0
library(Hmisc)
#10% difference
n1 = 30
n2 = 30
n = 60
p1 = seq(0.1, 0.9, 0.1)
p2 = p1 + 0.1
> bpower(p1, p2, n, n1, n2, alpha = 0.05)
Power1 Power2 Power3 Power4 Power5 Power6 Power7 Power8 Power9
0.9997976 0.9992461 0.9933829 0.9670958 0.8995984 0.7799309 0.6141349 0.4211642 0.2252629
#20% difference
n1 = 30
n2 = 30
n = 60
p1 = seq(0.1, 0.8, 0.1)
p2 = p1 + 0.2
> bpower(p1, p2, n, n1, n2, alpha = 0.05)
Power1 Power2 Power3 Power4 Power5 Power6 Power7 Power8
0.9997976 0.9992461 0.9933829 0.9670958 0.8995984 0.7799309 0.6141349 0.4211642
在這裏,我使用bpower
功能Hmisc
計算兩樣品二項式檢驗的功率。我的假設是:H0:p1 = p2與H1:p1!= p2。在第一種情況下,樣本比例相差0.1(即p2 - p1 = 0.1),在第二種情況下,樣本比例相差0.2(p2 - p1 = 0.2)。但是,當我計算兩個案例的權力時,這些值是完全一樣的?我的代碼中有錯誤嗎?R:如何使用bpower函數來計算2-樣品二項式檢驗功率
我明白了。非常感謝你。 – Adrian