用三种实验方法检验同一组样本,所得三组结果,用配对t检验检验两两之间的差异,过程结果如下:
(d11) | Fuchs Rosenthal-Foto_Screen | Jeromin-Mikroskopie | Fuchs Rosenthal-Mikrosopie | F1 | 5000 | 11267 | 16146 | F2 | 48958 | 174532 | 152604 | F3 | 625 | 3189 | 5208 | F4 | 625 | 1913 | 4687 | F5 | 2083 | 7547 | 13021 | F6 | 72708 | 232355 | 176875 | R script:
>FR_Foto<-c(5000,48958,625,625,2083,72708) >J_MkSk<-c(11267,174532,3189,1913,7547,232355) >FR_MkSk<-c(16146,152604,5208,4687,13021,176875)
>t.test(FR_Foto,FR_MkSk,paired = T) Pairedt-testdata: FR_Foto and FR_MkSkt =-1.9562, df = 5, p-value = 0.1078alternativehypothesis: true difference in means is not equal to 095 percent confidenceinterval: -91999.34 12485.34sampleestimates:mean of the differences -39757 > t.test(J_MkSk,FR_MkSk,paired = T) Pairedt-testdata: J_MkSk and FR_MkSkt =1.0415, df = 5, p-value = 0.3454alternativehypothesis: true difference in means is not equal to 095 percent confidenceinterval: -15235.75 35989.75sampleestimates:mean of the differences 10377 >t.test(FR_Foto,J_MkSk,paired = T) Pairedt-testdata: FR_Foto and J_MkSkt =-1.6947, df = 5, p-value = 0.1509alternativehypothesis: true difference in means is not equal to 095 percent confidenceinterval: -126176.96 25908.96sampleestimates:mean of the differences -50134
##所示结果并无显著性差异,可这足以说明,不同方法所得的结果就相同么? ##如果不能,还可以用别的什么方法么?
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