|
|
本帖最后由 Menuett 于 2013-12-22 15:59 编辑 4 m$ S1 N! G9 C6 p/ S3 E$ _
煮酒正熟 发表于 2013-12-20 12:05 ![]()
$ ?6 {# L0 I! o9 q2 |基本可以说是显著的。总的来说,在商界做统计学分析,95%信心水平是用得最多的,当95%上不显著时,都会去 ... 7 g! I7 \- |2 a& E9 x
2 s( N+ K8 @4 x- b/ p
这个其实是一种binomial response,应该用Contigency Table或者Logisitic Regression(In case there are cofactors)来做。只记比率丢弃了Number of trial的信息(6841和1217个客户)。 : x" I# V9 M3 }* a( n) R) M
- c' w/ e9 N& N
结果p=0.5731。 远远不显著。要在alpha level 0.05的水平上检验出76.42%和75.62%的区别,即使实验组和对照组各自样本大小相同,各自尚需44735个样本(At power level 80%)。see: Statistical Methods for Rates and Proportions by Joseph L. Fleiss (1981)
$ |, u+ M" G8 C* Y5 D) i: r1 b" f, U6 V4 q9 f
R example:
8 z! e% @' [' K/ B/ L9 ~* W! y9 G4 D) Q9 ?
> M<-as.table(rbind(c(1668,5173),c(287,930)))
& `* i' Q+ u2 Q& `+ p, X) P# ?> chisq.test(M)
) Z1 S5 k7 n w3 C! h* S3 m0 H h" L# |& d( n: q2 r' u
Pearson's Chi-squared test with Yates' continuity correction
5 s) U$ P6 P& z+ `
. v# H3 V0 I8 G" b6 D! ^. v( D, Rdata: M% x% `: O7 _/ T' \9 `+ D
X-squared = 0.3175, df = 1, p-value = 0.5731+ {, P' j$ k) R8 L; S
+ ? v( L1 r: p! Y9 P9 G- |Python example:; X0 @3 o/ u8 G( L
C% f! H$ T( _) I6 F, i) K7 n' l
>>> from scipy import stats4 }/ s/ E! o, g F$ K Y8 o$ U& r3 S& P1 c
>>> stats.chi2_contingency([[6841-5173,5173],[1217-930,930]])
- n$ n0 ?& @* S; s+ I(0.31748297614660292, 0.57312422493552839, 1, array([[ 1659.73628692, 5181.26371308],
3 t& B6 |, M! M3 i [ 295.26371308, 921.73628692]])) |
|