T,
+                      type = 3)
Warning message:
In log(det(U)) : NaNs produced

为什么我没有得到anova分析和mauchlys测试？

Answer 1

我试过了，我没有问题：

essays = c("NaCl1","NaCl2","NaCl3","NaCl1","NaCl2","NaCl3","NaCl1","NaCl2","NaCl3","NaCl1","NaCl2","NaCl3")
Ionic_Strength = c(0.01,0.01,0.01,0.02,0.02,0.02,0.03,0.03,0.03,0.04,0.04,0.04)
Water_Absorbancy = c(52.70,52.02,52.98,43.90,46.54,43.60,30.40,31.70,31.80,28.00,27.75,26.80)
longnacl = data.frame(essays, Ionic_Strength,Water_Absorbancy)
require(ez)  
anova <- ezANOVA(data = longnacl , 
                                    dv = Water_Absorbancy,
                                    wid= essays,
                                    within = Ionic_Strength,
                                    detailed = T,
                                    type = 3)

#Warning: "Ionic_Strength" will be treated as numeric.
#Warning: There is at least one numeric within variable, therefore aov() will be used for computation and no assumption checks will be obtained.
> anova
#$ANOVA
      Effect DFn DFd      SSn     SSd        F            p p<.05      ges
#1 Ionic_Strength   1   2 1175.634 0.24577 9566.946 0.0001045102     * 0.999791

Na中产生的NaNs

1 个答案: