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本帖最后由 Menuett 于 2013-12-22 15:59 编辑 ) M* f q9 K1 \! R5 f2 p
煮酒正熟 发表于 2013-12-20 12:05 ![]()
% C: l+ O% d7 L6 A: \" w5 z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 ... : O# E" p: o6 U+ a
+ K8 k' r# q6 G4 F3 C! V: o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)。
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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)
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R example:( f S! G) ^2 Q% t. C6 Y3 f" M
) x/ J! ^0 M1 J4 Q; Z> M<-as.table(rbind(c(1668,5173),c(287,930)))) e: Z# |' Z K& ^
> chisq.test(M)6 e1 d9 I; {, \8 y$ ^' r
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Pearson's Chi-squared test with Yates' continuity correction
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data: M) [4 o. d; ^% e o% m
X-squared = 0.3175, df = 1, p-value = 0.5731
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Python example:- G; q6 n+ {% ]8 J/ T) r
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>>> from scipy import stats
. r1 l! u" N' I; m>>> stats.chi2_contingency([[6841-5173,5173],[1217-930,930]])
6 _- f! ], h" X6 x6 |4 l. f& m(0.31748297614660292, 0.57312422493552839, 1, array([[ 1659.73628692, 5181.26371308],) q; _9 z' \, f) y& I9 N7 R
[ 295.26371308, 921.73628692]])) |
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